# Find an ordered basis $\beta$ such that $[T]_{\beta}^{\beta}$ is diagonal

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!

• What have you tried? You should link this to your previous question. Jan 7, 2021 at 17:55

$$[T]_\beta^\beta=M_{\epsilon\to\beta}[T]_\epsilon^\epsilon M_{\beta\to\epsilon}=M^{-1}[T]_\epsilon^\epsilon M$$ 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$$\{[2,1,0,0]^T,[0,0,2,3]^T,[0,0,1,1]^T,[-1,1,0,0]^T\}$$