MAT1041 · Week 7 · - Subspaces of Vector Spaces

Subspaces and the subspace test

You learn to check with a three-step test whether a set inside a vector space is a subspace; to recognize the typical subspaces and trap sets in Rⁿ, M₂₂ and P₂; and how intersection and union behave.

subspacesubspace testclosed under additionclosed under scalar multiplicationzero vectortrivial subspacesolution spacenull spaceintersectionunioncounterexample

Last week we defined a vector space with ten axioms. Fortunately, most of the time you don’t need to check ten axioms one by one: if the set you have is inside a known vector space (for example, a part of $\mathbb{R}^3$ ), you only look at three things. This is called the subspace test. It is a question type that almost certainly appears on the midterm and the final: “Determine whether $W$ is a subspace. Prove or give a counterexample.” Three or four choices, each a few minutes.

Golden rules:

The first thing to do is to check whether $\mathbf{0} \in W$ . If not, the answer is “no” and it ends in one line.
If $\mathbf{0}$ is in, you are not done: check closure under addition and scalar multiplication too.
If it is not a subspace, don’t write a general proof: give one concrete numerical counterexample.
If it is a subspace, write a general proof: take arbitrary $\mathbf{u}, \mathbf{v}, c$ and show the condition holds.

Definition and test

Let $V$ be a vector space and $W \subseteq V$ a nonempty subset. If $W$ is a vector space on its own with the operations of $V$ , then $W$ is called a subspace of $V$ .

The subspace test. $W$ is a subspace if and only if:

$\mathbf{0} \in W$ ,
$\mathbf{u}, \mathbf{v} \in W \Rightarrow \mathbf{u} + \mathbf{v} \in W$ (closed under addition),
$\mathbf{u} \in W$ , $c \in \mathbb{R} \Rightarrow c\mathbf{u} \in W$ (closed under scalar multiplication).

Why is it enough? Axioms such as commutativity, associativity and distributivity are already true in $V$ ; since the elements of $W$ are also elements of $V$ , they hold automatically. The only risk is the operations leaving $W$ .

Every vector space has two “boring” subspaces: $\{\mathbf{0}\}$ (the trivial subspace) and $V$ itself.

Subspaces of R³

The subspaces of $\mathbb{R}^3$ are exactly these: $\{\mathbf{0}\}$ , lines through the origin, planes through the origin, and $\mathbb{R}^3$ . A line or plane that does not pass through the origin cannot be a subspace because it does not contain the zero vector.

Example 1 (yes). $W = \{(x, y, z) : x - 2y + 5z = 0\}$ . The full write-up of the proof:

(1) 0 - 2·0 + 5·0 = 0              → (0,0,0) ∈ W
(2) u = (x1,y1,z1), v = (x2,y2,z2) ∈ W. For u + v:
    (x1+x2) - 2(y1+y2) + 5(z1+z2) = (x1 - 2y1 + 5z1) + (x2 - 2y2 + 5z2) = 0 + 0 = 0
(3) for cu:  cx1 - 2cy1 + 5cz1 = c(x1 - 2y1 + 5z1) = c·0 = 0
→ W is a subspace (a plane through the origin)

Example 2 (no, no zero). $W = \{(x, y, z) : x + y + z = 1\}$ : $0 + 0 + 0 \neq 1$ .

Example 3 (no, zero is there but not closed). $W = \{(x, y, z) : y = x^2\}$ . $(0, 0, 0) \in W$ . $(1, 1, 0) \in W$ , but for $2(1, 1, 0) = (2, 2, 0)$ , $2 \neq 2^2 = 4$ .

PITFALL · “the zero is in, so it’s a subspace”. The zero vector is necessary but not sufficient. In Example 3 and in Q3 the zero is in, but the set is not closed under scalar multiplication.

The solution set of Ax = 0

If $A$ is $m \times n$ , the set of all solutions of $A\mathbf{x} = \mathbf{0}$ is a subspace of $\mathbb{R}^n$ . It is called the null space (solution space) of $A$ :

$A\mathbf{0} = \mathbf{0}$ ,
$A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} = \mathbf{0} + \mathbf{0} = \mathbf{0}$ ,
$A(c\mathbf{u}) = cA\mathbf{u} = \mathbf{0}$ .

All the “yes” answers in Q1, Q8, Q9, Q12 are really special cases of this theorem: the condition is linear equations with a zero right-hand side.

If $\mathbf{b} \neq \mathbf{0}$ , the solution set of $A\mathbf{x} = \mathbf{b}$ is not a subspace: $A\mathbf{0} = \mathbf{0} \neq \mathbf{b}$ . Also the sum of two solutions gives $A(\mathbf{u} + \mathbf{v}) = 2\mathbf{b}$ . For example, for $x + 2y = 3$ (second equation $2x + 4y = 6$ ), $(3, 0)$ is a solution, but for $2(3, 0) = (6, 0)$ , $x + 2y = 6 \neq 3$ .

