Linear Algebra and Vector Geometry 201-SN4-RE • CEGEP, Montreal

Revision sheet: linear combinations, linear dependence and bases (201-SN4-RE)

Q: How do I know if three vectors are linearly independent?

Write a zero linear combination of the three vectors, turn it into a system of equations, one per component, and solve it. The set is linearly independent if the only solution is the one where every coefficient equals zero. In three-dimensional space, it is enough to check that the determinant of the matrix of their components is not zero.

Q: What is the difference between a linearly independent set and a spanning set?

An independent set contains no redundancy: no vector can be written from the others. A spanning set reaches the whole space: every vector of the space is a combination of the set. A basis is both at once, and that is what makes the expression of each vector unique.

Q: How do I find the coordinates of a vector relative to a basis?

Write the vector as a linear combination of the basis vectors, with unknown coefficients, then solve the resulting system component by component. The coefficients you find are the coordinates. Careful: they are not the usual components of the vector, unless the basis is the standard basis.

Q: Can two vectors form a basis of three-dimensional space?

No. A basis of three-dimensional space has exactly three vectors. Two non-collinear vectors form a basis of the plane they span, but a whole direction is missing for them to reach all of space. Counting the vectors is enough to answer, with no computation at all.

Q: Why are three vectors of the plane always linearly dependent?

Because the plane has dimension two: a linearly independent set in it has at most two vectors. The associated system then has three unknowns for only two equations, so it has infinitely many solutions, and in particular a non-zero one. The set is dependent, without the slightest computation.

Q: What is the determinant used for in this CEGEP chapter?

It gives an immediate verdict when the number of vectors equals the dimension of the space. A non-zero determinant means that the set is linearly independent, spanning, and therefore a basis, all in a single computation. It exists only for a square matrix: outside that case, you have to use Gaussian elimination.

This sheet does not repeat the course on sets of vectors: you have it in your notes. It deals with what the course never says and what still decides the grade: the exact error that costs two marks, the system to set up for each question, and the write-up the 201-SN4-RE (formerly 201-NYC) grader expects word for word.

It is written for CEGEP students in the Science program, and for Social Science students in the mathematics profile who meet the same chapter in 201-105. Once the sheet is read, the corrected exercise set of the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Every question of the chapter is the SAME linear system: what changes from one question to the next is not the method, it is the right-hand side. A linear combination is $A X=\vec{w}$ ; an independent set is $A X=\vec{0}$ ; coordinates relative to a basis are again $A X=\vec{v}$ .

This chapter is part of 201-SN4-RE, Linear Algebra and Vector Geometry

The essentials

One machine for four questions

• Put the vectors of the set IN COLUMNS in a matrix $A$ , and every question of the chapter becomes a system $AX=B$ . Only $B$ changes.
• “Is $\vec{w}$ a linear combination of the set?”: $B=\vec{w}$ , and the answer is yes if the system is consistent.
• “Is the set linearly independent?”: $B=\vec{0}$ , and the answer is yes if the ONLY solution is zero.
• “What are the coordinates of $\vec{v}$ relative to the basis $\mathcal{B}$ ?”: $B=\vec{v}$ , with the vectors of $\mathcal{B}$ in columns, and the unique solution gives the coordinates.
• “Does the set span $\mathbb{R}^{n}$ ?”: the system must be consistent for EVERY right-hand side, which amounts to $n$ pivots.

Two collinear vectors only reach a line; two non-collinear vectors tile the whole plane. Same number of vectors, a completely different span.

Writing “let $a\vec{u}+b\vec{v}=\vec{w}$ , which gives the system ...” before computing almost always earns half a mark for method, even if the rest is wrong.

