Corrected exercises: vector spaces and subspaces (201-SN4-RE)

An element of $P_{2}$ is written $ax^{2}+bx+c$. ZERO vector: the zero polynomial $0x^{2}+0x+0$, which does satisfy $p+0=p$ for every $p$. Negative of $p(x)=3x^{2}-x+5$: the polynomial $-p(x)=-3x^{2}+x-5$, whose sum with $p$ is the zero polynomial ✓. Closure under addition: the sum of two polynomials of degree at most 2 has degree at most 2, since the coefficients are added term by term and no higher degree is ever created ✓. Closure under scalar multiplication: $\lambda(ax^{2}+bx+c)=(\lambda a)x^{2}+(\lambda b)x+\lambda c$, still of degree at most 2 ✓. The other axioms (associativity, commutativity, distributivity) follow from those of the real numbers, applied coefficient by coefficient. So $P_{2}$ is indeed a vector space, and its “vectors” are polynomials.

Standard basis of $P_{2}$: $\left\{1,\;x,\;x^{2}\right\}$. It SPANS $P_{2}$, since every polynomial $ax^{2}+bx+c$ can be written $c\cdot 1+b\cdot x+a\cdot x^{2}$; it is LINEARLY INDEPENDENT, since $c+bx+ax^{2}=0$ for every $x$ forces $a=b=c=0$ (the zero polynomial has all its coefficients equal to zero). Hence $\dim P_{2}=3$. For $M_{2\times 2}$, the standard basis consists of the four matrices with a single 1 and 0 everywhere else, so $\dim M_{2\times 2}=4$. General rule to remember: $\dim P_{n}=n+1$ (watch the $+1$, forgotten very often) and $\dim M_{m\times n}=mn$.

NO. Counterexample: take $p(x)=x^{2}+x$ and $q(x)=-x^{2}+3$, both of degree exactly 2. Their sum is $p+q=x+3$, of degree 1: it does NOT belong to the set, so closure under addition fails. A second argument, even faster, settles the question on its own: the zero polynomial does not have degree exactly 2, so the zero vector is missing and the test fails at once. This is the decisive difference between “degree at most $n$” and “degree exactly $n$”: only the first wording gives a vector space, and exam questions play on this distinction all the time.

The axiom that fails is $1\cdot u=u$. Indeed $1\odot(x,y)=(1\cdot x,0)=(x,0)$, which differs from $(x,y)$ as soon as $y\neq 0$. For example $1\odot(3,7)=(3,0)\neq(3,7)$. This axiom looks like the most harmless of the eight, so much so that one is tempted to call it superfluous, and that is exactly why textbooks single it out: without it, scalar multiplication could destroy information, as it does here to the second coordinate. One can check that distributivity, $\lambda\odot(u+v)=\lambda\odot u+\lambda\odot v$, still holds, which shows that a single failed axiom is enough to disqualify the structure, even when all the others are satisfied.

Test: $W$ is a subspace of $V$ if 1) $\vec{0}\in W$; 2) $u,v\in W\Rightarrow u+v\in W$; 3) $u\in W$ and $\lambda\in\mathbb{R}\Rightarrow\lambda u\in W$. Checking $W$. 1) $(0,0,0)$ gives $0+0-0=0$ ✓. 2) If $u=(x_{1},y_{1},z_{1})$ and $v=(x_{2},y_{2},z_{2})$ each satisfy the equation, their sum gives $(x_{1}+x_{2})+2(y_{1}+y_{2})-(z_{1}+z_{2})=(x_{1}+2y_{1}-z_{1})+(x_{2}+2y_{2}-z_{2})=0+0=0$ ✓. 3) $\lambda u$ gives $\lambda x+2\lambda y-\lambda z=\lambda(x+2y-z)=\lambda\cdot 0=0$ ✓. So $W$ is a subspace: it is the PLANE through the origin with normal vector $(1,2,-1)$, and its dimension is 2.

