Exercise 0.1: Finding the coefficients of a linear combination: ten direct computations

For each line, compute the coefficients, the components, the determinant or the parameter asked for. Give fractions exactly, in the form $-\frac{3}{4}$ or $-3/4$ .

One single question hides behind every line: does the system $a\vec{u}+b\vec{v}\,(+c\vec{w})=\vec{t}$ have a solution, and which one? Write one equation per component, then solve by substitution, starting with the equation that has the fewest unknowns. The determinant of the vectors settles it in advance: non-zero, the set is linearly independent and the solution is unique; zero, the set is linearly dependent. In an orthogonal basis, each coordinate is read directly from a dot product.

a) $\vec{u}=(3, -2)$ and $\vec{v}=(1, 4)$ : find $a$ and $b$ such that $(5, -8)=a\vec{u}+b\vec{v}$ .
b) $\vec{u}=(2, 5)$ and $\vec{v}=(-1, 3)$ : find $a$ and $b$ such that $(4, 1)=a\vec{u}+b\vec{v}$ .
c) A vector has coordinates $(-2, 3)$ relative to the basis $\{(1, -2), (4, 1)\}$ : give its two components in the standard basis.
d) In the orthogonal basis $\vec{f_{1}}=(2, 3)$ , $\vec{f_{2}}=(-3, 2)$ , give the coordinates $a$ and $b$ of $\vec{w}=(4, 19)$ by two dot products.
e) Compute the determinant of $\vec{p}=(6, -9)$ and $\vec{q}=(-4, 6)$ , then say whether the set $\{\vec{p}, \vec{q}\}$ is linearly independent or dependent.
f) Find the values of $k$ for which $(k, 3)$ and $(4, k+4)$ are collinear.
g) In the basis $\{(1, 0, 0), (2, 1, 0), (-1, 3, 1)\}$ of $\mathbb{R}^{3}$ , find $a$ , $b$ , $c$ such that $(4, -1, 2)=a(1, 0, 0)+b(2, 1, 0)+c(-1, 3, 1)$ , then give $a$ and $b$ .
h) Find $a$ , $b$ , $c$ such that $(0, 6, 6)=a(1, -1, 2)+b(2, 1, 1)+c(0, 3, 1)$ , then give $a$ and $c$ .
i) Compute the determinant of the vectors $(2, -1, 1)$ , $(1, 3, -2)$ and $(5, 1, 1)$ , then say whether the set is linearly independent or dependent.
j) Find the value of $m$ for which the vectors $(1, m, 0)$ , $(0, 1, 2)$ and $(2, 1, m)$ are linearly dependent.

Type your answers, the page tells you right or wrong 0/19

a)

a

b

b)

a

b

c)

first component

second component

d)

a

b

e)

determinant

the set is

f)

the smaller value

the larger value

g)

a

b

h)

a

c

i)

determinant

the set is

j)

m

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Answers

a) $(5, -8)=2\vec{u}-\vec{v}$ ; $a=2$ , $b=-1$
b) $a=\frac{13}{11}$ , $b=-\frac{18}{11}$
c) $-2(1, -2)+3(4, 1)=(10, 7)$
d) $a=\frac{65}{13}=5$ , $b=\frac{26}{13}=2$
e) $\det=0$ ; dependent, $\vec{q}=-\frac{2}{3}\vec{p}$
f) $(k+6)(k-2)=0$ ; $k=-6$ or $k=2$
g) $c=2$ , $b=-7$ , $a=20$
h) $a=2$ , $b=-1$ , $c=3$
i) $\det=7$ ; independent
j) $5m-2=0$ ; $m=\frac{2}{5}$

The move that decides everything: write ONE equation per component before looking for anything. Coefficients are not guessed, they are solved for, and a solution is always checked by rebuilding the target vector.

a)

$3a+b=5$ and $-2a+4b=-8$ . The first gives $b=5-3a$ ; in the second, $-2a+20-12a=-8$ , so $-14a=-28$ , $a=2$ and $b=-1$ . Check: $2(3, -2)-(1, 4)=(5, -8)$ .

b)

$2a-b=4$ and $5a+3b=1$ . The first gives $b=2a-4$ ; in the second, $5a+6a-12=1$ , so $a=\frac{13}{11}$ and $b=\frac{26}{11}-\frac{44}{11}=-\frac{18}{11}$ . The determinant $2(3)-(-1)(5)=11$ announced that denominator. Pitfall: rounding along the way; keep the fractions to the end.

c)

The coordinates are the coefficients: $-2(1, -2)+3(4, 1)=(-2+12, 4+3)=(10, 7)$ . Pitfall: $-2\times(-2)=+4$ in the second component.

d)

$\vec{f_{1}}\cdot\vec{f_{2}}=-6+6=0$ : the basis is orthogonal, and $\|\vec{f_{1}}\|^{2}=\|\vec{f_{2}}\|^{2}=13$ . So $a=\frac{\vec{w}\cdot\vec{f_{1}}}{\|\vec{f_{1}}\|^{2}}=\frac{8+57}{13}=5$ and $b=\frac{\vec{w}\cdot\vec{f_{2}}}{\|\vec{f_{2}}\|^{2}}=\frac{-12+38}{13}=2$ . Pitfall: dividing by the norm $\sqrt{13}$ instead of its square.

e)

$\det(\vec{p},\vec{q})=6(6)-(-9)(-4)=36-36=0$ : the set is linearly dependent. Indeed, $\vec{q}=-\frac{2}{3}\vec{p}$ . Pitfall: $(-9)(-4)=+36$ , and it is subtracted.

f)

Collinear if and only if the determinant is zero: $k(k+4)-3(4)=k^{2}+4k-12=(k+6)(k-2)=0$ , so $k=-6$ or $k=2$ . Check: $(2, 3)$ and $(4, 6)$ , then $(-6, 3)$ and $(4, -2)$ , are indeed proportional. Pitfall: keeping only one of the two.

g)

The basis is triangular: work back up from the last component. $z$ : $c=2$ . $y$ : $b+3c=-1$ , so $b=-7$ . $x$ : $a+2b-c=4$ , so $a=4+14+2=20$ . Check: $20(1, 0, 0)-7(2, 1, 0)+2(-1, 3, 1)=(4, -1, 2)$ . Pitfall: forgetting the $-c$ in the first row, which gives $a=18$ .

h)

$a+2b=0$ , $-a+b+3c=6$ , $2a+b+c=6$ . The first gives $a=-2b$ ; the other two become $3b+3c=6$ and $-3b+c=6$ . Subtracting, $4b=-4$ , so $b=-1$ , $c=3$ and $a=2$ . Check: $2(1, -1, 2)-(2, 1, 1)+3(0, 3, 1)=(0, 6, 6)$ .

i)

Along the first row: $2\begin{vmatrix} 3 & -2 \\ 1 & 1 \end{vmatrix}-(-1)\begin{vmatrix} 1 & -2 \\ 5 & 1 \end{vmatrix}+1\begin{vmatrix} 1 & 3 \\ 5 & 1 \end{vmatrix}=2(5)+1(11)+1(-14)=7$ . Non-zero: the set is linearly independent. Pitfall: the minus sign in front of the second cofactor.

j)

$\begin{vmatrix} 1 & m & 0 \\ 0 & 1 & 2 \\ 2 & 1 & m \end{vmatrix}=1(m-2)-m(0-4)+0=5m-2$ . Dependent if and only if $5m-2=0$ , so $m=\frac{2}{5}$ . Pitfall: $-m(0-4)=+4m$ .

Check: on every line, rebuild the combination with the coefficients found; you must land exactly on the target vector. For a zero determinant, look for the multiple or the combination that makes it zero, such as $\vec{q}=-\frac{2}{3}\vec{p}$ in e).