Independent, dependent, spanning, basis: the four exact words

• INDEPENDENT: $a_{1}\vec{v_{1}}+\dots+a_{k}\vec{v_{k}}=\vec{0}$ implies $a_{1}=\dots=a_{k}=0$ . It is an implication, not an observation on one example.
• DEPENDENT: there is a relation whose coefficients are not all zero. In other words, AT LEAST ONE vector can be written in terms of the others, not necessarily all of them.
• SPANNING $E$ : every vector of $E$ can be written as a linear combination of the set.
• BASIS: independent AND spanning. In $\mathbb{R}^{n}$ , a basis has EXACTLY $n$ vectors, never one more nor one less.
• Relative to a basis, the coordinates of a vector are UNIQUE. In a set that spans but is dependent, they are not: that is exactly what separates the two notions.

In the plane: one vector alone is independent but only spans a line; two non-collinear vectors form a basis; a third one keeps the set spanning but makes it dependent, here

\vec{w}=\vec{u}+\vec{v}

A set containing $\vec{0}$ is always dependent, and two proportional vectors always make the set dependent: two free verdicts to look for before any computation.

The count that settles the question without computing

• In $\mathbb{R}^{n}$ : more than $n$ vectors, the set is DEPENDENT, without the slightest computation.
• In $\mathbb{R}^{n}$ : fewer than $n$ vectors, the set does NOT span, without the slightest computation.
• Exactly $n$ vectors of $\mathbb{R}^{n}$ : a single determinant decides everything. If it is non-zero, the set is independent, spanning, and therefore a basis.
• In $\mathbb{R}^{2}$ , two vectors are dependent if and only if they are collinear, that is, if $\det\begin{pmatrix} a_{1} & a_{2} \\ b_{1} & b_{2} \end{pmatrix}=0$ .

Two vectors of

\mathbb{R}^{2}

: the determinant is the signed area of their parallelogram. Non-zero, a real area and a basis; zero, the figure flattens and the set is dependent.

This count never says that a set IS independent: three vectors of $\mathbb{R}^{4}$ may be independent or dependent, and you have to compute.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading “the set is dependent” as “the vectors are all collinear”

2 marks, and the next question with them, since you then look for a collinearity that does not exist.

What not to write

“ $\{\vec{u}, \vec{v}, \vec{w}\}$ is dependent, so the three vectors are pairwise collinear.”

What to write

“The set is dependent: AT LEAST one vector can be written in terms of the others. Here $\vec{w}=2\vec{u}-\vec{v}$ , while $\vec{u}$ and $\vec{v}$ are not collinear.”

None of these three vectors is collinear with another, and yet the set is dependent:

\vec{w}

is built by following

2\vec{u}

then

-\vec{v}

, as the dashed path shows.

Why: Collinearity is about TWO vectors; dependence is about the whole set. Three vectors can be dependent without any pair being collinear.

2. Reading a linear combination by eye, without setting up the system

the whole question, since

3\vec{u}+\vec{v}=(7, 5)

and not

(7, 4)

What not to write

“ $\vec{w}=(7, 4)$ , $\vec{u}=(2, 1)$ , $\vec{v}=(1, 2)$ : I can see that $\vec{w}=3\vec{u}+\vec{v}$ .”

What to write

“Let $a\vec{u}+b\vec{v}=\vec{w}$ , that is $\begin{cases} 2a+b=7 \\ a+2b=4 \end{cases}$ , hence $b=\frac{1}{3}$ and $a=\frac{10}{3}$ .”

3\vec{u}+\vec{v}

(dashed path) ends at

(7, 5)

, in red, next to

\vec{w}

. The system gives

\frac{10}{3}\vec{u}

then

\frac{1}{3}\vec{v}

(the green path), which lands exactly on

\vec{w}=(7, 4)

Why: The first component comes out right by chance one time in two. The system imposes BOTH components at once, and that is what you are graded on.

3. Computing a determinant of a matrix that is not square

1 mark for rigour, plus ten minutes spent looking for a formula that does not exist.

What not to write

“Three vectors of $\mathbb{R}^{2}$ : I form the $3\times 2$ matrix and compute its determinant to test independence.”

