# Quiz 5: Example and Non-Example of Subspaces in 3-Dimensional Space

## Problem 304

Problem 1 Let $W$ be the subset of the $3$-dimensional vector space $\R^3$ defined by
$W=\left\{ \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\in \R^3 \quad \middle| \quad 2x_1x_2=x_3 \right\}.$

(a) Which of the following vectors are in the subset $W$? Choose all vectors that belong to $W$.
$(1) \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \qquad(2) \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \qquad(3)\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix} \qquad(4) \begin{bmatrix} 0 \\ 0 \end{bmatrix} \qquad(5) \begin{bmatrix} 1 & 2 & 4 \\ 1 &2 &4 \end{bmatrix} \qquad(6) \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}.$

(b) Determine whether $W$ is a subspace of $\R^3$ or not.

Problem 2 Let $W$ be the subset of $\R^3$ defined by
$W=\left\{ \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \R^3 \quad \middle| \quad x_1=3x_2 \text{ and } x_3=0 \right\}.$ Determine whether the subset $W$ is a subspace of $\R^3$ or not.

## Solution.

### Problem 1 (a) Which of the following vectors are in the subset $W$?

First of all, each element in $W$ is a $3$-dimensional vector. Thus (4), (5) can not be elements of $W$.
If $\mathbf{x}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$, then $x_1=0, x_2=0, x_3=0$ and these satisfy the defining relation $2x_1x_2=x_3$. Thus $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ is in $W$.
If $\mathbf{x}=\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$, then we have $x_1=1, x_2=2, x_3=2$, and thus $2x_1x_2=4$ but $x_3=2$. Hence $2x_1x_2 \neq x_3$, and $\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$ is not in $W$. For (3), we have $2x_1x_2=0=x_3$, and thus $\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}$ is in $W$. For (6), we have $2x_1x_2=-2=x_3$, and thus $\begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}$ is in $W$.
In summary, the vectors (1), (3), (6) are in $W$.

### Problem 1 (b) Determine whether $W$ is a subspace of $\R^3$ or not.

We claim that $W$ is not a subspace. For example, consider the vectors
$\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix} \text{ and } \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}.$ In part (a), we showed that these vectors are in $W$.
However, the sum
$\begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}+\begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}=\begin{bmatrix} 4 \\ -1 \\ -2 \end{bmatrix}$ is not in $W$.
In fact, if $x_1=4, x_2=-1, x_3=-2$, then $2x_1x_2=-8\neq -2=x_3$.
Thus the sum does not satisfy the defining relation for $W$, hence not in $W$.
Therefore, the subset $W$ is not closed under addition, hence $W$ is not a subspace of $\R^3$.

### Solution of Problem 2

We prove that $W$ is a subspace of $\R^3$. To do this, we check the following subspace criteria.

Subspace Criteria
(a) The zero vector $\mathbf{0} \in \R^3$ is in $W$.
(b) If $\mathbf{x}, \mathbf{y} \in W$, then $\mathbf{x}+\mathbf{y}\in W$.
(c) If $\mathbf{x} \in W$ and $c\in \R$, then $c\mathbf{x} \in W$.

Consider the zero vector $\mathbf{0}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\in \R^3$. Then $x_1=0, x_2=0, x_3=0$ satisfy the defining relations $x_1=3x_2$ and $x_3=0$. Thus the zero vector $\mathbf{0}$ is in $W$, hence condition (a) is satisfied.

To check condition (b), let us take $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ and $\mathbf{y}=\begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}$ from $W$. Then we want to show that the sum $\mathbf{x}+\mathbf{y}$ is in $W$
Since $\mathbf{x}\in W$, we have
$x_1=3x_2 \text{ and } x_3=0.$ Also, since $\mathbf{y}\in W$, we have
$y_1=3y_2 \text{ and } y_3=0.$ From these, we see that
$x_1+y_1=3x_2+3y_2=3(x_2+y_2) \text{ and } x_3+y_3=0.$ Or simply we have
$x_1+y_1=3(x_2+y_2) \text{ and } x_3+y_3=0.$ Note that this implies that the sum
$\mathbf{x}+\mathbf{y}=\begin{bmatrix} x_1+y_1 \\ x_2+y_2 \\ x_3+y_3\end{bmatrix}$ satisfies the defining relation, and the sum is in $W$. Thus condition (b) is satisfied.

Finally, to prove condition (c), take $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ from $W$ and let $c\in \R$. We need to show that the scalar product $c\mathbf{x} \in W$.
Since $\mathbf{x}\in W$, we have
$x_1=3x_2 \text{ and } x_3=0.$ Multiplying these by $c$, we have
$cx_1=3(cx_2) \text{ and } cx_3=0.$ These equalities tells that
the scalar product $c\mathbf{x}=\begin{bmatrix} cx_1 \\ cx_2 \\ cx_3 \end{bmatrix}$ satisfies the defining relation and hence it is in $W$. Thus condition (c) is met as well.

Since we have checked all subspace criteria, $W$ is a subspace of $\R^3$.

## Comment.

These are Quiz 5 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

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