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Exercise 1: Components, norm and operations in the plane

Consider the points $A(-2, 3)$ , $B(1, 7)$ and $C(-10, 9)$ of the plane.

a) Find the components of the vectors $\vec{AB}$ and $\vec{AC}$ , then compute their norms.
b) Find the components of the vector $2\vec{AB}-3\vec{AC}$ .
c) Are the vectors $\vec{AB}$ and $\vec{AC}$ collinear? Justify by computation.
d) Find the vector $\vec{w}$ of norm $15$ , parallel to $\vec{AB}$ but pointing the opposite way.

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Answers

a) $\vec{AB}=(3, 4)$ , norm 5; $\vec{AC}=(-8, 6)$ , norm 10
b) $(30, -10)$
c) No: determinant $50\neq 0$
d) $\vec{w}=(-9, -12)$

a)

$\vec{AB}=(1-(-2), 7-3)=(3, 4)$ and $\vec{AC}=(-10-(-2), 9-3)=(-8, 6)$ . The norms are $\|\vec{AB}\|=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5$ and $\|\vec{AC}\|=\sqrt{(-8)^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10$ .

b)

$2\vec{AB}=(6, 8)$ and $3\vec{AC}=(-24, 18)$ , so $2\vec{AB}-3\vec{AC}=(6-(-24), 8-18)=(30, -10)$ .

c)

Two vectors of the plane are collinear if and only if their determinant is zero: $\begin{vmatrix} 3 & -8 \\ 4 & 6 \end{vmatrix}=(3)(6)-(-8)(4)=18+32=50\neq 0$ . So they are NOT collinear. Here is the trap of the chapter: $\|\vec{AC}\|$ is exactly twice $\|\vec{AB}\|$ , which pushes many students to conclude too fast that $\vec{AC}=2\vec{AB}$ . The norm says nothing about the direction. Comparing the ratios of components: $\dfrac{-8}{3}\neq\dfrac{6}{4}$ , same conclusion.

d)

The unit vector of $\vec{AB}$ is $\dfrac{\vec{AB}}{5}=\left(\dfrac{3}{5}, \dfrac{4}{5}\right)$ . For a norm of $15$ and the opposite direction, multiply by $-15$ : $\vec{w}=-15\left(\dfrac{3}{5}, \dfrac{4}{5}\right)=(-9, -12)$ . Check: $\|\vec{w}\|=\sqrt{81+144}=\sqrt{225}=15$ , and $\vec{w}=-3\vec{AB}$ is indeed collinear with $\vec{AB}$ , with a negative coefficient, hence pointing the opposite way.

Exercise 2: A linear combination and the system behind it

Consider the vectors $\vec{u}=(2, 1)$ and $\vec{v}=(1, 3)$ of the plane.

a) Write $\vec{w}=(7, 11)$ as a linear combination of $\vec{u}$ and $\vec{v}$ , that is, find the scalars $a$ and $b$ such that $\vec{w}=a\vec{u}+b\vec{v}$ .
b) Do the same for $\vec{t}=(5, -5)$ .
c) Explain why EVERY vector of the plane can be written as a linear combination of $\vec{u}$ and $\vec{v}$ , and say in one sentence what this means for the set $\{\vec{u}, \vec{v}\}$ .
d) Now in space, let $\vec{p}=(1, 2, 1)$ and $\vec{q}=(0, 1, 2)$ . Is the vector $\vec{r}=(1, 5, 7)$ a linear combination of $\vec{p}$ and $\vec{q}$ ? What about the vector $\vec{s}=(2, 3, 1)$ ?

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Answers

a) $\vec{w}=2\vec{u}+3\vec{v}$
b) $\vec{t}=4\vec{u}-3\vec{v}$
c) Determinant 5: a basis of $\mathbb{R}^2$
d) $\vec{r}=\vec{p}+3\vec{q}$ ; $\vec{s}$ is not

a)

The equality $a(2, 1)+b(1, 3)=(7, 11)$ gives one equation per component: $\begin{cases} 2a+b=7 \\ a+3b=11 \end{cases}$ . From the first, $b=7-2a$ . Substituting: $a+3(7-2a)=11$ , that is $a+21-6a=11$ , so $-5a=-10$ and $a=2$ , then $b=7-4=3$ . Check: $2(2, 1)+3(1, 3)=(4+3, 2+9)=(7, 11)$ . So $\vec{w}=2\vec{u}+3\vec{v}$ .

b)

Same method: $\begin{cases} 2a+b=5 \\ a+3b=-5 \end{cases}$ . From the first, $b=5-2a$ , hence $a+15-6a=-5$ , that is $-5a=-20$ and $a=4$ , then $b=5-8=-3$ . Check: $4(2, 1)-3(1, 3)=(8-3, 4-9)=(5, -5)$ . So $\vec{t}=4\vec{u}-3\vec{v}$ . A negative coefficient is perfectly normal; it is not a sign of an error.

c)

For an arbitrary vector $(x, y)$ , the system reads $\begin{cases} 2a+b=x \\ a+3b=y \end{cases}$ . Its determinant is $\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix}=6-1=5\neq 0$ , so the system has a unique solution whatever the right-hand side. Every vector of the plane can therefore be written, in exactly one way, as a linear combination of $\vec{u}$ and $\vec{v}$ : the set $\{\vec{u}, \vec{v}\}$ spans $\mathbb{R}^{2}$ , and since it is linearly independent (non-zero determinant), it is a basis of $\mathbb{R}^{2}$ .

d)

For $\vec{r}$ : $a(1, 2, 1)+b(0, 1, 2)=(1, 5, 7)$ gives three equations. The first forces $a=1$ . The second gives $2(1)+b=5$ , so $b=3$ . The third, unused so far, must still be CHECKED: $a+2b=1+6=7$ , which is indeed the third component. So $\vec{r}=\vec{p}+3\vec{q}$ : yes.

For $\vec{s}$ : the first equation forces $a=2$ , the second gives $2(2)+b=3$ , so $b=-1$ . Test of the third: $a+2b=2-2=0$ , whereas the third component of $\vec{s}$ is $1$ . Contradiction: the system is inconsistent, and $\vec{s}$ is NOT a linear combination of $\vec{p}$ and $\vec{q}$ . This is the point not to miss in space: with three equations and two unknowns, there is one equation too many, and that is the one that decides. Never stop as soon as $a$ and $b$ are found.

Exercise 3: Linear dependence and independence

This exercise is about the test for linear independence in $\mathbb{R}^{3}$ . Recall that three vectors of $\mathbb{R}^{3}$ are linearly independent if and only if the determinant formed by their components is non-zero.

a) Let $\vec{u}=(1, 2, -1)$ , $\vec{v}=(2, -1, 3)$ and $\vec{w}=(4, 3, 1)$ . Show that $\vec{w}$ is a linear combination of $\vec{u}$ and $\vec{v}$ , and conclude about the set $\{\vec{u}, \vec{v}, \vec{w}\}$ .
b) Recover the conclusion of a) in a single line, by computing the determinant.
c) Is the set $\{(1, 0, 2), (0, 1, -1), (2, 1, 1)\}$ linearly independent or dependent? Justify.
d) Find the value of $m$ for which the set $\{(1, 2, -1), (2, -1, 3), (0, 5, m)\}$ is linearly dependent.

