Worked examples · Sheet 1

Subspace checks with the criterion

Theory used: Vector subspace Subspace criterion Homogeneous and non-homogeneous equations

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Determine whether the following subsets of $R^{3}$ are vector subspaces:

W_{1} = {(x, y, z) : 2 x - y + 3 z = 0}, W_{2} = {(x, y, z) : 2 x - y + 3 z = 4} .

W_{1} : 2 \cdot 0 - 0 + 3 \cdot 0 = 0 ⟹ (0, 0, 0) \in W_{1}

2 (u_{1} + v_{1}) - (u_{2} + v_{2}) + 3 (u_{3} + v_{3}) = (2 u_{1} - u_{2} + 3 u_{3}) + (2 v_{1} - v_{2} + 3 v_{3}) = 0 + 0 = 0

2 (λ u_{1}) - λ u_{2} + 3 (λ u_{3}) = λ (2 u_{1} - u_{2} + 3 u_{3}) = λ \cdot 0 = 0

W_{2} : 2 \cdot 0 - 0 + 3 \cdot 0 = 0 \neq = 4 ⟹ (0, 0, 0) \in / W_{2}

For $W_{1}$ we check the three conditions of the subspace criterion. The zero vector satisfies the equation. If $u = (u_{1}, u_{2}, u_{3})$ and $v = (v_{1}, v_{2}, v_{3})$ lie in $W_{1}$ , the expressions $2 u_{1} - u_{2} + 3 u_{3}$ and $2 v_{1} - v_{2} + 3 v_{3}$ equal $0$ ; since the left-hand side of the equation evaluated at $u + v$ is the sum of these two expressions, $u + v$ also satisfies the equation. In the same way the left-hand side evaluated at $λ u$ is $λ$ times the one evaluated at $u$ , so it equals $0$ . Therefore $W_{1}$ is a subspace: it is the plane through the origin formed by the vectors orthogonal to $(2, - 1, 3)$ , since the left-hand side of the equation is the dot product $(2, - 1, 3) \cdot (x, y, z)$ .

The set $W_{2}$ does not contain the zero vector, because at $(0, 0, 0)$ the left-hand side equals $0$ and not $4$ . Hence $W_{2}$ is not a subspace: it is the plane parallel to $W_{1}$ that does not pass through the origin.

Exercise 2

Determine whether $W = {(x, y) \in R^{2} : y = x^{2}}$ is a subspace of $R^{2}$ .

(0, 0) \in W since 0 = 0^{2}

u = (1, 1) \in W, 2 u = (2, 2), 2 \neq = 2^{2} = 4 ⟹ 2 u \in / W

y = x^{2}

The parabola

y = x^{2}

contains

u = (1, 1)

and does not contain

2 u = (2, 2)

The zero vector lies in $W$ , so the first condition of the criterion is satisfied and is not enough to decide. We test closure under scalar multiplication with a concrete vector: $u = (1, 1)$ belongs to $W$ because $1 = 1^{2}$ , while its double $(2, 2)$ does not belong to it, because $2 \neq = 4$ . Since the third condition is not satisfied, $W$ is not a subspace. A single counterexample is sufficient; geometrically, the parabola does not contain the line through the origin and $u$ .

Exercise 3

Determine whether the following subsets of $M_{2} (R)$ are subspaces:

W = {A \in M_{2} (R) : tr A = 0}, U = {A \in M_{2} (R) : det A = 0} .

tr O = 0 + 0 = 0 ⟹ O \in W

tr (A + B) = (a_{11} + b_{11}) + (a_{22} + b_{22}) = tr A + tr B = 0

tr (λ A) = λ a_{11} + λ a_{22} = λ tr A = 0

A = (1000) \in U, B = (0001) \in U, det (A + B) = det (1001) = 1 \neq = 0

The trace of the zero matrix $O$ is $0$ , so $O \in W$ . The trace is additive and homogeneous, that is $tr (A + B) = tr A + tr B$ and $tr (λ A) = λ tr A$ , as the entrywise computations show; therefore if $A$ and $B$ have zero trace, $A + B$ and $λ A$ also have zero trace. By the subspace criterion $W$ is a subspace.

The set $U$ contains the zero matrix and is closed under scalar multiplication, since $det (λ A) = λ^{2} det A$ . The sum, however, can leave $U$ : the matrices $A$ and $B$ have determinant $0$ , while their sum is the identity, with determinant $1$ . Hence $U$ is not a subspace.

Exercise 4

Determine whether the following subsets of $R_{2} [x]$ are subspaces:

W = {p \in R_{2} [x] : p (1) = 0}, U = {p \in R_{2} [x] : p (0) \cdot p (1) = 0} .

p (x) = 0 for every x ⟹ p (1) = 0, 0 \in W

(p + q) (1) = p (1) + q (1) = 0 + 0 = 0, (λ p) (1) = λ p (1) = λ \cdot 0 = 0

p (x) = x : p (0) \cdot p (1) = 0 \cdot 1 = 0 ⟹ p \in U

q (x) = x - 1 : q (0) \cdot q (1) = (- 1) \cdot 0 = 0 ⟹ q \in U

(p + q) (x) = 2 x - 1 : (p + q) (0) \cdot (p + q) (1) = (- 1) \cdot 1 = - 1 \neq = 0

The value of a polynomial at a point behaves linearly: the value of the sum is the sum of the values, and the value of $λ p$ is $λ$ times the value of $p$ . Hence the zero polynomial lies in $W$ and, if $p (1) = q (1) = 0$ , then $p + q$ and $λ p$ also vanish at $1$ . By the subspace criterion $W$ is a subspace: it consists of the polynomials divisible by $x - 1$ .

The condition that defines $U$ is a product, and the product is not linear. The polynomial $p (x) = x$ lies in $U$ because it vanishes at $0$ , and $q (x) = x - 1$ lies in $U$ because it vanishes at $1$ ; the sum $2 x - 1$ equals $- 1$ at $0$ and $1$ at $1$ , so the product of the two values is $- 1 \neq = 0$ . Therefore $U$ is not a subspace: it is the union of the two subspaces ${p (0) = 0}$ and ${p (1) = 0}$ .