Find two $2\times2$ matrices which have the same eigenvalues, but are not similar

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?

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.