4.2 Subspaces: Class Notes

Definition

A subset $W$ of a vector space $V$ is called a subspace of $V$ if $W$ is itself a vector space under the addition and scalar multiplication defined on $V$ .

Two diagrams: W = (0, infinity) inside V = R, crossed out; a subspace W inside V

Left (crossed out): $V = R$ drawn as a large region containing $W = (0, \infty)$ , with the zero vector $\vec{0}$ inside $V$ but outside $W$ . Because $\vec{0} \notin W$ , this $W$ is not a subspace. Right: a subspace $W$ drawn as a smaller region inside $V$ .

Theorem 4.2.1 Subspace Test

If $W$ is a nonempty set of vectors in a vector space $V$ , then $W$ is a subspace of $V$ if and only if the following conditions are satisfied.

a) If $𝐮$ and $𝐯$ are vectors in $W$ , then $𝐮 + 𝐯$ is in $W$ .

b) If $k$ is a scalar and $𝐮$ is a vector in $W$ , then $k 𝐮$ is in $W$ .

Tip: Because $W$ must be nonempty, we sometimes see a step 0 before (a): show $\vec{0} \in W$ .

Since any vector $\vec{w} \in W$ is also in $V$ and the operations are the same throughout $V$ , axioms 2, 3, and 7–10 are satisfied. (These are inherited.) To show $W \subset V$ is itself a subspace, we need to confirm axioms 1, 4, 5, and 6.

If Axiom 6 is satisfied ( $\vec{u} \in W \Rightarrow k \vec{u} \in W$ for all $k$ ), then it is true for $k = 0$ (giving Axiom 4) and $k = - 1$ (giving Axiom 5).

Thus $W \subset V$ is a subspace if Axioms 1 and 6 are satisfied.

#3

Use the Subspace Test to determine which of the sets are subspaces of $M_{n n}$ .

$M_{n n}$ is the set of $n \times n$ matrices with the standard operations of addition and scalar multiplication.

a. The set of all diagonal $n \times n$ matrices.

b. The set of all $n \times n$ matrices $A$ such that $det (A) = 0$ .

c. The set of all $n \times n$ matrices $A$ such that $t r (A) = 0$ .

d. The set of all symmetric $n \times n$ matrices.

Solution

a. (Optional first step) The zero matrix

$\vec{0} = [\begin{matrix} 0 & \dots & 0 \\ ⋮ & ⋱ & ⋮ \\ 0 & \dots & 0 \end{matrix}]$

is a diagonal matrix, so the set is nonempty.

Let

$U = [\begin{matrix} u_{1} & 0 & \dots & 0 \\ 0 & u_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & u_{n} \end{matrix}], V = [\begin{matrix} v_{1} & 0 & \dots & 0 \\ 0 & v_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & v_{n} \end{matrix}], k \in ℝ .$

$U + V = [\begin{matrix} u_{1} + v_{1} & 0 & \dots & 0 \\ 0 & u_{2} + v_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & u_{n} + v_{n} \end{matrix}]$

a diagonal matrix, so $U + V \in W$ .

$k U = [\begin{matrix} k u_{1} & 0 & \dots & 0 \\ 0 & k u_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & k u_{n} \end{matrix}]$

which is diagonal, so $k U \in W$ .

Answer: $a. This is a subspace.$

b. Not a subspace. Let

$U = [\begin{matrix} 1 & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & 0 \end{matrix}], V = [\begin{matrix} 0 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & 1 \end{matrix}]$

Then $det (U) = 0 = det (V)$ . But $U + V = I$ , so $det (U + V) = 1 \neq 0$ .

Answer: $b. Not a subspace.$

c. Let $U = [a_{i j}]$ and $V = [b_{i j}] \in W$ (so $t r (U) = 0 = t r (V)$ ):

$t r (U) = \sum_{i = 1}^{n} a_{i i} = 0 = \sum_{i = 1}^{n} b_{i i} = t r (V)$

$t r (U + V) = \sum_{i = 1}^{n} (a_{i i} + b_{i i}) = \sum_{i = 1}^{n} a_{i i} + \sum_{i = 1}^{n} b_{i i} = 0 + 0 = 0$

so $U + V$ is in the set. Let $k \in ℝ$ :

$t r (k U) = \sum_{i = 1}^{n} k a_{i i} = k \sum_{i = 1}^{n} a_{i i} = k (0) = 0$

so $k U$ is in the set.

