# Matrix of Linear Transformation with respect to a Basis Consisting of Eigenvectors

## Problem 314

Let $T$ be the linear transformation from the vector space $\R^2$ to $\R^2$ itself given by
$T\left( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \right)= \begin{bmatrix} 3x_1+x_2 \\ x_1+3x_2 \end{bmatrix}.$

(a) Verify that the vectors
$\mathbf{v}_1=\begin{bmatrix} 1 \\ -1 \end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ are eigenvectors of the linear transformation $T$, and conclude that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis of $\R^2$ consisting of eigenvectors.

(b) Find the matrix of $T$ with respect to the basis $B=\{\mathbf{v}_1, \mathbf{v}_2\}$.

## Solution.

### (a) $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis of $\R^2$ consisting of eigenvectors

We compute that
\begin{align*}
T(\mathbf{v}_1)=T\left( \begin{bmatrix}
1 \\
-1
\end{bmatrix} \right)= \begin{bmatrix}
2 \\
-2
\end{bmatrix}
=2\begin{bmatrix}
1 \\
-1
\end{bmatrix}=2\mathbf{v}_1
\end{align*}
and
\begin{align*}
T(\mathbf{v}_2)=T\left( \begin{bmatrix}
1 \\
1
\end{bmatrix} \right)= \begin{bmatrix}
4 \\
4
\end{bmatrix}
=4\begin{bmatrix}
1 \\
1
\end{bmatrix}=4\mathbf{v}_2.
\end{align*}

Thus, $\mathbf{v}_1$ is an eigenvector corresponding to the eigenvalue $2$ and $\mathbf{v}_2$ is an eigenvector corresponding to the eigenvalue $4$.
Since $\mathbf{v}_1, \mathbf{v}_2$ are eigenvectors corresponding to distinct eigenvalues, they are linearly independent, and thus $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis of $\R^2$.

### (b) Find the matrix of $T$ with respect to the basis $B$

From the computation in part (a), we have
\begin{align*}
T(\mathbf{v}_1)=2\mathbf{v}_1+0\mathbf{v}_2\\
T(\mathbf{v}_2)=0\mathbf{v}_1+4\mathbf{v}_2.
\end{align*}

Hence the coordinate vectors of $T(\mathbf{v}_1), T(\mathbf{v}_2)$ with respect to the basis $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis of $\R^2$ are
$[T(\mathbf{v}_1)]_B=\begin{bmatrix} 2 \\ 0 \end{bmatrix}, [T(\mathbf{v}_2)]_B=\begin{bmatrix} 0 \\ 4 \end{bmatrix}.$

Thus the matrix $A$ of the linear transformation $T$ with respect to the basis $B$ is
\begin{align*}
A=[\,[T(\mathbf{v}_1)]_B, [T(\mathbf{v}_2)]_B\,]=\begin{bmatrix}
2 & 0\\
0& 4
\end{bmatrix}.
\end{align*}

(a) Let $A=\begin{bmatrix} 1 & 2 & 1 \\ 3 &6 &4 \end{bmatrix}$ and let \[\mathbf{a}=\begin{bmatrix} -3 \\ 1 \\...