Homework 5

1. (Although this question is just copying definitions, this is important to understand the whole concepts and I expect students should remember these definitions)
Write down the definition of the following:

(a) $W$ is a subspace of a vector space $V$

$W$ is a subspace if:

• given vectors $\mathbf{u}, \mathbf{v}\in W$, then $\mathbf{u+v}\in W$; and
• given scalar $\alpha\in\R$ and vector $\mathbf{v}\in W$, then $\alpha\mathbf{v}\in W$.

(b) $\operatorname{span}\set{\mathbf{v}_1, ...., \mathbf{v}_n}$.

The set of all linear combinations $c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n$ of the vectors $\mathbf{v}_1, \ldots, \mathbf{v}_n$ is called their span.

$$\operatorname{span}\set{\mathbf{v}_1, \ldots, \mathbf{v}_n} = \set{c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n : c_1,\ldots,c_n\in \R}$$

(c) $\mathbf{v}_1, ...., \mathbf{v}_n$ are linearly independent.

…if and only if

$$\set{x_1\mathbf{v}_1 + \cdots + x_n\mathbf{v}_n : x_i\in\R}$$

are distinct vectors.

(d) $\mathbf{v}_1, ...., \mathbf{v}_n$ are linearly dependent.

…if and only if

$$\exists\mathbf{v}_j \in \operatorname{span}\set{\mathbf{v}_i: i\neq j}$$

(e) $\mathbf{v}_1, ...., \mathbf{v}_n$ forms a basis of $V$.

…if:

• $\set{\mathbf{v}_1, \ldots, \mathbf{v}_n}$ are linearly independent; and
• $\operatorname{span}\set{\mathbf{v}_1, \ldots, \mathbf{v}_n} = V$.

This means that every vector in $V$ is a unique linear combination of $\set{\mathbf{v}_1, \ldots, \mathbf{v}_n}$.

(f) The dimension of a vector space $V$.

…is the number of vectors that make up a basis of $V$.

2. Determine if the following sets are subspaces of $\R^3$. Justify your answer.

(i) $W_1 = \set{(x, y, z) : x = z + 2}$

Checking $\mathbf{0}\in W_1$

Clearly, $(0,0,0)\notin W_1$. Therefore $W_1$ is not a subspace of $\R^3$.

$$(x,y,z) = (0,0,0) \implies 0 \neq 0 + 2 \\ \therefore \mathbf{0}\notin W_1$$

(ii) $W_2 = \set{(x, y, z) : x = 3y \text{ and } z = -y}$

Checking if $W_2$ is closed under addition

Consider two vectors $\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}, \begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} \in W_2$.

Then,
$$\begin{array}{c} \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} \iff \begin{cases} x_1 = 3y_1 \\ z_1 = -y_1 \end{cases}, &\begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} \iff \begin{cases} x_2 = 3y_2 \\ z_2 = -y_2 \end{cases} \end{array}$$

And so, for $\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} + \begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} = \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \\ z_1 + z_2 \end{pmatrix}$, notice that:
$$\begin{cases} x_1 + x_2 = 3y_1 + 3y_2 = 3(y_1 + y_2) \\ z_1 + z_2 = -y_1 - y_2 = -(y_1 + y_2) \end{cases} \implies \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \\ z_1 + z_2 \end{pmatrix} \in W_2$$

As such, $W_2$ is closed under addition.

Checking if $W_2$ is closed under scalar multiplication

Consider $\alpha\in\R$ and $\mathbf{u}=\begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}\in W_2$. Then, for $\alpha\mathbf{u} = \begin{pmatrix} \alpha x_1 \\ \alpha y_1 \\ \alpha z_1 \end{pmatrix}$, notice that:

$$\begin{cases} \alpha x_1 = \alpha(3y_1) = 3(\alpha y_1) \\ \alpha z_1 = \alpha(-y_1) = -(\alpha y_1) \end{cases} \implies \begin{pmatrix} \alpha x_1 \\ \alpha y_1 \\ \alpha z_1 \end{pmatrix} \in W_2$$

As such, $W_2$ is closed under scalar multiplication.

Therefore, we can conclude that $W_2$ is a subspace of $\R^3$.

(iii) $W_3 = \set{(x, y, z) : z = x^2 + y^2}$

Checking if $W_3$ is closed under addition

Consider two vectors $\mathbf{u},\mathbf{v}\in W_3$ where: $\mathbf{u} = (1,1,2)$ and $\mathbf{v} = (2, 2, 8)$.

$$\mathbf{u} = (1,1,2) \implies 2 = 1^2 + 1^2 \\ \mathbf{v} = (2, 2, 8) \implies 8 = 2^2 + 2^2$$

Then, $\mathbf{u}+\mathbf{v} = (3, 3, 10)$. But clearly, $\mathbf{u}+\mathbf{v}\notin W_2$:

$$\mathbf{u}+\mathbf{v} = (3, 3, 10) \implies 10 \neq 3^2 + 3^2$$

$W_3$ is not closed under addition, therefore it is not a subspace of $\R^3$.