What to write

“A determinant exists only for a SQUARE matrix. Three vectors in a space of dimension $2$ always form a dependent set: the answer is immediate, with no computation.”

Why: The determinant test is reserved for the case “ $n$ vectors in $\mathbb{R}^{n}$ ”. Outside that case, the count of vectors answers, or Gaussian elimination does.

4. Giving the standard components as the coordinates relative to another basis

3 marks, that is the whole question, and often the next one that reuses the result.

What not to write

“ $\vec{v}=(3, 3)$ , so its coordinates relative to $\mathcal{B}=\{\vec{u_{1}}, \vec{u_{2}}\}$ are $(3, 3)$ .”

What to write

“ $(3, 3)$ are the components in the STANDARD basis. Relative to $\mathcal{B}$ , I solve $a\vec{u_{1}}+b\vec{u_{2}}=\vec{v}$ and get $(2, 1)$ , since $2\vec{u_{1}}+\vec{u_{2}}=(3, 3)$ .”

One vector, two readings:

(3, 3)

on the standard grid,

(2, 1)

relative to the oblique basis

\mathcal{B}

, as the dashed path

2\vec{u_{1}}

then

\vec{u_{2}}

shows.

Why: A pair of numbers means nothing on its own: you must know IN WHICH BASIS it is read. The same vector has as many expressions as there are bases.

5. Promoting an independent set of two vectors to a basis of space

2 marks, and the question on coordinates becomes impossible to finish.

What not to write

“ $\vec{u}$ and $\vec{v}$ are not collinear, so $\{\vec{u}, \vec{v}\}$ is a basis of $\mathbb{R}^{3}$ .”

What to write

“ $\{\vec{u}, \vec{v}\}$ is independent, so it is a basis of the PLANE $\text{span}\{\vec{u}, \vec{v}\}$ . It is not a basis of $\mathbb{R}^{3}$ , which needs exactly three vectors.”

The combinations of

\vec{u}

and

\vec{v}

stay in the shaded plane;

\vec{w}

leaves it and is reached by none of them. The set is independent, and yet it does not span

\mathbb{R}^{3}

Why: Independent and spanning are two separate requirements. Two vectors of $\mathbb{R}^{3}$ cannot span space: a whole direction is missing.

6. Hoping that a set containing the zero vector is independent

the whole proof question, often 4 marks.

What not to write

“ $\{\vec{0}, \vec{u}, \vec{v}\}$ : since $\vec{u}$ and $\vec{v}$ are not collinear, the set is independent.”

What to write

“Every set containing $\vec{0}$ is dependent: $1\cdot\vec{0}+0\cdot\vec{u}+0\cdot\vec{v}=\vec{0}$ is a relation whose coefficients are not all zero.”

Why: ONE non-trivial relation is enough to make the set dependent. The coefficient $1$ in front of $\vec{0}$ provides it for free, whatever the other vectors.

7. Testing independence with the wrong right-hand side

3 marks, the whole proof being about the wrong object.

What not to write

“The system $a\vec{u}+b\vec{v}=\vec{w}$ has a solution, so $\{\vec{u}, \vec{v}\}$ is independent.”

What to write

“This system says that $\vec{w}\in\text{span}\{\vec{u}, \vec{v}\}$ , nothing more. Independence is tested on the HOMOGENEOUS system $a\vec{u}+b\vec{v}=\vec{0}$ .”

a\vec{u}+b\vec{v}=\vec{w}

has solutions, and yet

\vec{v}=2\vec{u}

: the set is dependent. Consistent does not mean independent: only the right-hand side

\vec{0}

tests independence.

Why: This is the thread of the chapter: the right-hand side changes the question. A non-zero right-hand side asks about the span, a zero right-hand side asks about independence.

8. Claiming that coordinates are unique for any set

2 marks, and above all the understanding of what a basis adds.

What not to write

“The coordinates of a vector are always unique, so it does not matter whether the set is dependent.”

