Find two matrices $\mathbf{A}\in\mathbb{C}^{2\times2}$ and $\mathbf{B}\in\mathbb{C}^{2\times2}$ such that:
- they have the same eigenvalues
- but they are NOT similar
MY ATTEMPT
The characteristic polynomial of $\mathbf{A}$ is $p_{\mathbf{A}}(\lambda)=\lambda^{2}-\mathrm{tr}(\mathbf{A})+\mathrm{det}(\mathbf{A})$ and the characteristic polynomial of $\mathbf{B}$ is $p_{\mathbf{B}}(\lambda)=\lambda^{2}-\mathrm{tr}(\mathbf{B})+\mathrm{det}(\mathbf{B})$.
"The same eigenvalues" $\Longleftrightarrow P_{\mathbf{A}}(\lambda)=P_{\mathbf{B}}(\lambda)\Longleftrightarrow$ $\Longleftrightarrow\begin{cases}
\begin{array}{c}
\mathrm{tr}(\mathbf{A})=\mathrm{tr}(\mathbf{B})\\
\mathrm{det}(\mathbf{A})=\mathrm{det}(\mathbf{B})
\end{array} & \Longleftrightarrow\end{cases}\begin{cases}
\begin{array}{c}
a_{11}+a_{22}=b_{11}+b_{22}\\
a_{11}a_{22}-a_{21}a_{12}=b_{11}b_{22}-b_{21}b_{12}
\end{array}\end{cases}$
And now?
Best Answer
Consider the matrices \begin{align} A = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \ \ \text{ and } \ \ B = \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} \end{align} which clearly have the same eigenvalues, but they are not similar because $B$ is a Jordan block that can't be diagonalized.