NO: the zero vector $(0,0,0)$ gives $0+0-0=0\neq 3$, so $\vec{0}\notin U$ and the test fails on its very first condition. Geometrically, $U$ is a plane that does NOT pass through the origin, that is, a translate of $W$. The general rule shows here: a HOMOGENEOUS equation (right-hand side zero) defines a subspace, a non-homogeneous equation never does. This is the test to apply first to any set defined by an equation.

NO, even though it contains the zero vector. The condition $xy=0$ means $x=0$ OR $y=0$: geometrically, $S$ is the UNION of the two coordinate axes, a cross shape. Counterexample to closure under addition: $u=(1,0)\in S$ (because $1\times 0=0$) and $v=(0,1)\in S$, but $u+v=(1,1)$ gives $1\times 1=1\neq 0$, so $u+v\notin S$ ✓. The example is instructive: $S$ is closed under scalar multiplication (multiplying a point of an axis keeps it on that axis) and contains the zero vector, yet it fails on addition. The three conditions are therefore INDEPENDENT and must all be checked; passing the zero vector check guarantees nothing. More generally, the union of two subspaces is almost never a subspace, whereas their intersection always is.

Let $T=\left\{A\in M_{2\times 2}:\operatorname{tr}(A)=0\right\}$, where $\operatorname{tr}\begin{pmatrix}a&b\\c&d\end{pmatrix}=a+d$. 1) The zero matrix has trace zero ✓. 2) $\operatorname{tr}(A+B)=\operatorname{tr}(A)+\operatorname{tr}(B)=0+0=0$ ✓. 3) $\operatorname{tr}(\lambda A)=\lambda\operatorname{tr}(A)=0$ ✓. So it is a subspace, and the check is immediate because the trace is a LINEAR map: any set defined by “a linear map equals zero” is automatically a subspace, which is exactly the notion of kernel met later on. Dimension: the condition $d=-a$ leaves $a$, $b$ and $c$ free, so a basis is $\left\{\begin{pmatrix}1&0\\0&-1\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix},\begin{pmatrix}0&0\\1&0\end{pmatrix}\right\}$ and $\dim T=3$. This matches intuition: one non-trivial linear constraint removes exactly one dimension, here from 4 to 3.

$\operatorname{span}\left\{(1,2,1)\right\}$ is the set of multiples $\lambda(1,2,1)$: it is the LINE through the origin with direction vector $(1,2,1)$, of dimension 1. For the second one, the two vectors $(1,2,1)$ and $(0,1,1)$ are not scalar multiples of each other (the second has a zero first component, the first does not), so they are linearly independent and span a PLANE through the origin, of dimension 2. In general, $k$ linearly independent vectors span a subspace of dimension $k$: a line for 1, a plane for 2, the whole space for 3 in $\mathbb{R}^{3}$.

Test whether $(2,5,3)$ is a linear combination of the first two: look for $a$ and $b$ such that $a(1,2,1)+b(0,1,1)=(2,5,3)$. The first component gives $a=2$; the third gives $a+b=3$, so $b=1$; check on the second: $2a+b=4+1=5$ ✓. So $(2,5,3)=2(1,2,1)+1(0,1,1)$: the third vector is REDUNDANT, it adds nothing. The set spans the same plane as in part a), and $\dim=2$. Lesson: the number of vectors in a spanning set is NOT the dimension; the dimension is the number of vectors in a linearly INDEPENDENT spanning set, that is, the rank of the set.

Look for $a,b$ with $a(1,2,1)+b(0,1,1)=(3,7,4)$. First component: $a=3$. Third: $a+b=4$, so $b=1$. Check on the second: $2a+b=6+1=7$ ✓. The system is CONSISTENT, so $(3,7,4)=3(1,2,1)+(0,1,1)$ does belong to the subspace. A note on method: two equations find the unknowns, and the third serves as a consistency TEST. Had it failed, for example with $(3,7,9)$, which gives $a=3$, then $b=6$ and $2a+b=12\neq 7$, the vector would not have been in the subspace. This is the standard procedure for any membership question.

