# Are identical eigenvalues sufficient for matrix similarity of diagonizable matrices?

Given two diagonizable matrices $A$ and $B$ with identical eigenvalues (and therefore determinants, traces and ranks) that do not commute, is this sufficient to claim $\exists P: B = P^{-1}AP$

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No, see this question; I gave an explicit example in the comments to Arturo's answer. – J. M. Aug 16 '11 at 13:42
Repeating again my comment here: $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ are not similar despite having identical eigenvalue sets. – J. M. Aug 16 '11 at 13:43
@J.M.: good point, thank you. I realized the matrices I consider are also diagonalizable, I changed the question. Your second matrix is not diagonizable, does that change matters? (sorry for that edit, please don't delete your comment though) – Tobias Kienzler Aug 16 '11 at 13:46
If both are now diagonalizable matrices, then yes... just take the matrix of eigenvectors of $A$ and $B$ and form the appropriate product. – J. M. Aug 16 '11 at 13:50
Yes, Jordan blocks aren't diagonalizable... :) – J. M. Aug 16 '11 at 13:51

Consider the eigen-system of $A$, i.e. ${\lambda_i, v^{(A)}_i}$ such that $A.v^{(A)}_i = \lambda_i v^{(A)}_i$. Consider a matrix $T_A = \left\{v^{(A)}_1, \ldots, v^{(A)}_n\right\}$. Then $A.T_A = T_A.D$, where $D$ is the diagonal matrix is the equivalent notation for collection of equations $(A.T_A)_{j, i} = (A.v^{(A)}_i)_j = \lambda_i (v^{(A)}_i)_j = (T_A.D)_{j,i}$. Similarly $B.T_B = T_B.D$.

Assuming $T_B$ and $T_A$ are invertible, it follows $D = T_B^{-1}.B.T_B = T_A^{-1}.A.T_A$, and $B=(T_A.T_B^{-1})^{-1}.A.(T_A.T_B^{-1})$.

But not all matrices have the complete set of eigen-vectors, i.e. $T_A$ or $T_B$ can be degenerate.

Consider matrix $D_1 = \left( \begin{array}{ccc} 4 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$ and $D_2 = \mathrm{diag}(4,4,1)$. Also let $s_1=\left( \begin{array}{ccc} -1 & 1 & 1 \\ 1 & 1 & -1 \\ 0 & 2 & 2 \\ \end{array} \right)$, $s_2 = \left( \begin{array}{ccc} 1 & -1 & 0 \\ 2 & 1 & 0 \\ -2 & -2 & 2 \\ \end{array} \right)$.

Now define $A = s_1.D_1.s_1^{-1}$ and $B=s_2.D_2.s_2^{-1}$.

$$A = \left( \begin{array}{ccc} 5 & 1 & -\frac{3}{2} \\ -1 & 3 & \frac{3}{2} \\ 3 & 3 & 1 \\ \end{array} \right) \quad \quad B = \frac{1}{3} \left( \begin{array}{ccc} 10 & 1 & 0 \\ -4 & 14 & 0 \\ 10 & -14 & 3 \\ \end{array} \right)$$

These matrices have the same eigenvalues, i.e. $\left\{4, 4, 1\right\}$, they do not commute, but they are not similar, because their Jordan normal forms are different.

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Yes:

Say $A$ is diagonalized by $Q$ into the diagonal Matrix $\Lambda$, and $B$ by $R$ such that:

$$\begin{array}{rl} A &= Q\Lambda Q^{-1} \Rightarrow \Lambda=Q^{-1}\Lambda Q,\\ B &= R\Lambda R^{-1} = (QR^{-1})^{-1}A\underbrace{(QR^{-1})}_{=P}\end{array}$$

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