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

Linear Algebra Problems and Solutions

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 $V$.

 
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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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