It is enough to check the three conditions of the test, and each one follows from the very structure of linear combinations. 1) The zero vector is obtained by taking every coefficient equal to zero, $0v_{1}+\cdots+0v_{k}=\vec{0}$ ✓. 2) The sum of two linear combinations of the $v_{i}$ is again a linear combination of the $v_{i}$, by adding the coefficients term by term ✓. 3) A multiple of a linear combination is a linear combination, by multiplying each coefficient by $\lambda$ ✓. The conclusion therefore holds for any set at all, even a dependent one, even a single vector, and even the empty set (in which case the convention is $\operatorname{span}\emptyset=\left\{\vec{0}\right\}$, the zero subspace). This is what makes the span the universal construction tool: to build a subspace, pick some vectors and take all their linear combinations; what you get is in fact the SMALLEST subspace containing the original set.

Let $u=(x_{1},y_{1},z_{1})$, $v=(x_{2},y_{2},z_{2})$ and $\lambda\in\mathbb{R}$. Additivity: $T(u+v)=\left((x_{1}+x_{2})+(y_{1}+y_{2}),\;(y_{1}+y_{2})+(z_{1}+z_{2})\right)=\left((x_{1}+y_{1})+(x_{2}+y_{2}),\;(y_{1}+z_{1})+(y_{2}+z_{2})\right)=T(u)+T(v)$ ✓. Homogeneity: $T(\lambda u)=(\lambda x+\lambda y,\;\lambda y+\lambda z)=\lambda(x+y,\;y+z)=\lambda T(u)$ ✓. So $T$ is linear. Its matrix with respect to the standard bases is $\begin{pmatrix}1&1&0\\0&1&1\end{pmatrix}$, of size $2\times 3$, consistent with a map from $\mathbb{R}^{3}$ to $\mathbb{R}^{2}$.

$\ker T$ is the set of $(x,y,z)$ such that $T(x,y,z)=(0,0)$, that is, the system $x+y=0$ and $y+z=0$. The first equation gives $x=-y$; the second gives $z=-y$. Taking $y=t$ as a free parameter: $(x,y,z)=(-t,t,-t)=t(-1,1,-1)$. A basis of $\ker T$ is therefore $\left\{(-1,1,-1)\right\}$ and $\dim(\ker T)=1$. Geometrically, it is the LINE through the origin with direction $(-1,1,-1)$: the vectors on this line, and only those, are sent to the zero vector by $T$.

$\operatorname{im}\,T$ is spanned by the images of the standard basis vectors: $T(1,0,0)=(1,0)$, $T(0,1,0)=(1,1)$, $T(0,0,1)=(0,1)$. These three vectors live in $\mathbb{R}^{2}$, so they cannot be linearly independent; but $(1,0)$ and $(0,1)$ already span all of $\mathbb{R}^{2}$. So $\operatorname{im}\,T=\mathbb{R}^{2}$ and $\dim(\operatorname{im}\,T)=2$. The transformation is therefore ONTO: every vector of $\mathbb{R}^{2}$ is reached. Check with the rank-nullity theorem: $\dim(\ker T)+\dim(\operatorname{im}\,T)=1+2=3=\dim\mathbb{R}^{3}$ ✓, which confirms both computations at once.

Let $T:V\to W$ be linear. 1) Zero vector: linearity gives $T(\vec{0})=T(0\cdot\vec{0})=0\cdot T(\vec{0})=\vec{0}$, so $\vec{0}\in\ker T$ ✓. 2) Addition: if $u,v\in\ker T$, then $T(u+v)=T(u)+T(v)=\vec{0}+\vec{0}=\vec{0}$, so $u+v\in\ker T$ ✓. 3) Scalar multiplication: if $u\in\ker T$ and $\lambda\in\mathbb{R}$, then $T(\lambda u)=\lambda T(u)=\lambda\vec{0}=\vec{0}$, so $\lambda u\in\ker T$ ✓. All three conditions hold, so $\ker T$ is a subspace of $V$. Notice that each step uses ONLY the linearity of $T$: the proof is valid for every linear transformation, between any spaces, including spaces of polynomials or of matrices. This is the model of abstract proof that the exam asks you to reproduce.