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Answers

a) $\vec{w}=2\vec{u}+\vec{v}$ : dependent
b) Zero determinant
c) Determinant $-2$ : independent
d) $m=-5$

a)

We look for $a$ and $b$ such that $a\vec{u}+b\vec{v}=\vec{w}$ , that is $\begin{cases} a+2b=4 \\ 2a-b=3 \\ -a+3b=1 \end{cases}$ . The first two give $a=4-2b$ , then $2(4-2b)-b=3$ , that is $8-5b=3$ , so $b=1$ and $a=2$ . Test of the third equation: $-2+3(1)=1$ , it holds. So $\vec{w}=2\vec{u}+\vec{v}$ . Since one of the three vectors is expressed in terms of the other two, the set is linearly DEPENDENT. The dependence relation can also be written in its standard form: $2\vec{u}+\vec{v}-\vec{w}=\vec{0}$ , a zero combination whose coefficients are not all zero.

b)

$\begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 4 & 3 & 1 \end{vmatrix}=1\big((-1)(1)-(3)(3)\big)-2\big((2)(1)-(3)(4)\big)+(-1)\big((2)(3)-(-1)(4)\big)$ . This gives $1(-1-9)-2(2-12)-1(6+4)=-10+20-10=0$ . The determinant is zero, so the set is linearly dependent. One line is enough where a) needed a full system: on a timed exam, this is the method to use by reflex, keeping the first one for when you are explicitly asked for the dependence relation.

c)

$\begin{vmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 2 & 1 & 1 \end{vmatrix}=1\big((1)(1)-(-1)(1)\big)-0\big((0)(1)-(-1)(2)\big)+2\big((0)(1)-(1)(2)\big)=1(2)-0+2(-2)=2-4=-2$ . The determinant is $-2\neq 0$ , so the set is linearly INDEPENDENT. Being independent with three vectors in $\mathbb{R}^{3}$ , it is also a basis of $\mathbb{R}^{3}$ .

d)

$\begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 0 & 5 & m \end{vmatrix}=1\big((-1)(m)-(3)(5)\big)-2\big((2)(m)-(3)(0)\big)+(-1)\big((2)(5)-(-1)(0)\big)$ . This gives $(-m-15)-2(2m)-1(10)=-m-15-4m-10=-5m-25$ . The set is dependent when the determinant vanishes: $-5m-25=0$ , so $m=-5$ . Check: with $m=-5$ , the third vector is $(0, 5, -5)$ , and indeed $2(1, 2, -1)-(2, -1, 3)=(0, 5, -5)$ : the dependence relation is explicit.

Exercise 4: Bases and the coordinates of a vector

In the plane, consider the set $\mathcal{B}=\{\vec{e_{1}}, \vec{e_{2}}\}$ with $\vec{e_{1}}=(1, 1)$ and $\vec{e_{2}}=(1, -1)$ .

a) Show that $\mathcal{B}$ is a basis of $\mathbb{R}^{2}$ .
b) Find the coordinates of the vector $\vec{v}=(5, 1)$ relative to the basis $\mathcal{B}$ .
c) The vector $\vec{t}$ has coordinates $(-2, 4)$ relative to the basis $\mathcal{B}$ . Find its components in the standard basis.
d) In $\mathbb{R}^{3}$ , take for granted that $\mathcal{C}=\{(1, 0, 0), (1, 1, 0), (1, 1, 1)\}$ is a basis. Find the coordinates of $\vec{z}=(3, 5, 2)$ relative to this basis.

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Answers

a) Determinant $-2\neq 0$ : a basis
b) $(3, 2)$
c) $(2, -6)$
d) $(-2, 3, 2)$

a)

$\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}=(1)(-1)-(1)(1)=-2\neq 0$ , so the set is linearly independent. Since it has two vectors and $\mathbb{R}^{2}$ has dimension $2$ , a linearly independent set of two vectors automatically spans it: $\mathcal{B}$ is a basis of $\mathbb{R}^{2}$ . This is the argument to know; it saves you from proving separately that the set spans.

b)

We look for $a$ and $b$ such that $a(1, 1)+b(1, -1)=(5, 1)$ , that is $\begin{cases} a+b=5 \\ a-b=1 \end{cases}$ . Adding: $2a=6$ , so $a=3$ , then $b=2$ . Check: $3(1, 1)+2(1, -1)=(3+2, 3-2)=(5, 1)$ . The coordinates of $\vec{v}$ relative to $\mathcal{B}$ are therefore $(3, 2)$ . Note that $\vec{v}$ has not changed: it is the same vector of the plane, described by two different pairs of numbers depending on the basis. $(5, 1)$ in the standard basis, $(3, 2)$ relative to $\mathcal{B}$ .

c)

Here the work goes the other way, and it is a plain substitution: $\vec{t}=-2\vec{e_{1}}+4\vec{e_{2}}=-2(1, 1)+4(1, -1)=(-2+4, -2-4)=(2, -6)$ . No system to solve. Recognizing which way the question goes saves several minutes on an exam: going from coordinates to components is a direct computation, the reverse requires a system.

d)

We look for $c_{1}$ , $c_{2}$ , $c_{3}$ such that $c_{1}(1, 0, 0)+c_{2}(1, 1, 0)+c_{3}(1, 1, 1)=(3, 5, 2)$ , that is $\begin{cases} c_{1}+c_{2}+c_{3}=3 \\ c_{2}+c_{3}=5 \\ c_{3}=2 \end{cases}$ . The system is already in echelon form, so work back up: $c_{3}=2$ , then $c_{2}=5-2=3$ , then $c_{1}=3-3-2=-2$ . Check: $-2(1, 0, 0)+3(1, 1, 0)+2(1, 1, 1)=(-2+3+2, 3+2, 2)=(3, 5, 2)$ . The coordinates are $(-2, 3, 2)$ .

Exercise 5: Too many vectors: a dependent spanning set and the extraction of a basis

In the plane, consider the three vectors $\vec{v_{1}}=(1, 2)$ , $\vec{v_{2}}=(3, 1)$ and $\vec{v_{3}}=(1, -3)$ .

a) Without any computation, explain why the set $\{\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}\}$ is necessarily linearly dependent.
b) Express $\vec{v_{3}}$ as a linear combination of $\vec{v_{1}}$ and $\vec{v_{2}}$ , then write the dependence relation as a zero combination whose coefficients are not all zero.
c) Does the set $\{\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}\}$ span $\mathbb{R}^{2}$ ? Justify.
d) Extract from this set a basis of $\mathbb{R}^{2}$ , then give the coordinates of $\vec{v_{3}}$ relative to the basis you chose.

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Answers

a) Three vectors in dimension 2
b) $\vec{v_3}=-2\vec{v_1}+\vec{v_2}$
c) Yes: spanning and dependent
d) Basis $\{\vec{v_1}, \vec{v_2}\}$ ; $(-2, 1)$

a)

These three vectors live in $\mathbb{R}^{2}$ , which has dimension $2$ . A linearly independent set of $\mathbb{R}^{2}$ has at most two vectors: as soon as there are three, the set must be dependent. This is a counting argument; no computation is required. The reflex to keep: more vectors than the dimension always means dependence.

b)

We look for $a$ and $b$ such that $a\vec{v_{1}}+b\vec{v_{2}}=\vec{v_{3}}$ , that is $\begin{cases} a+3b=1 \\ 2a+b=-3 \end{cases}$ . Multiplying the first by $2$ and subtracting the second: $5b=5$ , so $b=1$ and $a=1-3=-2$ . Check: $-2(1, 2)+(3, 1)=(-2+3, -4+1)=(1, -3)$ . So $\vec{v_{3}}=-2\vec{v_{1}}+\vec{v_{2}}$ , and moving everything to one side, the dependence relation reads $-2\vec{v_{1}}+\vec{v_{2}}-\vec{v_{3}}=\vec{0}$ , a zero combination whose coefficients $-2$ , $1$ , $-1$ are not all zero.

c)

The vectors $\vec{v_{1}}$ and $\vec{v_{2}}$ are not collinear: $\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix}=(1)(1)-(3)(2)=-5\neq 0$ . On their own they already form a basis of $\mathbb{R}^{2}$ , so they span the whole plane. Adding $\vec{v_{3}}$ takes nothing away: the set $\{\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}\}$ therefore spans $\mathbb{R}^{2}$ . The same set can be both spanning and dependent: spanning and being independent are two separate properties, and it is precisely the redundant vector $\vec{v_{3}}$ that keeps the set from being a basis.

d)

Remove the redundant vector $\vec{v_{3}}$ and keep $\{\vec{v_{1}}, \vec{v_{2}}\}$ , independent by c), hence a basis of $\mathbb{R}^{2}$ . The coordinates of $\vec{v_{3}}$ relative to this basis are read directly from the relation in b): $\vec{v_{3}}=-2\vec{v_{1}}+1\vec{v_{2}}$ , so its coordinates relative to $\{\vec{v_{1}}, \vec{v_{2}}\}$ are $(-2, 1)$ .