3. For the following sets of vectors

(i) $\mathbf{v}_1 = (0, 1, 1)$, $\mathbf{v}_2 = (1, -1, 0)$ and $\mathbf{v}_3 = (3, -1, 2)$.
(ii) $\mathbf{v}_1 = (2, 1, 3)$, $\mathbf{v}_2 = (1, -2, 1)$, $\mathbf{v}_3 = (2, -3, 0)$ and $\mathbf{v}_4 = (0, -1, 4)$.
(iii) $\mathbf{v}_1 = (1, 0, 2, 1)$, $\mathbf{v}_2 = (-2, 3, -1, 1)$ and $\mathbf{v}_3 = (2, -2, 1, -1)$.

(a) Determine if the above set of vectors linearly dependent or linearly independent.

(i) $\mathbf{v}_1 = (0, 1, 1)$, $\mathbf{v}_2 = (1, -1, 0)$ and $\mathbf{v}_3 = (3, -1, 2)$.

$$\begin{array}{c} \begin{bmatrix} 0 & 1 & 3 \\ 1 & -1 & -1 \\ 1 & 0 & 2 \end{bmatrix} &\xrightarrow{R_2 + R_1} &\begin{bmatrix} 0 & 1 & 3 \\ 1 & 0 & 2 \\ 1 & 0 & 2 \end{bmatrix} &\xrightarrow{R_3 - R_2} &\begin{bmatrix} 0 & 1 & 3 \\ 1 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} &\xrightarrow{R_1 \leftrightarrow R_2} &\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix} \end{array}$$

The linear combination of the vectors has free variables. As such, they are linearly dependent.

(ii) $\mathbf{v}_1 = (2, 1, 3)$, $\mathbf{v}_2 = (1, -2, 1)$, $\mathbf{v}_3 = (2, -3, 0)$ and $\mathbf{v}_4 = (0, -1, 4)$.

The number of vectors (4) is greater than their dimensions (3). As such, they are linearly dependent.

(iii) $\mathbf{v}_1 = (1, 0, 2, 1)$, $\mathbf{v}_2 = (-2, 3, -1, 1)$ and $\mathbf{v}_3 = (2, -2, 1, -1)$.

$$\operatorname{rref} \left[ \begin{array}{ccc} | & | & | \\ \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \\ | & | & | \end{array} \right] = \operatorname{rref} \begin{bmatrix} 1 & -2 & 2 \\ 0 & 3 & -2 \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

The linear combination of the vectors do not have free variables. As such, they are linearly independent.

(b) For (i), determine if $\mathbf{w} = (1, 1, 1)$ lies in the span.

Where $\mathbf{v}_1 = (0, 1, 1)$, $\mathbf{v}_2 = (1, -1, 0)$, and $\mathbf{v}_3 = (3, -1, 2)$.

$$\operatorname{rref} \left[ \begin{array}{ccc|c} | & | & | & | \\ \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \mathbf{w} \\ | & | & | & | \end{array} \right] = \operatorname{rref} \left[ \begin{array}{ccc|c} 0 & 1 & 3 & 1 \\ 1 & -1 & -1 & 1 \\ 1 & 0 & 2 & 1 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]$$

The system has no solution. Therefore, $\mathbf{w} = (1, 1, 1)$ is not in the span.

(c) For (ii), express $\mathbf{v}_4$ as a linear combination of $\mathbf{v}_1$, $\mathbf{v}_2$ and $\mathbf{v}_3$.

Where $\mathbf{v}_1 = (2, 1, 3)$, $\mathbf{v}_2 = (1, -2, 1)$, $\mathbf{v}_3 = (2, -3, 0)$, and $\mathbf{v}_4 = (0, -1, 4)$.

$$\operatorname{rref} \left[ \begin{array}{ccc|c} | & | & | & | \\ \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \mathbf{v}_4 \\ | & | & | & | \end{array} \right] = \operatorname{rref} \left[ \begin{array}{ccc|c} 2 & 1 & 2 & 0 \\ 1 & -2 & -3 & -1 \\ 3 & 1 & 0 & 4 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 0 & 0 & \frac{2}{11} \\[.4em] 0 & 1 & 0 & \frac{38}{11}\\[.4em] 0 & 0 & 1 & -\frac{21}{11} \end{array} \right] \\[1em] \therefore \mathbf{v}_4 = \frac{2}{11}\mathbf{v}_1 + \frac{38}{11}\mathbf{v}_2 -\frac{21}{11}\mathbf{v}_3$$

4. Expand the kernel of the following matrices as span of vectors and then compute the dimension.

Assuming the question is asking for the dimension of the kernel i.e., the nullity.

(a) $$\begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3\end{bmatrix}$$

$$\operatorname{rref} \left[ \begin{array}{cc|c} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 3 & 0 \end{array} \right] = \left[ \begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \\[1.2em] \therefore\ker\begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{bmatrix} = \Set{\begin{pmatrix} 0 \\ 0 \end{pmatrix}} = \operatorname{span}\Set{ \begin{pmatrix} 0 \\ 0 \end{pmatrix} }$$

The nullity is zero.