What to write

“Uniqueness holds RELATIVE TO A BASIS. With $\vec{u}=(1, 0)$ , $\vec{v}=(0, 1)$ and $\vec{w}=(1, 1)$ , we have $(2, 2)=2\vec{u}+2\vec{v}$ and also $(2, 2)=\vec{u}+\vec{v}+\vec{w}$ : two expressions for the same vector.”

With

\vec{w}=\vec{u}+\vec{v}

, two paths reach

(2, 2)

2\vec{u}+2\vec{v}

in blue,

\vec{u}+\vec{v}+\vec{w}

in orange. A dependent set: the coordinates are no longer unique.

Why: Uniqueness of the expression is EXACTLY what independence adds to spanning. That is why a basis is useful and a dependent spanning set is not.

Which method to choose

Which system to set up, depending on what the question asks

The vectors always go in columns in $A$ . The only choice to make is the RIGHT-HAND SIDE, and the verb of the question dictates it.

If “Is $\vec{w}$ a linear combination of $\vec{v_{1}},\dots,\vec{v_{k}}$ ?” → solve $AX=\vec{w}$ ; the answer is yes if the system is consistent

Example: $\begin{cases} 2a+b=7 \\ a+2b=4 \end{cases}$ is consistent, so $\vec{w}\in\text{span}\{\vec{u}, \vec{v}\}$
If “is the set linearly independent?” or “show that the set is linearly independent” → solve the HOMOGENEOUS system $AX=\vec{0}$ ; independent if the only solution is $X=\vec{0}$

Example: $\det A=3\neq 0$ is enough when the matrix is square
If “what are the coordinates of $\vec{v}$ relative to the basis $\mathcal{B}$ ?” → solve $AX=\vec{v}$ with the vectors of $\mathcal{B}$ in columns; the unique solution IS the pair of coordinates

Example: $2\vec{u_{1}}+\vec{u_{2}}=(3, 3)$ gives the coordinates $(2, 1)$
If “does the set span $\mathbb{R}^{n}$ ?” → row reduce $A$ and count the pivots: spanning if and only if there are $n$ pivots

No need to carry a general right-hand side: counting the pivots is enough.
If “is it a basis of $\mathbb{R}^{n}$ ?” → first check that there are EXACTLY $n$ vectors, then a single non-zero determinant concludes

If the number of vectors is not $n$ , the answer is no and there is nothing to compute.

No branch applies? Then the set lives in $P_{n}$ or in $M_{m\times n}$ : write each vector in the standard basis of that space (the monomials, the elementary matrices) and you are back to an ordinary system.

What counting alone settles, before any computation

Compare the number $k$ of vectors with the dimension $n$ of the space, and look at the set without computing.

If $k>n$ in $\mathbb{R}^{n}$ → the set is DEPENDENT, the question is settled

Example: three vectors of $\mathbb{R}^{2}$ , four vectors of $\mathbb{R}^{3}$
If $k<n$ in $\mathbb{R}^{n}$ → the set does NOT span, so it is not a basis

Example: two vectors of $\mathbb{R}^{3}$ only span a plane
If the set contains $\vec{0}$ → it is DEPENDENT, whatever the other vectors
If two vectors of the set are proportional → it is DEPENDENT, without looking at the others

Example: $\vec{v_{2}}=-3\vec{v_{1}}$ provides the relation $3\vec{v_{1}}+\vec{v_{2}}=\vec{0}$
If $k=n$ in $\mathbb{R}^{n}$ → compute $\det A$ : non-zero gives independent, spanning and a basis in one go

It is the only case where a determinant is legitimate, and it saves the whole Gaussian elimination.

This count NEVER proves that a set is independent when $k<n$ : three vectors of $\mathbb{R}^{4}$ may be independent or dependent, and you have to row reduce.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Showing that a set is linearly independent

When to use it: The question says “show that the set is linearly independent”, “show that the vectors are linearly independent”, or “show that $\mathcal{B}$ is a basis”.