Exercise 6: Proofs: independent sets and the zero vector

This exercise asks for almost no numerical computation: you reason directly from the definition of a linearly independent set. Recall the definition: $\{\vec{v_{1}}, \dots, \vec{v_{n}}\}$ is linearly independent if the equality $a_{1}\vec{v_{1}}+\dots+a_{n}\vec{v_{n}}=\vec{0}$ implies $a_{1}=\dots=a_{n}=0$ .

a) Let $\{\vec{u}, \vec{v}\}$ be a linearly independent set. Prove that the set $\{\vec{u}+\vec{v}, \vec{u}-\vec{v}\}$ is linearly independent too.
b) Prove that every set containing the zero vector is linearly dependent, whatever the other vectors are.
c) Prove that if $\vec{w}$ is a linear combination of $\vec{u}$ and $\vec{v}$ , then the set $\{\vec{u}, \vec{v}, \vec{w}\}$ is linearly dependent.
d) True or false: if three vectors of $\mathbb{R}^{3}$ are pairwise non-collinear, then the set they form is linearly independent. Justify, and give an explicit counterexample if the statement is false.

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Answers

a) Coefficients $a+b=a-b=0$ : independent
b) $1\cdot\vec{0}=\vec{0}$ : dependent
c) Coefficient $-1$ is non-zero: dependent
d) False: $(1, 0, 0)$ , $(0, 1, 0)$ , $(1, 1, 0)$

a)

Start from a zero combination of the new set: $a(\vec{u}+\vec{v})+b(\vec{u}-\vec{v})=\vec{0}$ . Group the terms along $\vec{u}$ and $\vec{v}$ : $(a+b)\vec{u}+(a-b)\vec{v}=\vec{0}$ . Since $\{\vec{u}, \vec{v}\}$ is independent by hypothesis, both coefficients are zero: $\begin{cases} a+b=0 \\ a-b=0 \end{cases}$ . Adding, $2a=0$ so $a=0$ , then $b=0$ . The only zero combination is the trivial one, so $\{\vec{u}+\vec{v}, \vec{u}-\vec{v}\}$ is linearly independent. The mechanism of the proof is always the same: start from a zero combination, regroup on the set whose independence you know, and conclude.

b)

Suppose the set is $\{\vec{0}, \vec{v_{2}}, \dots, \vec{v_{n}}\}$ . We can write the combination $1\cdot\vec{0}+0\cdot\vec{v_{2}}+\dots+0\cdot\vec{v_{n}}=\vec{0}$ . It is indeed a zero combination, and yet its first coefficient is $1$ , so the coefficients are not all zero. The definition of an independent set fails: the set is linearly dependent. Practical consequence on an exam: as soon as a zero vector appears in a set, the answer is immediate, with no determinant to compute.

c)

By hypothesis, there are scalars $a$ and $b$ such that $\vec{w}=a\vec{u}+b\vec{v}$ . Move everything to one side: $a\vec{u}+b\vec{v}-\vec{w}=\vec{0}$ . This is a zero linear combination of the three vectors, with coefficients $a$ , $b$ and $-1$ . The coefficient of $\vec{w}$ is $-1$ , which is not zero, so the coefficients are not all zero, whatever the values of $a$ and $b$ . The set $\{\vec{u}, \vec{v}, \vec{w}\}$ is therefore linearly dependent. The coefficient $-1$ carries the whole proof.

d)

The statement is FALSE. Pairwise collinearity only tests pairs, whereas linear dependence is about the whole set: a vector can depend on the OTHER TWO without being collinear with either of them taken alone.

Explicit counterexample: $\vec{a}=(1, 0, 0)$ , $\vec{b}=(0, 1, 0)$ and $\vec{c}=(1, 1, 0)$ . No pair is collinear, as the ratios of components show: $\vec{a}$ and $\vec{b}$ are clearly not proportional, and $\vec{c}$ is proportional neither to $\vec{a}$ (second component) nor to $\vec{b}$ (first component). Yet $\vec{c}=\vec{a}+\vec{b}$ , so the set is linearly dependent. The determinant confirms it: $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{vmatrix}=0$ . Geometrically, the three vectors are coplanar, all in the plane $z=0$ .

The converse, on the other hand, is true: if a set is linearly independent, then its vectors are pairwise non-collinear. It is a one-way implication, and exams often test it in the wrong direction.

Exercise 7: Problem: an oblique basis of space and the uniqueness of coordinates

In a rectangular coordinate system of space, consider the three vectors $\vec{u_{1}}=(2, 1, 0)$ , $\vec{u_{2}}=(0, 1, 1)$ and $\vec{u_{3}}=(1, 0, 2)$ .

The goal is to show that these three vectors form a basis of $\mathbb{R}^{3}$ , to read the coordinates of given vectors relative to it, and to establish that these coordinates are always unique.

a) Show that $\mathcal{B}=\{\vec{u_{1}}, \vec{u_{2}}, \vec{u_{3}}\}$ is a basis of $\mathbb{R}^{3}$ .
b) Find the coordinates of the vector $\vec{v}=(5, 3, 8)$ relative to the basis $\mathcal{B}$ .
c) Interpret geometrically the absolute value of the determinant computed in a).
d) Is the basis $\mathcal{B}$ orthogonal? Justify with a computation, and say what this changes when reading coordinates.
e) Without solving a system, give the coordinates of $\vec{w}=(4, 2, 0)$ relative to the basis $\mathcal{B}$ and justify your answer in one sentence.
f) Prove that relative to any basis of $\mathbb{R}^{3}$ , the coordinates of a vector are unique.

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Answers

a) Determinant 5: a basis
b) $(1, 2, 3)$
c) Volume 5
d) No: $\vec{u_1}\cdot\vec{u_2}=1$
e) $(2, 0, 0)$
f) Zero difference, independent set: uniqueness

a)

Compute the determinant of the matrix whose columns are the components of the three vectors: $\begin{vmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 2 \end{vmatrix}=2\big((1)(2)-(0)(1)\big)-0\big((1)(2)-(0)(0)\big)+1\big((1)(1)-(1)(0)\big)=2(2)-0+1(1)=5$ . The determinant is $5\neq 0$ , so the set is linearly independent. Three independent vectors in $\mathbb{R}^{3}$ , which has dimension $3$ , form a basis. So $\mathcal{B}$ is indeed a basis of $\mathbb{R}^{3}$ .

b)

We look for $a$ , $b$ , $c$ such that $a\vec{u_{1}}+b\vec{u_{2}}+c\vec{u_{3}}=(5, 3, 8)$ , that is $\begin{cases} 2a+c=5 \\ a+b=3 \\ b+2c=8 \end{cases}$ . From the first, $c=5-2a$ . Substituting into the third: $b+2(5-2a)=8$ , that is $b=8-10+4a=4a-2$ . Substituting into the second: $a+4a-2=3$ , so $5a=5$ and $a=1$ . Then $b=4-2=2$ and $c=5-2=3$ . Full check: $1(2, 1, 0)+2(0, 1, 1)+3(1, 0, 2)=(2+0+3, 1+2+0, 0+2+6)=(5, 3, 8)$ . The coordinates of $\vec{v}$ relative to $\mathcal{B}$ are therefore $(1, 2, 3)$ .