Write each polynomial as its coefficient vector: $p_{1}=(1, 1, 0)$, $p_{2}=(0, 1, 1)$, $p_{3}=(1, 0, 1)$. The set is linearly independent if the determinant of these three rows is non-zero: $\begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix}=1(1-0)-1(0-1)+0=1+1=2\neq 0$. The determinant is $2$, so the set is linearly independent: a vanishing combination $ap_{1}+bp_{2}+cp_{3}=0$ has only the trivial solution $a=b=c=0$.

$P_{2}$ has dimension 3. A linearly independent set of three vectors in a space of dimension 3 is automatically a basis: it necessarily spans the space, with nothing more to check. This is the counting argument, valid in every finite-dimensional space, and here it saves solving a general system.

Look for $a$, $b$, $c$ such that $ap_{1}+bp_{2}+cp_{3}=q$, that is, in coefficients, $a(1, 1, 0)+b(0, 1, 1)+c(1, 0, 1)=(2, 3, 1)$. The system is $\begin{cases} a+c=2 \\ a+b=3 \\ b+c=1 \end{cases}$. Subtracting the third equation from the second: $a-c=2$; together with the first, $a+c=2$, this gives $a=2$ and $c=0$, then $b=3-a=1$. Check on the third: $b+c=1+0=1$ ✓. The coordinates of $q$ with respect to the basis are therefore $(2, 1, 0)$, that is, $q=2p_{1}+p_{2}$. Check: $2(1+x)+(x+x^{2})=2+3x+x^{2}$ ✓.

In coefficients: $r_{1}=(1, 1, 0)$, $r_{2}=(0, 1, 1)$, $r_{3}=(1, 2, 1)$. Notice that $r_{1}+r_{2}=(1, 2, 1)=r_{3}$, so $r_{3}=r_{1}+r_{2}$: one of the three polynomials is a combination of the other two, and the set is linearly dependent. The dependence relation is $r_{1}+r_{2}-r_{3}=0$, a vanishing combination with coefficients $(1, 1, -1)$, not all zero. The determinant confirms it: $\begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix}=1(1-2)-1(0-1)+0=-1+1=0$. The reasoning on polynomials is exactly the reasoning in $\mathbb{R}^{3}$: that is the payoff of translating each polynomial into its coefficient vector.

RANK-NULLITY THEOREM: for $T:V\to W$ linear with $V$ finite-dimensional, $\dim(\ker T)+\dim(\operatorname{im}\,T)=\dim V$. The number $\dim(\operatorname{im}\,T)$ is called the RANK of $T$ and $\dim(\ker T)$ its NULLITY. Illustration on the projection $P(x,y,z)=(x,y,0)$: the kernel is the set of $(x,y,z)$ such that $(x,y,0)=(0,0,0)$, that is, $x=y=0$ with $z$ free, which is the $z$-AXIS, of dimension 1. The image is the set of all $(x,y,0)$, which is the $xy$-PLANE, of dimension 2. Result: $1+2=3=\dim\mathbb{R}^{3}$ ✓. The interpretation is geometric and vivid: the projection flattens one direction (that of the kernel, lost for good) and keeps the other two (the image). What is lost and what is kept share out exactly the dimensions of the domain.

Let $U$ and $W$ be two subspaces of $V$. 1) Zero vector: $\vec{0}\in U$ and $\vec{0}\in W$ since each is a subspace, so $\vec{0}\in U\cap W$ ✓. 2) Addition: let $u,v\in U\cap W$. Then $u,v\in U$, and since $U$ is a subspace, $u+v\in U$; likewise $u,v\in W$ gives $u+v\in W$. So $u+v$ belongs to both, that is, $u+v\in U\cap W$ ✓. 3) Scalar multiplication: let $u\in U\cap W$ and $\lambda\in\mathbb{R}$. Then $\lambda u\in U$ and $\lambda u\in W$ by the same argument, so $\lambda u\in U\cap W$ ✓. The intersection is therefore a subspace. The argument extends without change to the intersection of any number of subspaces, even infinitely many.

