6.2 Subspaces

Subspaces appear as soon as we notice that many linear problems do not use the whole ambient vector space. The solution set of a homogeneous system, the set of symmetric matrices, and the set of polynomials vanishing at a chosen point are all smaller collections inside larger vector spaces.

A subset is a subspace when its inherited addition and scalar multiplication still satisfy the vector-space laws.

Why only a few axioms need checking

Suppose $V$ is already a vector space and $W \subseteq V$. If we use on $W$ the same addition and scalar multiplication already defined on $V$, then we do not need to reprove associativity, commutativity, or the distributive laws from the ground up. Those identities are already true in $V$.

The real question is narrower:

• does addition of two vectors in $W$ stay inside $W$?
• does scalar multiplication stay inside $W$?
• does $W$ still contain the vectors needed for the vector-space structure, especially the zero vector?

That is why the subspace test is a closure test.

Definition 6.2.1

(Subspace definition)

Let $V$ be a vector space. A subset $W \subseteq V$ is called a subspace of $V$ if:

1. $W$ is nonempty;
2. $u + v \in W$ for all $u, v \in W$;
3. $\alpha u \in W$ for all $u \in W$ and all $\alpha \in R$.

The word “inherited” is essential. A subset with newly invented operations may be a vector space in its own right without being a subspace of the original space. Here all sums and scalar multiples are calculated in $V$. Closure says these already defined operations also have outputs in $W$; only then can they serve as operations on $W$.

There is one further logical point in reducing ten axioms to three conditions. Identities such as associativity restrict automatically to elements of $W$, but the existence of a zero and inverses inside $W$ needs proof. Choose $w\in W$, using nonemptiness. Scalar closure gives $0w=0_V\in W$. For any $u\in W$, it also gives $(-1)u=-u\in W$. These are exactly the ambient zero and inverse, so their identity equations continue to hold with all required elements inside $W$. Together with the inherited identities, this proves that the three conditions give a vector space.

The nonempty condition cannot be deleted. If $W=\varnothing$, the statements “for every $u,v\in W$” have no inputs to test, so both closure assertions are vacuously true. Nevertheless there is no zero vector in $W$. One actual member is enough to start the proof above; checking the ambient zero directly is usually the most convenient way to supply it.

A single linear-combination condition packages the two closure checks into a convenient form.

Theorem 6.2.2

(Equivalent subspace test)

Let $V$ be a vector space and $W \subseteq V$. Then $W$ is a subspace of $V$ if and only if:

1. $0 \in W$;

2. for every $u, v \in W$ and every $\alpha, \beta \in R$, the vector

   $$\alpha u + \beta v$$

   also belongs to $W$.

Proof

(Why the equivalent test works)

Assume first that $W$ is a subspace. Because $W$ is nonempty, choose some $w \in W$. Since $W$ is closed under scalar multiplication,

$$0w = 0 \in W.$$

Now let $u, v \in W$ and let $\alpha, \beta \in R$. Closure under scalar multiplication gives $\alpha u \in W$ and $\beta v \in W$, and then closure under addition gives

$$\alpha u + \beta v \in W.$$

Conversely, suppose $0 \in W$ and $\alpha u + \beta v \in W$ whenever $u, v \in W$ and $\alpha, \beta \in R$. Then $W$ is nonempty because it contains $0$. If $u, v \in W$, choose $\alpha = \beta = 1$ to get

$$u + v \in W.$$

If $u \in W$ and $\kappa \in R$, choose $v = 0$ and $\alpha = \kappa$, $\beta = 0$ to get

$$\kappa u + 0 \cdot 0 = \kappa u \in W.$$

So the original three-part definition is satisfied.

Standard examples from equations

Linear equations provide the first examples: their zero right-hand sides explain why addition and scalar multiplication preserve membership.

Worked example 6.2.3

(A line through the origin in $R^2$)

Let

$$W = \left\{(x, y) \in R^2 : y = 2x\right\}.$$

To prove that $W$ is a subspace of $R^2$, check the three defining conditions.

1. $W$ is nonempty because $(0, 0)$ satisfies $0 = 2 \cdot 0$.

2. If $(x_1, y_1), (x_2, y_2) \in W$, then $y_1 = 2x_1$ and $y_2 = 2x_2$. Therefore

   $$(y_1 + y_2) = 2(x_1 + x_2),$$

   so $(x_1 + x_2, y_1 + y_2) \in W$.

3. If $(x, y) \in W$ and $\alpha \in R$, then $y = 2x$, so

   $$\alpha y = 2(\alpha x),$$

   which means $(\alpha x, \alpha y) \in W$.

Hence $W$ is a subspace of $R^2$.