Answer: $c. This is a subspace.$

d. For you.

#6

Use the Subspace Test to determine which of the sets are subspaces of $P_{3}$ .

$P_{3}$ is the set of polynomials of degree $\leq 3$ .

a. All polynomials of the form $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}$ in which $a_{0}$ , $a_{1}$ , $a_{2}$ , and $a_{3}$ are rational numbers.

b. All polynomials of the form $a_{0} + a_{1} x$ in which $a_{0}$ and $a_{1}$ are real numbers.

Solution

a. Not a subspace. If $k = π$ , then $k \vec{p} \notin$ the set (for example, $π \cdot 1 = π$ has an irrational coefficient).

Answer: $a. Not a subspace.$

b. Let $\vec{p} = p_{0} + p_{1} x$ and $\vec{q} = q_{0} + q_{1} x$ , where $p_{0}, p_{1}, q_{0}, q_{1} \in ℝ$ , and let $k \in ℝ$ .

$\vec{p} + \vec{q} = p_{0} + q_{0} + (p_{1} + q_{1}) x,$

which has the form $a_{0} + a_{1} x$ with $a_{0}, a_{1} \in ℝ$ .

$k \vec{p} = k p_{0} + k p_{1} x,$

where $k p_{0}, k p_{1} \in ℝ$ .

Answer: $b. This is a subspace.$

#11

Use the Subspace Test to determine which of the sets are subspaces of $M_{22}$ .

a. All matrices of the form $[\begin{matrix} a & 0 \\ b & 0 \end{matrix}]$ .

b. All matrices of the form $[\begin{matrix} a & 1 \\ b & 1 \end{matrix}]$ .

c. All $2 \times 2$ matrices $A$ such that $A [\begin{matrix} 1 \\ - 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \end{matrix}]$ .

Solution

b. Not a subspace:

$[\begin{matrix} x_{1} & 1 \\ x_{2} & 1 \end{matrix}] + [\begin{matrix} y_{1} & 1 \\ y_{2} & 1 \end{matrix}] = [\begin{matrix} x_{1} + y_{1} & 2 \\ x_{2} + y_{2} & 2 \end{matrix}]$

is not of the form $[\begin{matrix} a & 1 \\ b & 1 \end{matrix}]$ , because $1 + 1 = 2$ .

Answer: $b. Not a subspace.$

c. Let $A$ and $B$ be in the set. Then $A [\begin{matrix} 1 \\ - 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \end{matrix}]$ and $B [\begin{matrix} 1 \\ - 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \end{matrix}]$ , and

$(A + B) [\begin{matrix} 1 \\ - 1 \end{matrix}] = A [\begin{matrix} 1 \\ - 1 \end{matrix}] + B [\begin{matrix} 1 \\ - 1 \end{matrix}] = [\begin{matrix} 2 \\ 0 \end{matrix}] + [\begin{matrix} 2 \\ 0 \end{matrix}] = [\begin{matrix} 4 \\ 0 \end{matrix}] \neq [\begin{matrix} 2 \\ 0 \end{matrix}]$

Answer: $c. Not a subspace.$

#14

Use the Subspace Test to determine which of the sets are subspaces of $R^{4}$ .

a. All vectors $𝐱$ in $R^{4}$ such that $A 𝐱 = [\begin{matrix} 0 \\ 1 \end{matrix}]$ , where $A = [\begin{matrix} 0 & - 1 & 0 & 2 \\ - 1 & 1 & 0 & 1 \end{matrix}]$

b. All vectors $𝐱$ in $R^{4}$ such that $A 𝐱 = [\begin{matrix} 0 \\ 0 \end{matrix}]$ , where $A$ is as in part (a).