In $\mathbb{R}^{2}$, take the two lines $U=\operatorname{span}\left\{(1,0)\right\}$ (the $x$-axis) and $W=\operatorname{span}\left\{(0,1)\right\}$ (the $y$-axis), which are indeed two subspaces. Their union $U\cup W$ is the cross formed by the two axes. Now $(1,0)\in U\subseteq U\cup W$ and $(0,1)\in W\subseteq U\cup W$, but their sum $(1,1)$ lies on NEITHER axis, so $(1,1)\notin U\cup W$: closure under addition fails ✓. This is exactly the set $S=\{xy=0\}$ of Exercise 2, in different clothing. The general principle: the union of two subspaces is a subspace only in the degenerate case where one contains the other. The asymmetry with the intersection makes sense: the intersection RESTRICTS (the conditions pile up, and closure is preserved), whereas the union ENLARGES without any control over what the cross sums produce.

$T$ is onto, so $\operatorname{im}\,T=\mathbb{R}^{3}$ and $\dim(\operatorname{im}\,T)=3$. The rank-nullity theorem gives $\dim(\ker T)+3=\dim\mathbb{R}^{5}=5$, so $\dim(\ker T)=2$. $T$ CANNOT be one-to-one: a linear transformation is one-to-one if and only if its kernel contains only the zero vector, that is, has dimension 0; here the kernel has dimension 2, strictly positive. There is therefore a whole plane of non-zero vectors sent to $\vec{0}$, and, for example, two distinct vectors of this plane have the same image. The general argument is worth remembering in its broadest form: a linear map from a space of dimension 5 to a space of dimension 3 can never be one-to-one, because the rank-nullity theorem forces $\dim(\ker T)\geq 5-3=2>0$. You cannot squeeze a space into a smaller one without losing information, and the rank-nullity theorem is the precise statement of that impossibility.

The solution set of a HOMOGENEOUS system (all right-hand sides zero) is always a subspace. Three equivalent arguments. First the direct one: $\vec{0}$ is a solution ✓, a sum of solutions is a solution (by linearity of each equation) ✓, a multiple of a solution is a solution ✓. Next the structural one: $W$ is exactly the KERNEL of the linear transformation with matrix $A=\begin{pmatrix}1&2&-1\\2&4&-2\end{pmatrix}$, and a kernel is always a subspace by Exercise 4. Finally the geometric one: each homogeneous equation defines a plane through the origin, and an intersection of subspaces is a subspace by Exercise 6. This also explains why a NON-homogeneous system never has a subspace as its solution set, since $\vec{0}$ is not a solution.

The second equation is exactly TWICE the first: it carries no information and disappears during row reduction. The system reduces to the single equation $x+2y-z=0$, that is, $x=-2y+z$. With two free parameters $y=s$ and $z=t$: $(x,y,z)=(-2s+t,\;s,\;t)=s(-2,1,0)+t(1,0,1)$. A basis of $W$ is therefore $\left\{(-2,1,0),\;(1,0,1)\right\}$, two vectors that are clearly not scalar multiples of each other, and $\dim W=2$. Geometric nature: a PLANE through the origin, with normal vector $(1,2,-1)$. Check: both basis vectors are indeed orthogonal to the normal, since $(-2)(1)+(1)(2)+(0)(-1)=0$ ✓ and $(1)(1)+(0)(2)+(1)(-1)=0$ ✓.