The geometry matches the algebra: $W$ is a line through the origin. The phrase "through the origin" is not decorative. It is the visible sign that the zero vector belongs to the set.

Worked example 6.2.4

(A plane cut out by one homogeneous equation)

Now let

$$W = \left\{(x, y, z) \in R^3 : x + 2y + 3z = 0\right\}.$$

Again we check the definition.

1. $W$ is nonempty because $(0, 0, 0)$ satisfies the equation.

2. If $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$ lie in $W$, then

   $$u_1 + 2u_2 + 3u_3 = 0, \qquad v_1 + 2v_2 + 3v_3 = 0.$$

   Adding these equations gives

   $$(u_1 + v_1) + 2(u_2 + v_2) + 3(u_3 + v_3) = 0,$$

   so $u + v \in W$.

3. If $u \in W$ and $\alpha \in R$, then

   $$\alpha u_1 + 2\alpha u_2 + 3\alpha u_3 = \alpha(u_1 + 2u_2 + 3u_3) = \alpha \cdot 0 = 0,$$

   so $\alpha u \in W$.

Thus $W$ is a subspace of $R^3$.

This example is the model case for a set defined by a homogeneous linear equation. The zero right-hand side is what makes the closure argument work.

In the checker, compare which of the zero-vector and closure conditions hold for each set.

Consequences of the subspace test

Once $W$ is known to be a subspace, several facts follow immediately.

The earlier argument already gave $0_V\in W$ and $-u\in W$ whenever $u\in W$. These elements are inherited from $V$; the subspace does not choose new ones.

Two immediate examples now become unavoidable:

• $\{0\}$ is a subspace of every vector space, so it is the smallest subspace;
• $V$ itself is a subspace of $V$, so it is the largest subspace.

Null spaces are subspaces

The first important family of subspaces coming from matrix theory is the null space.

Theorem 6.2.5

(The null space of a matrix is a subspace)

Let $A$ be an $m \times n$ matrix, and let

$$N(A) = \left\{x \in R^n : Ax = 0\right\}.$$

Then $N(A)$ is a subspace of $R^n$.

Proof

(Why $N(A)$ is a subspace)

First, $0 \in N(A)$ because

$$A0 = 0.$$

Now let $u, v \in N(A)$ and let $\alpha \in R$. Then $Au = 0$ and $Av = 0$, so

$$A(u + v) = Au + Av = 0 + 0 = 0,$$

and

$$A(\alpha u) = \alpha(Au) = \alpha \cdot 0 = 0.$$

Therefore $u + v \in N(A)$ and $\alpha u \in N(A)$, so $N(A)$ is a subspace of $R^n$.

This theorem explains why the phrase null space is not just convenient terminology. The solution set to a homogeneous system really is a vector space.

Worked example 6.2.6

(A homogeneous equation defines a null space)

Let

$$A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}.$$

Then

$$N(A) = \left\{ \begin{bmatrix} x \\ y \\ z \end{bmatrix} : x + 2y + 3z = 0 \right\}.$$

So the plane from the earlier example is exactly the null space of a matrix. Solving for $x$ gives

$$x = -2y - 3z,$$

and therefore every vector in the null space has the form

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = y \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + z \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}.$$

So this subspace is a plane through the origin generated by two vectors.

Worked example 6.2.7

(Symmetry conditions on matrices and polynomials)

Let $S_n$ be the set of real symmetric $n\times n$ matrices. The zero matrix is symmetric. If $X^T=X$, $Y^T=Y$ and $\alpha,\beta\in\mathbb R$, then

$$(\alpha X+\beta Y)^T=\alpha X^T+\beta Y^T =\alpha X+\beta Y.$$

Thus $S_n$ is a subspace of the fixed ambient matrix space. We used linearity of transpose and the two membership equations, not a claim that arbitrary matrix products preserve symmetry. Products are irrelevant to the subspace test.

Similarly, let $E$ consist of the even polynomials in $P_n$: those satisfying $p(-t)=p(t)$ for every real $t$. The zero polynomial belongs to $E$. For $p,q\in E$ and real $\alpha,\beta$,

$$(\alpha p+\beta q)(-t)=\alpha p(-t)+\beta q(-t) =\alpha p(t)+\beta q(t)=(\alpha p+\beta q)(t).$$

The identity holds for every $t$, so the resulting polynomial remains even; it remains in $P_n$ by the ambient vector-space operations. Hence $E$ is a subspace. In both examples the defining equality survives every linear combination. Neither a matrix entry nor one sample function value alone would verify that whole equality.

A broader pattern: matrix conditions that preserve subspaces

We can generalize the null-space condition. Instead of requiring $Ax=0$, require $Ax$ to belong to a known subspace; closure in that target space will supply the proof.

