Consider the basis $\epsilon$ for $V=M_{2 \times 2}(\mathbb{R})$ with the following basis vectors:

$$ e_1 = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right) $$

$$ e_2 = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) $$

$$ e_3 = \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right) $$

$$ e_4 = \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right) $$

And consider the linear transformation $T: V \rightarrow V$:

$T(\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)) = \left( \begin{matrix} 2a-2b & -a+3b \\ 4c-2d & 3c-d \end{matrix} \right) $

Can somebody help me to find an ordered basis $\beta$ such that $[T]_{\beta}^{\beta}$ is diagonal? Thank you!


1 Answer 1


$[T]_\beta^\beta=M_{\epsilon\to\beta}[T]_\epsilon^\epsilon M_{\beta\to\epsilon}=M^{-1}[T]_\epsilon^\epsilon M$ where $M=M_{\beta\to\epsilon}$ is the basis change matrix from basis $\beta$ to $\epsilon$.

You want $M^{-1}[T]_\epsilon^\epsilon M=\delta$, a diagonal matrix. So $\delta$ must contain the eigenvalues of $[T]_\epsilon^\epsilon$ along its diagonals and $M$ is the matrix whose column vectors are the corresponding linearly independent eigenvectors.

Once we have obtained $M=M_{\beta\to\epsilon}$, we can determine the $\beta$ basis vectors like this: the first vector in the basis has coordinates $[1,0,0,0]^T$ in the same basis, so its coordinates in the $\epsilon$ basis are given by $M[1,0,0,0]^T$ which is the first column vector of $M$, i.e. the first eigenvector of $[T]_\epsilon^\epsilon$. Similarly the $i^\text{th}$ basis vector is the $i^\text{th}$ column of $M$, i.e. the $i^\text{th}$ eigenvector of $[T]_\epsilon^\epsilon$.

For $[T]_\epsilon^\epsilon=\begin{bmatrix}2&-2&0&0\\-1&3&0&0\\0&0&4&-2\\0&0&3&-1\end{bmatrix}$ the eigenvalues and corresponding eigenvectors are $(1,[2,1,0,0]^T),(1,[0,0,2,3]^T),(2,[0,0,1,1]^T),(4,[-1,1,0,0]^T)$. You may verify that the eigenvectors are linearly independent. Thus, one option for $\beta$ is$$\{[2,1,0,0]^T,[0,0,2,3]^T,[0,0,1,1]^T,[-1,1,0,0]^T\}$$

  • 1
    $\begingroup$ You can change the order of the eigenvectors. It will remain a valid basis. $\endgroup$ Jan 7, 2021 at 18:53

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