Matrix and polynomial subspaces

The same test, the same logic. Don’t be scared because the elements are matrices or polynomials; write the condition in terms of the coefficients.

Set	Subspace?	Reason
Symmetric $2 \times 2$ ( $A^T = A$ )	Yes	$(A + B)^T = A + B$ , $(cA)^T = cA$
Trace 0 ( $a_{11} + a_{22} = 0$ )	Yes	homogeneous linear equation
Upper triangular	Yes	$a_{21} = 0$
$\det A = 0$	No	$\operatorname{diag}(1, 0) + \operatorname{diag}(0, 1) = I$
Invertible matrices	No	the zero matrix is not invertible
$A^2 = A$	No	$I$ is in, $(2I)^2 = 4I \neq 2I$
$P_2$ , $p(1) = 0$	Yes	$a + b + c = 0$
$P_2$ , $p(0) = 1$	No	for the zero polynomial $0 \neq 1$
$P_2$ , integer coefficients	No	$\frac12 \cdot 1$ is not in the set
Continuous $f$ , $f(0) = 0$	Yes	$(f + g)(0) = 0$ , $(cf)(0) = 0$

Quick intuition: If the condition is homogeneous linear equation(s) (every term is a constant multiple of a variable, right side 0), the set is a subspace. If the condition has a product, a square, an absolute value, an inequality, a nonzero constant, “integer” or a determinant, it most likely is not. Intuition lets you decide quickly; on the exam still write the justification.

Ready-made counterexamples

xy = 0       : (1,0) + (0,1) = (1,1),   1·1 ≠ 0
x ≥ 0        : (-1)·(1,0) = (-1,0)
integers     : (1/2)·(1,0) = (1/2,0)
y = x²       : 2·(1,1) = (2,2),  2 ≠ 4
det = 0      : diag(1,0) + diag(0,1) = I
right side ≠ 0: the zero vector is not in the set

Intersection and union

If $W_1$ and $W_2$ are subspaces:

$W_1 \cap W_2$ is always a subspace. $\mathbf{0}$ is in both sets; the sum of two vectors that are in both is in both.
$W_1 \cup W_2$ is generally not a subspace. In the union of the $x$ -axis and the $y$ -axis, $(1, 0) + (0, 1) = (1, 1)$ is not there. (If one is inside the other, the union equals the larger one, and then it is a subspace.)

To find the intersection of two planes, solve the two equations together (Q12). The result is usually a line through the origin.

Common mistakes

Wrong	Right
Saying subspace because $\mathbf{0} \in W$	Check closure too
Using a vector that is not an element of $W$ in a counterexample	First show that $\mathbf{u}, \mathbf{v} \in W$
Proving “subspace” with a single numerical example	A general proof with arbitrary $\mathbf{u}, \mathbf{v}, c$
Trying only positive $c$ for scalar multiplication	$c = -1$ and $c = \frac12$ break most sets
Thinking a plane not through the origin is a subspace	There is no zero vector
Thinking a union is a subspace	Intersection yes, union generally no
Thinking the condition $p(0) = 1$ in $P_2$ gives a subspace	The zero polynomial has $p(0) = 0$
Thinking the set $\det A = 0$ is “like a null space”	The determinant is not linear

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

Is $W = \{(x, y, z) : x - 2y + 5z = 0\}$ a subspace of $\mathbb{R}^3$ ?

Show answer

Yes

(1) $\mathbf{0}$ : $0 - 0 + 0 = 0$ , it is in. (2) $\mathbf{u}, \mathbf{v} \in W$ : $(x_1 + x_2) - 2(y_1 + y_2) + 5(z_1 + z_2) = 0 + 0 = 0$ . (3) $c\mathbf{u}$ : $c(x - 2y + 5z) = c \cdot 0 = 0$ .
A plane through the origin. A homogeneous linear equation (right side 0) always gives a subspace.

Question 2easy

Is $W = \{(x, y, z) : x + y + z = 1\}$ a subspace of $\mathbb{R}^3$ ?

Show answer

$(0, 0, 0)$ : $0 + 0 + 0 = 0 \neq 1$ . The zero vector is not in the set, done.
Geometrically a plane that does not pass through the origin.

Question 3medium

Is $W = \{(x, y) : y = |x|\}$ a subspace of $\mathbb{R}^2$ ?

Show answer

$\mathbf{0} \in W$ but it is not closed under scalar multiplication: $(1, 1) \in W$ , $(-1)(1, 1) = (-1, -1)$ and $|-1| = 1 \neq -1$ .
"The zero is there, so it's a subspace" is a trap.

Question 4medium

Is $W = \{(a, b, a + 2b) : a, b \in \mathbb{R}\}$ a subspace of $\mathbb{R}^3$ ?

Show answer

Yes

$(a, b, a + 2b) = a(1, 0, 1) + b(0, 1, 2)$ . $W$ is the set of all linear combinations of two vectors, that is, a span: always a subspace.
Direct test: $a = b = 0$ gives zero; sums and scalar multiples again obey the rule "third = first + 2·second".