1 Open with the set-up sentence: “Let $a$ , $b$ , $c$ be real numbers such that $a\vec{u}+b\vec{v}+c\vec{w}=\vec{0}$ .” Without it, what follows has no logical status.
2 Translate this vector equation into a system, ONE EQUATION PER COMPONENT, and write it out in full.
3 Announce the method used, determinant or Gaussian elimination, and carry it through to the end.
4 Conclude on the solution of the system, saying explicitly that it is the ONLY one.
5 If the question asked for a basis of $\mathbb{R}^{n}$ , add the count: the set is independent and has $n$ vectors, so it is a basis.

Concluding sentence

The only solution of the system is $a=b=c=0$ , so the set $\{\vec{u}, \vec{v}, \vec{w}\}$ is linearly independent; since it has $3$ vectors in $\mathbb{R}^{3}$ , it is a basis of $\mathbb{R}^{3}$ .

The trap: Writing “suppose the set is dependent” and then looking for a contradiction. It is not wrong, but it is three times longer and the conclusion is almost always badly quantified. The direct proof is the one expected.

Marking: 0.5 mark for the set-up sentence, 1 mark for the correct system, 1 mark for solving it, 0.5 mark for the conclusion with the word “only”.

Finding the coordinates of a vector relative to a basis

When to use it: The question gives a non-standard basis $\mathcal{B}$ and asks for “the coordinates of $\vec{v}$ relative to $\mathcal{B}$ ”.

1 First check that $\mathcal{B}$ really is a basis, with the determinant, and SAY it: what follows only makes sense if the expression is unique.
2 Write $\vec{v}=a\vec{u_{1}}+b\vec{u_{2}}$ with named unknowns, then write the system component by component.
3 Solve, keeping exact values rather than decimals.
4 Check by recomputing $a\vec{u_{1}}+b\vec{u_{2}}$ and getting $\vec{v}$ back, then write this check on your paper.
5 Conclude with the coordinate notation, recalling the basis they are read in.

Concluding sentence

The coordinates of $\vec{v}$ relative to the basis $\mathcal{B}$ are $(2, 1)$ , in other words $\vec{v}=2\vec{u_{1}}+\vec{u_{2}}$ .

The trap: Missing the reverse question. When the question gives the coordinates and asks for the components, there is NO system to solve: multiply and add, that is all.

Marking: 0.5 mark for checking that $\mathcal{B}$ is a basis, 1.5 marks for the system and its solution, 0.5 mark for the conclusion written in the right basis.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Substitute the coefficients back in

Take the $a$ and $b$ you found and recompute $a\vec{u}+b\vec{v}$ component by component. The result must be exactly the right-hand side, on ALL the components.

$a=\frac{10}{3}$ and $b=\frac{1}{3}$ give back $\left(\frac{20}{3}+\frac{1}{3}, \frac{10}{3}+\frac{2}{3}\right)=(7, 4)$ . Getting $(7, 5)$ signals that the second equation was forgotten.
Count before believing the result

Compare the number of vectors with the dimension. A set of $4$ vectors of $\mathbb{R}^{3}$ announced as independent is certainly wrong, without rereading a single line of the computation.

A paper concludes “these four vectors of $\mathbb{R}^{3}$ are independent”: that is impossible, the error is in the row reduction.
Recompute the determinant along another row

Expand the $3\times 3$ determinant along ANOTHER row or another column, preferably the one with the most zeros. The two values must agree.

On the calculator, TI: $\texttt{MATRIX} \to \texttt{MATH} \to \texttt{det(}$ ; Casio: menu $\texttt{MAT}$ then $\texttt{Det}$ . A sign difference betrays a cofactor with the wrong sign.
Check the size of the system you set up

The system must have as many EQUATIONS as the vectors have components, and as many UNKNOWNS as there are vectors in the set. A square system when the set is not square is the symptom of a set-up error.