c)

The absolute value of the determinant of three vectors of space is the volume of the parallelepiped built on these three vectors. Here this volume is $5$ cubic units. The reading is then immediate: a zero determinant would mean a flattened parallelepiped, hence three coplanar vectors, hence a dependent set that cannot span all of space. The algebraic criterion and the geometric criterion say exactly the same thing.

d)

Test orthogonality pair by pair: $\vec{u_{1}}\cdot\vec{u_{2}}=(2)(0)+(1)(1)+(0)(1)=1\neq 0$ . A single non-zero dot product is enough to conclude: the basis is NOT orthogonal, it is an oblique basis. What this changes: in an orthonormal basis, each coordinate is read directly from a dot product, $a_{i}=\vec{v}\cdot\vec{e_{i}}$ . That shortcut is invalid here, and using it anyway is a classic error. In an oblique basis, you must solve the system, as in b). Let us check on $\vec{v}$ : $\vec{v}\cdot\vec{u_{1}}=10+3+0=13$ , whereas the first coordinate is $1$ . The shortcut would give nonsense.

e)

Notice that $\vec{w}=(4, 2, 0)=2(2, 1, 0)=2\vec{u_{1}}$ . The vector $\vec{w}$ is collinear with $\vec{u_{1}}$ , so its coordinates relative to $\mathcal{B}$ are $(2, 0, 0)$ . No system is needed: spotting a collinearity with a basis vector gives the result directly, and by uniqueness of coordinates (part f), it is necessarily the right answer.

f)

Let $\mathcal{B}=\{\vec{e_{1}}, \vec{e_{2}}, \vec{e_{3}}\}$ be a basis of $\mathbb{R}^{3}$ and $\vec{v}$ any vector. Suppose $\vec{v}$ can be written in two ways: $\vec{v}=a_{1}\vec{e_{1}}+a_{2}\vec{e_{2}}+a_{3}\vec{e_{3}}$ and $\vec{v}=b_{1}\vec{e_{1}}+b_{2}\vec{e_{2}}+b_{3}\vec{e_{3}}$ . Subtracting side by side: $(a_{1}-b_{1})\vec{e_{1}}+(a_{2}-b_{2})\vec{e_{2}}+(a_{3}-b_{3})\vec{e_{3}}=\vec{0}$ . A basis is in particular a linearly independent set, so all the coefficients of this zero combination are zero: $a_{1}-b_{1}=a_{2}-b_{2}=a_{3}-b_{3}=0$ , that is $a_{1}=b_{1}$ , $a_{2}=b_{2}$ and $a_{3}=b_{3}$ . The two expressions are therefore identical: the coordinates are unique.

It is exactly linear independence that carries uniqueness, and spanning that carries existence. A basis serves both at once, and that is why the definition demands both properties.

Exercise 8: Problem: change of basis and coordinates

In $\mathbb{R}^{2}$ , consider the two bases $\mathcal{B_{1}}=\{\vec{f_{1}}, \vec{f_{2}}\}$ and $\mathcal{B_{2}}=\{\vec{g_{1}}, \vec{g_{2}}\}$ , with $\vec{f_{1}}=(1, 1)$ , $\vec{f_{2}}=(1, 2)$ , $\vec{g_{1}}=(2, 1)$ and $\vec{g_{2}}=(1, 1)$ .

A vector $\vec{v}$ has coordinates $(3, -1)$ relative to the basis $\mathcal{B_{1}}$ .

a) Check that $\mathcal{B_{1}}$ and $\mathcal{B_{2}}$ are indeed two bases of $\mathbb{R}^{2}$ , then find the components of $\vec{v}$ in the standard basis.
b) Find the coordinates of $\vec{v}$ relative to the basis $\mathcal{B_{2}}$ .
c) Express each of the vectors $\vec{f_{1}}$ and $\vec{f_{2}}$ as a linear combination of $\vec{g_{1}}$ and $\vec{g_{2}}$ .
d) Recover the result of b) starting from $\vec{v}=3\vec{f_{1}}-\vec{f_{2}}$ and the expressions found in c), without going back through the standard basis. What do you notice?

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Answers

a) Two bases; $\vec{v}=(2, 1)$
b) $(1, 0)$
c) $\vec{f_1}=\vec{g_2}$ ; $\vec{f_2}=-\vec{g_1}+3\vec{g_2}$
d) Same result by substitution

a)

For $\mathcal{B_{1}}$ : $\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix}=2-1=1\neq 0$ , it is a basis. For $\mathcal{B_{2}}$ : $\begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix}=2-1=1\neq 0$ , it is one too. The coordinates $(3, -1)$ relative to $\mathcal{B_{1}}$ mean $\vec{v}=3\vec{f_{1}}-\vec{f_{2}}=3(1, 1)-(1, 2)=(3-1, 3-2)=(2, 1)$ . These are the components of $\vec{v}$ in the standard basis.

b)

We look for $a$ and $b$ such that $a\vec{g_{1}}+b\vec{g_{2}}=(2, 1)$ , that is $\begin{cases} 2a+b=2 \\ a+b=1 \end{cases}$ . Subtracting the second from the first: $a=1$ , then $b=0$ . The coordinates of $\vec{v}$ relative to $\mathcal{B_{2}}$ are therefore $(1, 0)$ ; in other words $\vec{v}=\vec{g_{1}}$ , which is confirmed by the fact that $\vec{g_{1}}=(2, 1)=\vec{v}$ .

c)

For $\vec{f_{1}}=(1, 1)$ : $\begin{cases} 2a+b=1 \\ a+b=1 \end{cases}$ gives $a=0$ and $b=1$ , so $\vec{f_{1}}=\vec{g_{2}}$ . For $\vec{f_{2}}=(1, 2)$ : $\begin{cases} 2a+b=1 \\ a+b=2 \end{cases}$ gives $a=-1$ and $b=3$ , so $\vec{f_{2}}=-\vec{g_{1}}+3\vec{g_{2}}$ . Check: $-(2, 1)+3(1, 1)=(-2+3, -1+3)=(1, 2)$ .

d)

Substituting: $\vec{v}=3\vec{f_{1}}-\vec{f_{2}}=3\vec{g_{2}}-(-\vec{g_{1}}+3\vec{g_{2}})=\vec{g_{1}}+3\vec{g_{2}}-3\vec{g_{2}}=\vec{g_{1}}$ . We do find the coordinates $(1, 0)$ relative to $\mathcal{B_{2}}$ , without going through the standard basis. The change of coordinates can be done in two equivalent ways: either by going back through the standard components (b), or by substituting directly the expression of one basis in terms of the other (d). It is this substitution mechanism, once made systematic, that gives the transition matrix between two bases. On the figure, the same vector $\vec{v}$ reads $(2, 1)$ in the standard basis, $(3, -1)$ relative to $\mathcal{B_{1}}$ (the dashed path $3\vec{f_{1}}$ then $-\vec{f_{2}}$ ) and $(1, 0)$ relative to $\mathcal{B_{2}}$ , since it is $\vec{g_{1}}$ itself: the coordinates change, never the vector.

Exercise 9: Proofs: subsets, supersets and unions

We work in an arbitrary vector space. Recall that a subset of a set of vectors $F$ is obtained by removing vectors from $F$ , and a superset by adding vectors to it.

a) Prove that every subset of a linearly independent set is linearly independent.
b) Prove that every superset of a linearly dependent set is linearly dependent.
c) Deduce that in a linearly independent set, no vector is zero and no two distinct vectors are collinear.
d) True or false: the union of two linearly independent sets is always linearly independent. Justify, with an explicit counterexample if it is false.