Question 5medium

Is $W = \{(x, y, z) : xz = 0\}$ a subspace of $\mathbb{R}^3$ ?

Show answer

Not closed under addition: $(1, 0, 0) \in W$ , $(0, 0, 1) \in W$ , but their sum $(1, 0, 1)$ has $xz = 1 \neq 0$ .
If the variables are multiplied in the condition, it is most likely not a subspace.

Question 6medium

Is the set of all $2 \times 2$ upper triangular matrices a subspace of $M_{2,2}$ ?

Show answer

Yes

Upper triangular: $a_{21} = 0$ . The zero matrix is upper triangular. In the sum of two upper triangular matrices the $(2,1)$ entry is $0 + 0 = 0$ ; in the $c$ multiple, $c \cdot 0 = 0$ .
The condition is a single homogeneous linear equation ( $a_{21} = 0$ ).

Question 7medium

Is $W = \{A \in M_{2,2} : \det A = 0\}$ a subspace of $M_{2,2}$ ?

Show answer

$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ are singular (det 0), their sum is $I$ and $\det I = 1$ .
Not closed under addition. (It is closed under scalar multiplication, but that is not enough.)

Question 8medium

Is $W = \{p \in P_2 : p(2) = 0\}$ a subspace of $P_2$ ?

Show answer

Yes

The zero polynomial: $0(2) = 0$ . $(p + q)(2) = p(2) + q(2) = 0$ . $(cp)(2) = c \cdot 0 = 0$ .
For $p = a + bx + cx^2$ the condition is $a + 2b + 4c = 0$ : a homogeneous linear equation.

Question 9hard

Is $W = \{p \in P_2 : p'(1) = p(0)\}$ a subspace of $P_2$ ?

Show answer

Yes

$p = a + bx + cx^2$ : $p'(x) = b + 2cx$ , $p'(1) = b + 2c$ , $p(0) = a$ . The condition is $b + 2c - a = 0$ .
A homogeneous linear equation in the coefficients, that is, a null space: a subspace.
Alternatively: differentiation and evaluation are linear operations, $(p + q)' = p' + q'$ , $(cp)(0) = cp(0)$ .

Question 10hard

For which value of $k$ is $W = \{(x, y, z) : x + 2y - z = k\}$ a subspace of $\mathbb{R}^3$ ?

Show answer

k = 0

For $\mathbf{0} \in W$ we need $0 + 0 - 0 = k$ , so $k = 0$ is required.
When $k = 0$ it is a homogeneous plane and all three conditions hold. If $k \neq 0$ it is a shifted parallel plane, not a subspace.

Question 11hard

Let $W_1$ be the $x$ -axis and $W_2$ the $y$ -axis in $\mathbb{R}^2$ . Is $W_1 \cup W_2$ a subspace?

Show answer

$(1, 0) \in W_1$ , $(0, 1) \in W_2$ , but $(1, 0) + (0, 1) = (1, 1)$ is on neither axis.
A union is generally not a subspace. An intersection ( $\{\mathbf{0}\}$ ) is always a subspace.

Question 12medium

Find the intersection of the subspaces $x + y - z = 0$ and $2x - y + z = 0$ of $\mathbb{R}^3$ .

Show answer

\operatorname{span}\{(0, 1, 1)\}

Solve the two equations together: $\begin{bmatrix} 1 & 1 & -1 \\ 2 & -1 & 1 \end{bmatrix}$ , $R_2 \leftarrow R_2 - 2R_1$ : $(0, -3, 3)$ , so $y = z$ .
Equation 1: $x + z - z = 0 \Rightarrow x = 0$ . $z = t$ : $(0, t, t) = t(0, 1, 1)$ .
The intersection of two planes is a line through the origin.

Verification: For the sets that are subspaces, closure was tested in node with 200 random integer vector pairs and scalars from each set using BigInt fractions; for the sets that are not subspaces, the given counterexamples were computed one by one; the null space in the intersection question was found with the exact RREF (verify/MAT1041/07.mjs).

Review card

W ⊆ V is a subspace ⇔ (1) 0 ∈ W (2) u,v ∈ W → u+v ∈ W (3) u ∈ W → cu ∈ W Check 0 first: if it is not there you are DONE, not a subspace. Having 0 is not enough: check (2) and (3) too. To say "no", write ONE numerical counterexample. YES signs: linear equation(s) with right side 0, span{...}, solutions of Ax=0, symmetric, trace 0, upper triangular, diagonal matrices, p(a)=0, p'(a)=p(b). NO signs: right side ≠ 0, product (xy=0), square (y=x²), absolute value, inequality (x≥0), integers, det=0, invertible matrices, A²=A, p(0)=1, degree EXACTLY n. Subspaces of R³: {0}, lines through the origin, planes through the origin, R³. The solution set of Ax=0 is a subspace (null space); of Ax=b, b≠0, it is not. Intersection is ALWAYS a subspace; union generally is NOT.