Three vectors of $\mathbb{R}^{2}$ give $2$ equations in $3$ unknowns, so at least one free unknown, so a dependent set: the count confirms the result.
Look at the picture in the plane

In dimension $2$ , sketch the vectors freehand. Two collinear vectors show up immediately, and a third vector can never escape the plane.

$\vec{u}=(2, 1)$ and $\vec{v}=(4, 2)$ lie on the same line: no need to compute the determinant, the set is dependent.

The typical problem, taken apart

An oblique basis of the plane, in both directions

Let $\vec{u_{1}}=(2, 1)$ , $\vec{u_{2}}=(-1, 1)$ and $\vec{v}=(3, 3)$ in $\mathbb{R}^{2}$ with its standard basis.

1) Show that $\mathcal{B}=\{\vec{u_{1}}, \vec{u_{2}}\}$ is a basis of $\mathbb{R}^{2}$ .

2) Find the coordinates of $\vec{v}$ relative to the basis $\mathcal{B}$ .

3) The vector $\vec{w}$ has coordinates $(1, -2)$ relative to $\mathcal{B}$ . Give its components in the standard basis.

The three vectors of the problem on the standard grid: question 2 asks how to reach

\vec{v}

by walking only along

\vec{u_{1}}

and

\vec{u_{2}}

Step 1

$\det\begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}=2\times 1-(-1)\times 1=3\neq 0$ , so $\{\vec{u_{1}}, \vec{u_{2}}\}$ is linearly independent.

Why

Two vectors of $\mathbb{R}^{2}$ : the case $k=n$ , the only one where a determinant is legitimate. It answers in one line where the homogeneous system would take five.

Step 2

The set is independent and has $2$ vectors in $\mathbb{R}^{2}$ , which has dimension $2$ : it is therefore a basis.

Why

This is the counting argument, and it is worth half a mark on its own. Without it, you have proved independence but not spanning, so not the basis.

Step 3

Let $\vec{v}=a\vec{u_{1}}+b\vec{u_{2}}$ , which gives $\begin{cases} 2a-b=3 \\ a+b=3 \end{cases}$ .

Why

One equation per component, never just one. This is where question 2 is decided: the right-hand side is $\vec{v}$ , not $\vec{0}$ .

Step 4

Adding the two equations: $3a=6$ , so $a=2$ , then $b=3-a=1$ .

Why

Adding eliminates $b$ with no manipulation: it is the elementary row operation $R_{1}\to R_{1}+R_{2}$ , the one Gaussian elimination would perform anyway.

Step 5

Check: $2\vec{u_{1}}+\vec{u_{2}}=(4, 2)+(-1, 1)=(3, 3)=\vec{v}$ .

Why

Substituting back costs ten seconds and catches every sign error, the most frequent accident when the basis contains a $-1$ .

Step 6

Question 3: $\vec{w}=1\times\vec{u_{1}}+(-2)\times\vec{u_{2}}=(2, 1)-(-2, 2)=(4, -1)$ .

Why

The reverse direction, so NO system: the coordinates are given, you only multiply and add. Recognizing this direction saves five minutes.

Step 7

Check: solving $2a-b=4$ and $a+b=-1$ gives $3a=3$ , that is $a=1$ and $b=-2$ , the starting coordinates.

Why

Going there and back is the only check that tests both questions at once, and it uses the same system, so it costs almost nothing.

The conclusion, written out

$\mathcal{B}$ is a basis of $\mathbb{R}^{2}$ ; the coordinates of $\vec{v}$ relative to $\mathcal{B}$ are $(2, 1)$ , and the vector $\vec{w}$ has components $(4, -1)$ in the standard basis.

The classic mistake on this problem: Answering $(3, 3)$ to question 2, that is, copying the standard components, and treating question 3 as a system when it only asks for a direct computation.