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Answers

a) Complete with zero coefficients
b) Contrapositive of a)
c) Subsets $\{\vec 0\}$ and a collinear pair are dependent
d) False: $\{(1, 0)\}$ and $\{(2, 0)\}$

a)

Let $F=\{\vec{v_{1}}, \dots, \vec{v_{n}}\}$ be linearly independent and $G=\{\vec{v_{1}}, \dots, \vec{v_{k}}\}$ a subset with $k\le n$ (after renumbering if needed). Start from a zero combination of $G$ : $a_{1}\vec{v_{1}}+\dots+a_{k}\vec{v_{k}}=\vec{0}$ . Complete it into a zero combination of $F$ by adding the missing vectors with zero coefficients: $a_{1}\vec{v_{1}}+\dots+a_{k}\vec{v_{k}}+0\,\vec{v_{k+1}}+\dots+0\,\vec{v_{n}}=\vec{0}$ . Since $F$ is independent, all the coefficients are zero, in particular $a_{1}=\dots=a_{k}=0$ . So $G$ is linearly independent.

b)

This is the contrapositive of a). Let $F$ be dependent and $H$ a superset of $F$ . Since $F$ is dependent, there is a zero combination of $F$ whose coefficients are not all zero. Adding the extra vectors of $H$ with zero coefficients gives a zero combination of $H$ whose coefficients are not all zero (those inherited from $F$ prove it). So $H$ is linearly dependent. It can also be said in one breath: if the superset $H$ were independent, its subset $F$ would be too by a), which would contradict the hypothesis.

c)

Apply a). If a vector $\vec{v_{i}}$ of the set were zero, the subset $\{\vec{v_{i}}\}=\{\vec{0}\}$ would be dependent, since $1\cdot\vec{0}=\vec{0}$ is a non-trivial zero combination; that would contradict the independence of the set, by a). Likewise, if two vectors $\vec{v_{i}}$ and $\vec{v_{j}}$ were collinear, the subset $\{\vec{v_{i}}, \vec{v_{j}}\}$ would be dependent, the same contradiction. So in a linearly independent set, no vector is zero and the vectors are pairwise non-collinear.

d)

FALSE. Take in $\mathbb{R}^{2}$ the sets $F_{1}=\{(1, 0)\}$ and $F_{2}=\{(2, 0)\}$ . Each one, made of a single non-zero vector, is linearly independent. But their union $\{(1, 0), (2, 0)\}$ is dependent, since $(2, 0)=2(1, 0)$ : the two vectors are collinear. Independence is therefore not preserved by union. What remains true, by a), is that every subset of an independent set is still independent, never the reverse.

Exercise 10: True or false: dimension, sets and coordinates

For each statement, say whether it is true or false. A true statement must be justified; a false statement must come with an explicit counterexample. We work in $\mathbb{R}^{n}$ with its usual structure.

a) In $\mathbb{R}^{3}$ , every set of four vectors is linearly dependent.
b) In $\mathbb{R}^{3}$ , every set of two vectors is linearly independent.
c) If a set spans $\mathbb{R}^{3}$ , then it contains at least three vectors.
d) If $\{\vec{u}, \vec{v}, \vec{w}\}$ is a basis of $\mathbb{R}^{3}$ and $\lambda\neq 0$ , then $\{\lambda\vec{u}, \vec{v}, \vec{w}\}$ is still a basis of $\mathbb{R}^{3}$ .

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Answers

a) True
b) False: $(1, 2, 3)$ , $(2, 4, 6)$
c) True
d) True: the determinant is multiplied by $\lambda$

a)

TRUE. $\mathbb{R}^{3}$ has dimension $3$ : a linearly independent set in it contains at most three vectors. Four vectors exceed this ceiling, so the set is necessarily dependent, whatever the vectors chosen. For example $\{(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)\}$ is dependent, since $(1, 1, 1)$ is the sum of the first three.

b)

FALSE. Two collinear vectors form a dependent set. Counterexample: $\{(1, 2, 3), (2, 4, 6)\}$ , where the second vector is twice the first; the set is dependent. A set of two vectors is independent only if the two vectors are not collinear.

c)

TRUE. A spanning set of $\mathbb{R}^{3}$ contains at least as many vectors as the dimension, that is three. With only two vectors, the set of linear combinations is at best a plane (or a line, or the origin), never the whole space. For example $\{(1, 0, 0), (0, 1, 0)\}$ only produces vectors whose third component is zero: $(0, 0, 1)$ is never reached.

d)

TRUE. Multiplying a vector of a basis by a non-zero scalar keeps it a basis. The determinant shows it: that of $\{\lambda\vec{u}, \vec{v}, \vec{w}\}$ is $\lambda$ times that of $\{\vec{u}, \vec{v}, \vec{w}\}$ , which is non-zero; since $\lambda\neq 0$ , the new determinant stays non-zero, the set stays independent, hence a basis. If on the other hand $\lambda=0$ , the first vector becomes zero and the set is immediately dependent: the hypothesis $\lambda\neq 0$ is essential.

Exercise 11: Resolving a force along two given directions

A chandelier weighing $50\ \mathrm{N}$ hangs from two cables. In a coordinate system where $x$ is horizontal and $y$ vertical, the left cable pulls along the direction $\vec{u}=(-3, 4)$ and the right cable along $\vec{v}=(4, 3)$ .

At equilibrium, the tensions in the cables balance the weight: there are $a$ and $b$ such that $a\vec{u}+b\vec{v}=(0, 50)$ , and a cable can only PULL, so $a$ and $b$ must be positive.

The figure is not drawn to the scale of the forces: it shows the directions.

a) Show that $\{\vec{u}, \vec{v}\}$ is a basis of $\mathbb{R}^{2}$ .
b) Find $a$ and $b$ .
c) Compute the magnitude of the tension in each cable, in newtons.
d) The anchor of the right cable is moved: its direction becomes $\vec{w}=(6, -8)$ . Is equilibrium still possible? Justify with linear algebra.
e) Instead, the right cable is replaced by a horizontal cable, with direction $(1, 0)$ . Find the new coefficients and the tension in the left cable. Comment.

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Answers

a) Determinant $-25$
b) $a=8$ , $b=6$
c) 40 N and 30 N
d) No: $\vec{w}=-2\vec{u}$
e) $a=12.5$ , $b=37.5$ ; $62.5$ N

a)

$\det(\vec{u},\vec{v})=\begin{vmatrix} -3 & 4 \\ 4 & 3 \end{vmatrix}=-9-16=-25\neq 0$ : the set is linearly independent, and two independent vectors of $\mathbb{R}^{2}$ form a basis. The weight therefore resolves in a unique way along the two cables.

b)

$\begin{cases} -3a+4b=0 \\ 4a+3b=50 \end{cases}$ . The first gives $a=\frac{4b}{3}$ , then $\frac{16b}{3}+3b=50$ , that is $\frac{25b}{3}=50$ and $b=6$ , $a=8$ . Check: $8(-3, 4)+6(4, 3)=(-24+24, 32+18)=(0, 50)$ . Both coefficients are positive: the equilibrium can physically be achieved.

c)

$\|\vec{u}\|=\|\vec{v}\|=5$ . Left tension: $\|8\vec{u}\|=40\ \mathrm{N}$ . Right tension: $\|6\vec{v}\|=30\ \mathrm{N}$ . The sum of the magnitudes, $70\ \mathrm{N}$ , exceeds the weight: only the horizontal components cancel, the vertical ones add up.

d)

$\vec{w}=-2\vec{u}$ : the two directions are collinear, $\det(\vec{u},\vec{w})=(-3)(-8)-(6)(4)=0$ . The combinations of $\vec{u}$ and $\vec{w}$ stay on the line directed by $\vec{u}$ , and $(0, 50)$ is not on it, since $\det\big(\vec{u},(0, 50)\big)=-150\neq 0$ . Equilibrium is impossible: two aligned cables cannot hold a load that is not on their line.

e)

$a(-3, 4)+b(1, 0)=(0, 50)$ gives $4a=50$ , so $a=12.5$ , then $b=3a=37.5$ . Tension in the left cable: $12.5\times 5=62.5\ \mathrm{N}$ , more than the weight itself. The closer a cable is to horizontal, the harder it must pull: the useful vertical component is a small share of its tension.

