# Prove a Given Subset is a Subspace and Find a Basis and Dimension

## Problem 270

Let

\[A=\begin{bmatrix}

4 & 1\\

3& 2

\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.

\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]

**(a)** Prove that the subset $V$ is a subspace of $\R^2$.

**(b)** Find a basis for $V$ and determine the dimension of $V$.

Contents

## Proof.

### (a) $V$ is a subspace of $\R^2$

Note that the equality $A\mathbf{x}=5\mathbf{x}$ can be written as

\[(A-5I)\mathbf{x}=\mathbf{0},\]
where $I$ is the $2 \times 2$ matrix.

If we define $B=A-5I$, then the subset $V$ becomes

\begin{align*}

V&=\{\mathbf{x}\in \R^2 \mid B\mathbf{x}=\mathbf{0}\}\\

&=\calN(B),

\end{align*}

the null space of the matrix $B$.

Since any null space is a vector space, this shows that $V$ is a subspace of $\R^2$.

### (b) A basis for $V$ and the dimension of $V$

From part (a), we have obtained that

\[V=\calN(B),\]
where

\[B=A-5I=\begin{bmatrix}

-1 & 1\\

3& -3

\end{bmatrix}.\]
Thus, $V$ is a set of solutions of the system $B\mathbf{x}=\mathbf{0}$.

The augmented matrix for this system is reduced as follows.

\[\left[\begin{array}{rr|r}

-1 & 1 & 0 \\

3 &-3 &0

\end{array} \right]
\xrightarrow{R_2+3R_1}

\left[\begin{array}{rr|r}

-1 & 1 & 0 \\

0 & 0 & 0

\end{array} \right]
\xrightarrow{-R_1}

\left[\begin{array}{rr|r}

1 & -1 & 0 \\

0 & 0 & 0

\end{array} \right].

\]

From this we see that the general solution $\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2

\end{bmatrix}$ satisfies

\[x_1=x_2,\]
and thus the general solution is

\[\mathbf{x}=\begin{bmatrix}

x_2 \\

x_2

\end{bmatrix}=x_2\begin{bmatrix}

1 \\

1

\end{bmatrix}.\]

Therefore we have

\begin{align*}

V&=\left \{ \mathbf{x} \in \R^2 \quad \middle| \quad \mathbf{x}=x_2\begin{bmatrix}

1 \\

1

\end{bmatrix} \text{ for any } x_2 \in \R \right \}\\

&=\Span\left\{ \begin{bmatrix}

1 \\

1

\end{bmatrix} \right \}.

\end{align*}

Thus the set

\[\left\{ \begin{bmatrix}

1 \\

1

\end{bmatrix} \right \}\]
is a spanning set of $V$ and it is linearly independent as it consists of only a single vector.

Therefore

\[\left\{ \begin{bmatrix}

1 \\

1

\end{bmatrix} \right \}\]
is a basis for the subspace $V$, and thus the dimension of $V$ is $1$.

(Recall that the dimension of a vector space is the number of vectors in a basis for the vector space.)

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