Worked example 6.2.8

(Preimage of a subspace under matrix multiplication)

Let $A$ be a $4 \times 6$ matrix, let $V$ be a subspace of $R^4$, and define

$$W = \left\{x \in R^6 : Ax \in V\right\}.$$

Then $W$ is a subspace of $R^6$.

The proof is a direct application of the subspace test:

1. $0 \in W$ because $A0 = 0$, and $0 \in V$;

2. if $x, y \in W$, then $Ax, Ay \in V$, so

   $$A(x + y) = Ax + Ay \in V,$$

   which means $x + y \in W$;

3. if $x \in W$ and $\alpha \in R$, then $Ax \in V$, so

   $$A(\alpha x) = \alpha(Ax) \in V,$$

   which means $\alpha x \in W$.

The null space is the special case where $V = \{0\}$.

This viewpoint is worth remembering. Many subspaces are best described not by listing their elements, but by writing a condition that is preserved by linear operations.

Common ways a subset fails

The subspace test is short, but it is unforgiving. A set can fail in more than one way.

Common mistake

(A translated line loses the zero vector)

The set $\{(x,y)\in\mathbb R^2:x+y=1\}$ is not a subspace, because $(0,0)$ does not satisfy its equation. Its shape as a line does not supply the missing algebraic condition. Conversely, containing zero is only one part of the test, as the next example shows.

Counterexample mode

Zero and scalar closure do not imply a subspace

False claim. Every subset of a real vector space that contains zero and is closed under every real scalar multiple is a subspace.

Consider the actual real symmetric matrix and its quadratic zero set

$$C=\operatorname{diag}(1,-1,-1,-1),\qquad S=\{x\in\mathbb R^4:x^TCx=0\}.$$

Writing $q(x)=x^TCx$, we have $q(x)=x_1^2-x_2^2-x_3^2-x_4^2$. Thus $q(0)=0$. Also, $q(\alpha x)=\alpha^2q(x)$ for every real $\alpha$, so $S$ really is closed under all scalar multiples, including negative ones and zero.

Now take $u=(1,1,0,0)^T$ and $v=(1,-1,0,0)^T$. Direct substitution gives $q(u)=1-1=0$ and $q(v)=1-1=0$. Both vectors belong to $S$, but $u+v=(2,0,0,0)^T$ has $q(u+v)=4$. Therefore $u+v\notin S$: closure under addition is the precise failed axiom.

The expansion explains the failure. For any real square $C$ of matching size,

$$(u+v)^TC(u+v)=u^TCu+v^TCv+u^TCv+v^TCu.$$

For the displayed matrix, $C^T=C$. Since $u^TCv$ is a scalar, $u^TCv=(u^TCv)^T=v^TC^Tu=v^TCu$. Only with this justification can the two cross terms be combined as $2u^TCv$. Here $u^TCv=2$, so the expansion gives $0+0+2\cdot2=4$, agreeing with the direct calculation. For an unspecified nonsymmetric matrix the four-term expansion remains valid, but replacing the cross terms by twice one of them need not be valid.

Repaired statement. A subset $W$ of a real vector space, using the inherited operations, is a subspace exactly when it contains zero and is closed under every combination $\alpha u+\beta v$ of its members. Equivalently, nonemptiness together with both addition closure and scalar closure suffices. In this example the missing addition condition cannot be inferred from the other two: each was explicitly verified and the sum still failed.

Separating the closure conditions

Different examples reveal different missing conditions. The half-space $H=\{x\in\mathbb R^n:x_1\geq0\}$, for $n\geq1$, contains zero and is closed under addition: two nonnegative first coordinates have a nonnegative sum. But the first standard coordinate vector $e_1$ belongs to $H$ while $-e_1$ does not. Positive-scalar closure is insufficient when the field is all of $\mathbb R$.

By contrast, $Q=\{(x,y):xy\geq0\}$ is closed under all real scalar multiples because $(\alpha x)(\alpha y)=\alpha^2xy\geq0$. Yet $(1,0),(0,-1)\in Q$ and $(1,-1)\notin Q$, so addition fails. This simpler example isolates the same distinction as the quadratic zero set above.

Failure need not be geometric. The singular real $2\times2$ matrices contain the zero matrix, and every scalar multiple of a singular matrix remains singular. Nevertheless $\operatorname{diag}(1,0)$ and $\operatorname{diag}(0,1)$ are singular and their sum is the invertible identity. Thus the singular matrices are not a subspace. “Being a collection of matrices” does not decide the issue; the defining condition must survive the inherited operations.