The thread: resolving a force means finding its coordinates relative to the basis of the given directions; the sign of the coordinates tells whether the resolution can be achieved.

Exercise 12: Orthonormal basis: coordinates by dot products

In $\mathbb{R}^{2}$ , consider $\vec{e_{1}}=\left(\frac{3}{5}, \frac{4}{5}\right)$ and $\vec{e_{2}}=\left(-\frac{4}{5}, \frac{3}{5}\right)$ , and the vector $\vec{v}=(10, 5)$ .

On the figure, $\vec{e_{1}}$ and $\vec{e_{2}}$ are drawn twice as long, for readability.

a) Show that $\{\vec{e_{1}}, \vec{e_{2}}\}$ is an orthonormal basis of $\mathbb{R}^{2}$ .
b) Compute the coordinates of $\vec{v}$ relative to this basis using two dot products.
c) Check the result by rebuilding $\vec{v}$ from its coordinates.
d) Compute the norm of $\vec{v}$ from its coordinates relative to the new basis, to two decimal places, and compare with the one obtained in the standard basis. Why do they agree?
e) Now take the orthogonal but not normalized basis $\vec{f_{1}}=(3, 4)$ , $\vec{f_{2}}=(-4, 3)$ . Find the coordinates of $\vec{v}$ relative to this basis without solving a system.

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Answers

a) Orthogonal, of norm 1
b) $(10, -5)$
c) $10\vec{e_1}-5\vec{e_2}=(10, 5)$
d) $\sqrt{125}\approx 11.18$ in both bases
e) $(2, -1)$

a)

$\vec{e_{1}}\cdot\vec{e_{2}}=-\frac{12}{25}+\frac{12}{25}=0$ : the vectors are orthogonal. $\|\vec{e_{1}}\|^{2}=\frac{9}{25}+\frac{16}{25}=1$ and likewise $\|\vec{e_{2}}\|=1$ . Two non-zero orthogonal vectors are linearly independent, so they form a basis of $\mathbb{R}^{2}$ , and it is orthonormal.

b)

In an orthonormal basis, the coordinate along $\vec{e_{i}}$ is $\vec{v}\cdot\vec{e_{i}}$ . $\vec{v}\cdot\vec{e_{1}}=6+4=10$ and $\vec{v}\cdot\vec{e_{2}}=-8+3=-5$ . The coordinates are $(10, -5)$ .

c)

$10\vec{e_{1}}-5\vec{e_{2}}=(6, 8)-(-4, 3)=(10, 5)$ : we do get $\vec{v}$ back.

d)

$\sqrt{10^{2}+(-5)^{2}}=\sqrt{125}\approx 11.18$ , and in the standard basis $\sqrt{10^{2}+5^{2}}=\sqrt{125}$ as well. In an ORTHONORMAL basis, the norm is computed from the coordinates exactly as from the components, because the cross terms $2ab\,\vec{e_{1}}\cdot\vec{e_{2}}$ are zero and the basis vectors have norm $1$ . In an oblique basis, this computation would be wrong.

e)

The basis is orthogonal: if $\vec{v}=a\vec{f_{1}}+b\vec{f_{2}}$ , then $\vec{v}\cdot\vec{f_{1}}=a\|\vec{f_{1}}\|^{2}$ , hence $a=\frac{\vec{v}\cdot\vec{f_{1}}}{\|\vec{f_{1}}\|^{2}}=\frac{30+20}{25}=2$ and $b=\frac{-40+15}{25}=-1$ . Check: $2(3, 4)-(-4, 3)=(10, 5)$ . The trap is to divide by the norm instead of its SQUARE, which would give $10$ and $-5$ , the coordinates relative to the normalized basis.

The thread: orthogonality replaces solving a system with dot products; normalization sets the denominator to $1$ .

Exercise 13: The set with a parameter that changes rank

For every real number $t$ , consider in $\mathbb{R}^{3}$ the vectors $\vec{v_{1}}=(1, 1, t)$ , $\vec{v_{2}}=(1, t, 1)$ and $\vec{v_{3}}=(t, 1, 1)$ .

a) Compute the determinant $D(t)$ of the set $\{\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}\}$ and factor it completely.
b) For which values of $t$ is the set linearly dependent?
c) For $t=1$ , what is the dimension of the subspace spanned by the set?
d) For $t=-2$ , give a dependence relation, the dimension of the spanned subspace and a Cartesian equation of this subspace.
e) For $t=0$ , find the coordinates of the vector $(2, 2, 2)$ relative to the basis $\{\vec{v_{1}}, \vec{v_{2}}, \vec{v_{3}}\}$ .

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Answers

a) $D(t)=-(t-1)^2(t+2)$
b) $t=1$ or $t=-2$
c) Dimension 1
d) $\vec{v_1}+\vec{v_2}+\vec{v_3}=\vec 0$ ; plane $x+y+z=0$
e) $(1, 1, 1)$

a)

Expanding along the first row: $D(t)=1(t-1)-1(1-t)+t(1-t^{2})=2(t-1)-t(t-1)(t+1)$ . Factor out $(t-1)$ : $D(t)=(t-1)(2-t-t^{2})=-(t-1)(t^{2}+t-2)=-(t-1)^{2}(t+2)$ . Check: $D(0)=-2$ , and the factored form gives $-(1)(2)=-2$ .

b)

The set is dependent if and only if $D(t)=0$ , that is $t=1$ or $t=-2$ . For every other value, it is a basis of $\mathbb{R}^{3}$ .

c)

For $t=1$ , the three vectors equal $(1, 1, 1)$ : they are identical, and the spanned subspace is the line directed by $(1, 1, 1)$ , of dimension $1$ . The double root of the determinant signals here a drop in rank of two steps.

d)

For $t=-2$ : $\vec{v_{1}}=(1, 1, -2)$ , $\vec{v_{2}}=(1, -2, 1)$ , $\vec{v_{3}}=(-2, 1, 1)$ , and $\vec{v_{1}}+\vec{v_{2}}+\vec{v_{3}}=\vec{0}$ . $\vec{v_{1}}$ and $\vec{v_{2}}$ are not collinear, so the spanned subspace is a plane, of dimension $2$ . Each vector has components summing to zero, and the plane has equation $x+y+z=0$ .

e)

For $t=0$ : $\vec{v_{1}}=(1, 1, 0)$ , $\vec{v_{2}}=(1, 0, 1)$ , $\vec{v_{3}}=(0, 1, 1)$ , and $D(0)=-2\neq 0$ . Solve $a+b=2$ , $a+c=2$ , $b+c=2$ : adding, $2(a+b+c)=6$ , so $a+b+c=3$ and $c=b=a=1$ . The coordinates are $(1, 1, 1)$ .

The thread: the determinant detects the critical values, but only a look at the vectors at those values tells by how much the rank drops.

Exercise 14: Problem: the colours a screen can display

A screen mixes three diodes. Lit at full intensity, each diode excites the three colour sensors of the eye according to a vector of $\mathbb{R}^{3}$ : $\vec{R}=(4, 1, 0)$ , $\vec{G}=(1, 5, 1)$ and $\vec{B}=(0, 1, 4)$ .