Solution

a. Not a subspace. Consider $k = 5$ :

$A (5 \vec{x}) = 5 A \vec{x} = 5 [\begin{matrix} 0 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 5 \end{matrix}] \neq [\begin{matrix} 0 \\ 1 \end{matrix}] .$

Or: let $\vec{x}, \vec{y} \in R^{4}$ with $A \vec{x} = [\begin{matrix} 0 \\ 1 \end{matrix}]$ and $A \vec{y} = [\begin{matrix} 0 \\ 1 \end{matrix}]$ . Then

$A (\vec{x} + \vec{y}) = A \vec{x} + A \vec{y} = [\begin{matrix} 0 \\ 1 \end{matrix}] + [\begin{matrix} 0 \\ 1 \end{matrix}] \neq [\begin{matrix} 0 \\ 1 \end{matrix}] .$

Answer: $a. Not a subspace.$

b. $A \vec{0} = \vec{0}$ for every $A$ , so $\vec{0} \in$ the set.

Let $\vec{x}, \vec{y}$ be in the set, so $A \vec{x} = [\begin{matrix} 0 \\ 0 \end{matrix}] = A \vec{y}$ . Then

$A (\vec{x} + \vec{y}) = A \vec{x} + A \vec{y} = [\begin{matrix} 0 \\ 0 \end{matrix}] + [\begin{matrix} 0 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

$\Rightarrow \vec{x} + \vec{y}$ is in the set.

Let $k \in ℝ$ . Then $A (k \vec{x}) = k A \vec{x} = k [\begin{matrix} 0 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$ , so $k \vec{x}$ is in the set.

Answer: $b. The set is a subspace.$

Theorem 4.2.3

The solution set of a homogeneous system $A 𝐱 = 𝟎$ of $m$ equations in $n$ unknowns is a subspace of $R^{n}$ .

This is proved by generalizing #14(b).

Definition

The solution set of a homogeneous system in $n$ unknowns is a subspace of $R^{n}$ , called the solution space of the system.

Definition

Let $T_{A} : R^{n} \to R^{m}$ be multiplication by the coefficient matrix $A$ . The solution space of $A 𝐱 = 𝟎$ is the set of vectors in $R^{n}$ that $T_{A}$ maps into the zero vector in $R^{m}$ . This set is called the kernel of the transformation.

Theorem 4.2.4

If $A$ is an $m \times n$ matrix, then the kernel of the matrix transformation $T_{A} : R^{n} \to R^{m}$ is a subspace of $R^{n}$ .

Theorem 4.2.2

If $W_{1}, W_{2}, \dots, W_{r}$ are subspaces of a vector space $V$ , then the intersection of these subspaces is also a subspace of $V$ .

Proof

Let $\vec{u}, \vec{v} \in W_{1} \cap W_{2} \cap \dots \cap W_{r}$ .

Work on this.

Examples of Subspaces

${𝟎}$ is a subspace of every vector space $V$
Any vector space $V$ is a subspace of itself

Subspace of $R^{2}$
Lines through the origin

Subspaces of $R^{3}$
Lines through the origin
Planes through the origin

The solution space of a homogeneous system in $n$ unknowns is a subspace of $R^{n}$

Subspaces of $M_{n n}$
Symmetric matrices
Triangular matrices
Diagonal matrices

Subspaces of $F (- \infty, \infty)$ (The following is actually a sequence of nested subspaces)
$C (- \infty, \infty)$ , the set of functions continuous on $R$
$C^{1} (- \infty, \infty)$ , the set of functions with continuous first-order derivatives on $R$
$C^{n} (- \infty, \infty)$ , the set of functions with continuous $n^{t h}$ -order derivatives on $R$
$C^{\infty} (- \infty, \infty)$ , the set of functions with derivatives of all orders on $R$
$P_{\infty}$ , the set of polynomials
$P_{n}$ , polynomials of degree $\leq n$

Nested ovals: F contains C, which contains C1, Cn, C infinity, P infinity, and Pn

Nested regions, from outside in: $F (- \infty, \infty)$ , $C (- \infty, \infty)$ , $C^{1} (- \infty, \infty)$ , $C^{n} (- \infty, \infty)$ , $C^{\infty} (- \infty, \infty)$ , $P_{\infty}$ , $P_{n}$ . Each set is a subspace of every set that contains it.