(b) $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{bmatrix}$$


$$\ \operatorname{rref} \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \end{array} \right] = \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \\ \therefore x_1 = -x_2 - x_3 \\ x_2, x_3 \in\R \\[1.2em] \begin{align*} \therefore\ker\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} &= \Set{ \begin{pmatrix} -x_2 - x_3 \\ x_2 \\ x_3 \end{pmatrix} : x_2, x_3 \in \R } \\ &= \Set{ x_2 \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} : x_2, x_3 \in \R } \\ &= \operatorname{span}\Set{ \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} } \end{align*}$$

The nullity is two.

(c) $$\begin{bmatrix} 1 & -1 & -1 & 1 & 1 \\ -1 & 1 & 0 & -2 & 2 \\ 1 & -1 & -2 & 0 & 3 \\ 2 & -2 & -1 & 3 & 4\end{bmatrix}$$

$$\operatorname{rref} \left[ \begin{array}{ccccc|c} 1 & -1 & -1 & 1 & 1 & 0 \\ -1 & 1 & 0 & -2 & 2 & 0 \\ 1 & -1 & -2 & 0 & 3 & 0 \\ 2 & -2 & -1 & 3 & 4 & 0 \end{array} \right] = \left[ \begin{array}{ccccc|c} 1 & -1 & 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \\ \therefore x_5 = 0 \\ x_3 = -x_4 \\ x_1 = x_2 - 2x_4 \\ x_2, x_4 \in \R \\[1.2em] \begin{align*} \therefore\ker\begin{bmatrix} 1 & -1 & -1 & 1 & 1 \\ -1 & 1 & 0 & -2 & 2 \\ 1 & -1 & -2 & 0 & 3 \\ 2 & -2 & -1 & 3 & 4 \end{bmatrix} &= \Set{ \begin{pmatrix} x_2 - 2x_4 \\ x_2 \\ -x_4 \\ x_4 \\ 0 \end{pmatrix} : x_2, x_4 \in\R } \\ &= \Set{ x_2 \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} -2 \\ 0 \\ -1 \\ 1 \\ 0 \end{pmatrix} : x_2, x_4 \in\R } \\ &= \operatorname{span}\Set{ \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -2 \\ 0 \\ -1 \\ 1 \\ 0 \end{pmatrix} } \end{align*}$$

The nullity is two.

5. Let $W = \set{(x_1, x_2, x_3, x_4) : x_1 - x_2 + 2x_3 - x_4 = 0}$. Find a basis for the subspace $W$.

$$x_1 - x_2 + 2x_3 - x_4 = 0 \implies x_1 = x_2 - 2x_3 + x_4 \\ \begin{align*} W &= \Set{ \begin{pmatrix} x_2 - 2x_3 + x_4 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} : x_2, x_3, x_4 \in \R } \\ &= \Set{ x_2 \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} -2 \\0 \\ 1 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} : x_2, x_3, x_4 \in \R } \end{align*} \\[2em] \therefore \operatorname{basis}(W) = \Set{ \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} , \begin{pmatrix} -2 \\0 \\ 1 \\ 0 \end{pmatrix} , \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} }$$

6. Let $W = \Set{(x_1, x_2, x_3, x_4) :\begin{cases}x_1 + x_2 - x_3 + x_4 = 0, \\2x_1 + 2x_2 - 2x_3 + x_4 = 0,\end{cases}}$. Find a basis for the subspace $W$ and what is its dimension?

$$\begin{array}{c} \begin{cases} x_1 + x_2 - x_3 + x_4 = 0 \\ 2x_1 + 2x_2 - 2x_3 + x_4 = 0 \end{cases} &\iff& \left[ \begin{array}{cccc|c} 1 & 1 & -1 & 1 & 0 \\ 2 & 2 & -2 & 1 & 0 \end{array} \right] \end{array} \\[2em] \begin{array}{c} \left[ \begin{array}{cccc|c} 1 & 1 & -1 & 1 & 0 \\ 2 & 2 & -2 & 1 & 0 \end{array} \right] &\xrightarrow{R_2 - 2R_1} &\left[ \begin{array}{cccc|c} 1 & 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & -1 & 0 \end{array} \right] &\xrightarrow[-R_2]{R_1 + R_2} &\left[ \begin{array}{cccc|c} 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right] \end{array} \\ \therefore x_4 = 0 \\ x_1 = -x_2 + x_3 \\ x_2, x_3 \in \R \\ \begin{align*} W &= \Set{ \begin{pmatrix} -x_2 + x_3 \\ x_2 \\ x_3 \\ 0 \end{pmatrix} : x_2, x_3 \in \R } \\ &= \Set{ x_2 \begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}: x_2, x_3 \in \R } \end{align*} \\[2em] \therefore\operatorname{basis}(W) = \Set{ \begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} }$$

The nullity is two.