Lit at the fraction $a$ of its power, the red diode contributes $a\vec{R}$ , and so on. A colour $\vec{T}$ can be displayed if there are $a$ , $b$ , $c$ between $0$ and $1$ such that $a\vec{R}+b\vec{G}+c\vec{B}=\vec{T}$ .

a) Show that $\{\vec{R}, \vec{G}, \vec{B}\}$ is a basis of $\mathbb{R}^{3}$ .
b) Can the colour $\vec{T_{1}}=(5, 7, 5)$ be displayed? Give the three intensities.
c) Same question for $\vec{T_{2}}=(2, 3, 2)$ , intensities to three decimal places.
d) Same question for $\vec{T_{3}}=(1, 6, 1)$ .
e) Since the set is a basis, every colour can be written as a combination of the three diodes. Why do some colours remain impossible to display? What is the volume of the set of displayable colours?

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Answers

a) Determinant 72
b) $(1, 1, 1)$ : displayable
c) $a=c\approx 0.389$ , $b\approx 0.444$ : displayable
d) $b=\frac{11}{9}>1$ : no
e) Coefficients outside $[0, 1]$ ; volume 72

a)

$\begin{vmatrix} 4 & 1 & 0 \\ 1 & 5 & 1 \\ 0 & 1 & 4 \end{vmatrix}=4(20-1)-1(4-0)+0=76-4=72\neq 0$ : the set is linearly independent, hence a basis of $\mathbb{R}^{3}$ .

b)

The system reads $4a+b=5$ , $a+5b+c=7$ , $b+4c=5$ . The first and the third give $a=c=\frac{5-b}{4}$ , then the second gives $\frac{5-b}{2}+5b=7$ , that is $9b=9$ : $b=1$ and $a=c=1$ . All three diodes at full power: $\vec{T_{1}}$ can be displayed, it is the brightest white of the screen.

c)

Likewise $a=c=\frac{2-b}{4}$ and $\frac{2-b}{2}+5b=3$ , that is $2-b+10b=6$ : $b=\frac{4}{9}\approx 0.444$ and $a=c=\frac{7}{18}\approx 0.389$ . The three intensities lie between $0$ and $1$ : $\vec{T_{2}}$ can be displayed.

d)

$a=c=\frac{1-b}{4}$ and $\frac{1-b}{2}+5b=6$ , that is $1-b+10b=12$ : $b=\frac{11}{9}\approx 1.222$ and $a=c=-\frac{1}{18}$ . The green diode would have to be lit beyond its maximum power and the other two at a NEGATIVE intensity: $\vec{T_{3}}$ , a green that is too saturated, cannot be displayed.

e)

Being a basis guarantees the EXISTENCE and uniqueness of the coefficients, not that they fall in $[0, 1]$ . The displayable colours are the combinations with coefficients in $[0, 1]$ : they fill the parallelepiped built on $\vec{R}$ , $\vec{G}$ , $\vec{B}$ , whose volume is $|\det|=72$ . This is the “gamut” of the screen; a colour outside the parallelepiped exists mathematically but not on the screen.

The thread: a basis says the coordinates exist; a real problem adds constraints on those coordinates.

Exercise 15: Problem: a crystal lattice and its bases

In a plane crystal, the atoms sit exactly at the points $m\vec{u}+n\vec{v}$ , where $m$ and $n$ are INTEGERS, with $\vec{u}=(2, 1)$ and $\vec{v}=(1, 3)$ . These points form a lattice, and a point is a lattice point if its coordinates relative to the basis $\{\vec{u}, \vec{v}\}$ are integers.

a) Is the point $P=(7, 11)$ a lattice point? Give its coordinates.
b) Same question for $Q=(4, 5)$ .
c) Compute the area of the unit cell, the parallelogram built on $\vec{u}$ and $\vec{v}$ .
d) Consider the basis $\vec{u'}=\vec{u}+\vec{v}=(3, 4)$ , $\vec{v'}=\vec{v}$ . Compute the coordinates of $P$ relative to this basis, then the area of its unit cell. Show that this basis generates exactly the same lattice.
e) Finally, consider the basis $\{2\vec{u}, \vec{v}\}$ . Compute the area of its unit cell and the coordinates of $\vec{u}$ relative to this basis. Does it generate the same lattice?

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Answers

a) $(2, 3)$ : lattice point
b) $(1.4, 1.2)$ : not a lattice point
c) Area 5
d) $(2, 1)$ , area 5: same lattice
e) Area 10; $\left(\frac12, 0\right)$ : a sparser lattice

a)

$a(2, 1)+b(1, 3)=(7, 11)$ : $2a+b=7$ and $a+3b=11$ . From the first $b=7-2a$ , then $a+21-6a=11$ , so $a=2$ and $b=3$ . Integer coordinates: $P=2\vec{u}+3\vec{v}$ is a lattice point.

b)

$2a+b=4$ and $a+3b=5$ give $b=4-2a$ , then $a+12-6a=5$ , that is $a=\frac{7}{5}=1.4$ and $b=\frac{6}{5}=1.2$ . The coordinates exist, since the set is a basis, but they are not integers: $Q$ is not a lattice point, it is a point between the atoms.

c)

$|\det(\vec{u},\vec{v})|=|2\times 3-1\times 1|=5$ .

d)

$3a+b=7$ and $4a+3b=11$ give $b=7-3a$ , then $4a+21-9a=11$ , that is $a=2$ , $b=1$ : $P=2\vec{u'}+\vec{v'}$ , coordinates still integers. $\det(\vec{u'},\vec{v'})=9-4=5$ , the same area. Same lattice: $\vec{u'}$ and $\vec{v'}$ are integer combinations of $\vec{u}$ and $\vec{v}$ , and conversely $\vec{u}=\vec{u'}-\vec{v'}$ and $\vec{v}=\vec{v'}$ are integer combinations of $\vec{u'}$ and $\vec{v'}$ . Every lattice point of one is a lattice point of the other.

e)

$\det(2\vec{u},\vec{v})=4\times 3-1\times 2=10$ : the unit cell is twice as large. $\vec{u}=\frac{1}{2}(2\vec{u})+0\,\vec{v}$ : coordinates $\left(\frac{1}{2}, 0\right)$ , not integers. The lattice point $\vec{u}$ is no longer reached: this basis generates a sparser lattice, which keeps only some of the atoms. Two bases generate the same lattice exactly when they have the same unit-cell area and each one is expressed in terms of the other with integer coefficients.

The thread: in a lattice, coordinates must be integers; changing basis without changing the lattice requires a change with integer coefficients and a preserved unit-cell area.

Corrected exercises: linear combinations, linear dependence and bases (201-SN4-RE)

Course recap

Computations: coefficients, coordinates and determinants

Exercise 0.1: Finding the coefficients of a linear combination: ten direct computations

Part A: the basics (/50)

Exercise 1: Components, norm and operations in the plane

Exercise 2: A linear combination and the system behind it

Exercise 3: Linear dependence and independence

Exercise 4: Bases and the coordinates of a vector

Exercise 5: Too many vectors: a dependent spanning set and the extraction of a basis

Part B: exam level (/50)

Exercise 6: Proofs: independent sets and the zero vector

Exercise 7: Problem: an oblique basis of space and the uniqueness of coordinates

Exercise 8: Problem: change of basis and coordinates

Exercise 9: Proofs: subsets, supersets and unions

Exercise 10: True or false: dimension, sets and coordinates

Part C: the classics (/50)

Exercise 11: Resolving a force along two given directions

Exercise 12: Orthonormal basis: coordinates by dot products

Exercise 13: The set with a parameter that changes rank

Exercise 14: Problem: the colours a screen can display

Exercise 15: Problem: a crystal lattice and its bases

See also

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