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


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?

  • 2
    $\begingroup$ You don't need to do fancy math. All you need to do is find an example. Here, I'll give you some numbers to build two matrices out of: $0~0~0~0~0~0~0~1$. Have fun. $\endgroup$ – Anon Oct 5 '16 at 23:51
  • $\begingroup$ You could pick rotation matrices; they have the same eigenvalues (both are $1$), but they are not similar if they are not the same rotation. $\endgroup$ – Larry B. Oct 5 '16 at 23:52
  • $\begingroup$ @LarryB. Rotation matrices do have unit eigenvalues, but they are generally not $1$. The only rotation matrix with eigenvalues of $1$ is the identity matrix. $\endgroup$ – Anon Oct 5 '16 at 23:54
  • $\begingroup$ @LarryB. For $2\times2$ rotation matrices trace would be $2\cos\theta$ and this is not $1+1$ in general, and so eigen values are not 1. $\endgroup$ – P Vanchinathan Oct 5 '16 at 23:55

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.

  • 2
    $\begingroup$ Here's an even simpler example: $\begin{pmatrix}0&1\\0&0\end{pmatrix}\not\sim\begin{pmatrix}0&0\\0&0\end{pmatrix}$ $\endgroup$ – Anon Oct 5 '16 at 23:55
  • $\begingroup$ @McFry. True, but I wanted matrices that were invertible. $\endgroup$ – Jacky Chong Oct 5 '16 at 23:56
  • 6
    $\begingroup$ @JackyChong: Jordan block would be usually bit advanced for the question under discussion. It is easier to see that identity matrix can be similar to no other matrix as $CIC^{-1}= I$ $\endgroup$ – P Vanchinathan Oct 5 '16 at 23:59
  • $\begingroup$ @PVanchinathan I agree. $\endgroup$ – Jacky Chong Oct 6 '16 at 0:00

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