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

Linear Algebra Problems and Solutions

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 $B=\{\mathbf{v}_1, \mathbf{v}_2\}$.

 
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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*}


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