Likewise, polynomials of degree exactly $n$ are not a subspace of $P_n$: multiplication of $t^n$ by zero produces the zero polynomial, which does not have degree exactly $n$. This argument also works for $n=0$. For $n\geq1$, $t^n$ and $-t^n+1$ additionally witness addition failure, since their sum is one. The degree restriction must therefore be checked at boundary scalars as well as at generic ones.

Quick checks

Quick check

(Why is $E = \{f(x) \in P_n : f(1) = 0\}$ a subspace of $P_n$?)

Use the value at $x = 1$ as the quantity that has to stay unchanged under the subspace test.

Solution

The zero polynomial belongs to $E$ because $0(1) = 0$. If $f, g \in E$, then $f(1) = g(1) = 0$, so

$$(f + g)(1) = f(1) + g(1) = 0.$$

If $\alpha \in R$ and $f \in E$, then

$$(\alpha f)(1) = \alpha f(1) = \alpha \cdot 0 = 0.$$

So $E$ is closed under addition and scalar multiplication, hence it is a subspace of $P_n$.

Fix a real matrix $A\in M_{p,p}(\mathbb R)$ for the following check.

Quick check

(Let $W = \{X \in M_{p,p}(R) : AX - XA = O_{p \times p}\}$. Why does this set form a subspace?)

Treat the defining equation $AX - XA = O_{p \times p}$ the same way you treated $Ax = 0$.

Solution

The zero matrix belongs to $W$ because

$$A O_{p \times p} - O_{p \times p} A = O_{p \times p}.$$

If $X, Y \in W$ and $\kappa, \lambda \in R$, then

$$A(\kappa X + \lambda Y) - (\kappa X + \lambda Y)A = \kappa(AX - XA) + \lambda(AY - YA) = O_{p \times p}.$$

So $\kappa X + \lambda Y$ also belongs to $W$, and the equivalent subspace test applies.

Quick check

(Why is $\{(x, y) \in R^2 : x + y = 1\}$ not a subspace, even though it is still a line?)

Answer this without drawing a graph.

Solution

It fails at the first necessary step: the zero vector is not in the set, because

$$0 + 0 \neq 1.$$

Since every subspace must contain $0$, this line cannot be a subspace.

Exercise: recover the ordinary subspace test

Suppose $W\subseteq\mathbb R^4$, $k=(1,0,3,0)^T\in W$, and the following conditions hold: whenever $u,v\in W$, we have $u-v\in W$; whenever $t\in W$ and $a\in\mathbb R$, we have $at+k\in W$. Prove that $W$ is a subspace using the inherited operations. The expression $at+k$ itself is not the ordinary scalar multiple, so scalar closure still needs an argument.

Solution · Remove the fixed vector before applying the test

Since $k\in W$, difference closure gives $k-k=0\in W$. Now take arbitrary $x\in W$ and $b\in\mathbb R$. The second condition gives $bx+k\in W$; subtracting $k\in W$ using difference closure yields $(bx+k)-k=bx\in W$. Thus genuine scalar closure has been established, including multiplication by $-1$. If $x,y\in W$, we consequently know $-y\in W$, so difference closure gives $x-(-y)=x+y\in W$. All three conditions of the subspace test now hold.

The given nonzero vector supplied nonemptiness but was not a proposed new zero. Each use of difference closure had two previously justified members of $W$. Writing $bx\in W$ directly from $bx+k\in W$ without justifying the subtraction would omit the central step of the proof.

Exercise: an entry condition and a curved condition

Decide whether each set is a subspace with the usual operations: $L=\{\begin{bmatrix}a&b\\c&d\end{bmatrix}:a+b=c+d\}$ in $M_{2,2}(\mathbb R)$, and $P=\{(x,y)\in\mathbb R^2:y=x^2\}$. Prove a positive answer for arbitrary members; for a negative answer, give an actual member whose required operation leaves the set.

Solution · Follow the defining condition through the operation

For $L$, define the entry expression $r(X)=a+b-c-d$. The zero matrix has $r(0)=0$. If $X,Y\in L$ and $\alpha,\beta$ are real, ordinary entrywise operations give $r(\alpha X+\beta Y)=\alpha r(X)+\beta r(Y)=0$. Thus every such combination belongs to $L$, proving it is a subspace.

For $P$, the vector $(1,1)$ satisfies the defining equation, but its triple $(3,3)$ does not: $3\ne3^2$. Therefore scalar closure fails and $P$ is not a subspace, even though it contains zero. These arguments depend on how the condition behaves under the operations, rather than on whether the elements are presented as matrices or pairs.

Read this first

This section depends on 6.1 Vector spaces for the axioms and on 4.1 Homogeneous systems and null space for the meaning of $Ax = 0$.

Continue with

The next section, 6.3 Linear combinations and span, uses subspaces as the natural output of a generating process.