Here the two equations are not proportional: they are INDEPENDENT. Solving: adding the two equations gives $2x+y=0$, so $y=-2x$; substituting into the second, $x-(-2x)+z=0$, that is, $3x+z=0$ and $z=-3x$. With $x=t$: $(x,y,z)=t(1,-2,-3)$. A basis is $\left\{(1,-2,-3)\right\}$ and $\dim W=1$: it is a LINE through the origin. Comparison: two equations gave a plane (dimension 2) in the first case and a line (dimension 1) in the second. The difference does not come from the NUMBER of equations, which is the same, but from the number of INDEPENDENT equations: only one in part b), two here. A redundant equation constrains nothing more, and it is the rank of the coefficient matrix that measures the real constraint.

In $\mathbb{R}^{3}$: one independent constraint gives a plane (dimension 2), two independent constraints give a line (dimension 1), three independent constraints leave only the origin (dimension 0). It therefore takes TWO independent constraints to get a line, and THREE to keep only the origin. General rule: $\dim W=n-r$, where $n$ is the number of unknowns and $r$ the RANK of the system, that is, its number of independent equations. Each independent constraint costs exactly one dimension. The connection with the rank-nullity theorem is direct: $W$ is the kernel of the transformation $T$ with matrix $A$, the rank of $A$ is $\dim(\operatorname{im}\,T)=r$, and the theorem reads $\dim(\ker T)+r=n$, which is exactly $\dim W=n-r$ ✓. The two formulations, the one about systems and the one about transformations, are therefore the same statement seen from two sides: it is the central result of the chapter, and it pays to be able to switch from one reading to the other.

The two vectors of $U$, $(1, 1, 0)$ and $(1, 0, 1)$, are not scalar multiples of each other (the ratio of the first components is $1$, that of the second components is $0$), so they are linearly independent and $\dim U=2$: $U$ is a plane through the origin. Likewise $(0, 1, 1)$ and $(1, 1, 1)$ are not scalar multiples of each other, so $\dim W=2$: $W$ is a plane too.

A vector of $U\cap W$ can be written both as $a(1, 1, 0)+b(1, 0, 1)=(a+b, a, b)$ and as $c(0, 1, 1)+d(1, 1, 1)=(d, c+d, c+d)$. Equating component by component gives $\begin{cases} a+b=d \\ a=c+d \\ b=c+d \end{cases}$. The last two equations give $a=b$; then the first gives $d=2a$, and $a=c+d=c+2a$ implies $c=-a$. The common vector is therefore $(a+b, a, b)=(2a, a, a)=a(2, 1, 1)$. A basis of $U\cap W$ is $\left\{(2, 1, 1)\right\}$ and $\dim(U\cap W)=1$: the two planes meet along a line. Check: $(2, 1, 1)=(1, 1, 0)+(1, 0, 1)\in U$ and $(2, 1, 1)=-(0, 1, 1)+2(1, 1, 1)\in W$ ✓.

$U+W$ is spanned by the four vectors. The first three, $(1, 1, 0)$, $(1, 0, 1)$, $(0, 1, 1)$, have determinant $\begin{vmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix}=1(0-1)-1(1-0)+0=-2\neq 0$: they already form a basis of $\mathbb{R}^{3}$. So $\dim(U+W)=3$ and $U+W=\mathbb{R}^{3}$. Grassmann's formula checks out: $\dim U+\dim W-\dim(U\cap W)=2+2-1=3=\dim(U+W)$ ✓. It is a safe way to find $\dim(U\cap W)$ when the intersection is tedious to compute directly: $\dim(U\cap W)=\dim U+\dim W-\dim(U+W)$.

Criterion: a sum $U+W$ is DIRECT, written $U\oplus W$, if and only if $U\cap W=\left\{\vec{0}\right\}$, that is, $\dim(U\cap W)=0$. Here the intersection is a line of dimension $1$, so the sum is NOT direct: although $U+W$ fills all of $\mathbb{R}^{3}$, splitting a vector into a part in $U$ and a part in $W$ is not unique, since any vector of the common line can be moved from one side to the other. For a direct sum that fills $\mathbb{R}^{3}$, you would need, for example, a plane and a line not contained in that plane, whose dimensions $2+1=3$ add up without overlap.