Learn by heart

• The vectors go IN COLUMNS in $A$ . After that, only the right-hand side changes from one question to the next.
• Independence is tested on $AX=\vec{0}$ . Membership in the span and coordinates are tested on $AX=\vec{b}$ .
• Dependent means that AT LEAST one vector depends on the others, not that all of them are collinear.
• In $\mathbb{R}^{n}$ : more than $n$ vectors, always DEPENDENT. Fewer than $n$ , never SPANNING. Exactly $n$ , one determinant decides.
• A basis of $\mathbb{R}^{n}$ has EXACTLY $n$ vectors, and it is independent and spanning.
• Every set containing $\vec{0}$ is dependent, without computation.
• Coordinates are unique RELATIVE TO A BASIS, and only there.
• Coordinates known, components wanted: multiply. Components known, coordinates wanted: solve.
• $\det\begin{pmatrix} a & b \\ c & d \end{pmatrix}=ad-bc$ , and it is written only for a SQUARE matrix.

Frequently asked questions

How do I know if three vectors are linearly independent?

Write a zero linear combination of the three vectors, turn it into a system of equations, one per component, and solve it. The set is linearly independent if the only solution is the one where every coefficient equals zero. In three-dimensional space, it is enough to check that the determinant of the matrix of their components is not zero.

What is the difference between a linearly independent set and a spanning set?

An independent set contains no redundancy: no vector can be written from the others. A spanning set reaches the whole space: every vector of the space is a combination of the set. A basis is both at once, and that is what makes the expression of each vector unique.

How do I find the coordinates of a vector relative to a basis?

Write the vector as a linear combination of the basis vectors, with unknown coefficients, then solve the resulting system component by component. The coefficients you find are the coordinates. Careful: they are not the usual components of the vector, unless the basis is the standard basis.

Can two vectors form a basis of three-dimensional space?

No. A basis of three-dimensional space has exactly three vectors. Two non-collinear vectors form a basis of the plane they span, but a whole direction is missing for them to reach all of space. Counting the vectors is enough to answer, with no computation at all.

Why are three vectors of the plane always linearly dependent?

Because the plane has dimension two: a linearly independent set in it has at most two vectors. The associated system then has three unknowns for only two equations, so it has infinitely many solutions, and in particular a non-zero one. The set is dependent, without the slightest computation.

What is the determinant used for in this CEGEP chapter?

It gives an immediate verdict when the number of vectors equals the dimension of the space. A non-zero determinant means that the set is linearly independent, spanning, and therefore a basis, all in a single computation. It exists only for a square matrix: outside that case, you have to use Gaussian elimination.

Practise it

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

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

16 corrected exercises
160 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Revision sheet: linear combinations, linear dependence and bases (201-SN4-RE)

The essentials

One machine for four questions

Independent, dependent, spanning, basis: the four exact words

The count that settles the question without computing

The mistakes that cost marks

1. Reading “the set is dependent” as “the vectors are all collinear”

2. Reading a linear combination by eye, without setting up the system

3. Computing a determinant of a matrix that is not square

4. Giving the standard components as the coordinates relative to another basis

5. Promoting an independent set of two vectors to a basis of space

6. Hoping that a set containing the zero vector is independent

7. Testing independence with the wrong right-hand side

8. Claiming that coordinates are unique for any set

Which method to choose

Which system to set up, depending on what the question asks

What counting alone settles, before any computation

How the answer is expected to be written

Showing that a set is linearly independent

Finding the coordinates of a vector relative to a basis

Check before you hand in

The typical problem, taken apart

An oblique basis of the plane, in both directions

Learn by heart

Frequently asked questions

How do I know if three vectors are linearly independent?

What is the difference between a linearly independent set and a spanning set?

How do I find the coordinates of a vector relative to a basis?

Can two vectors form a basis of three-dimensional space?

Why are three vectors of the plane always linearly dependent?

What is the determinant used for in this CEGEP chapter?

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

See also

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