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Let matrices $$A=\begin{bmatrix} -3&3&-2\\ -7&6&-3\\ 1&-1&2 \end{bmatrix},\quad B=\begin{bmatrix} 0&1&-1\\ -4&4&-2\\ -2&1&1 \end{bmatrix},\quad C=\begin{bmatrix} 0&-1&-1\\ -3&-1&-2\\ 7&5&6 \end{bmatrix},\quad D=\begin{bmatrix} 0&1&2\\ 0&1&1\\ 0&0&2 \end{bmatrix} $$ My goal is to determine which of them are similar to each other. First, we observe that $$\det(A)=\det(B)=\det(C)=4\neq\det(D)=0$$ Thus, $D$ cannot be similar to any of $A,B,C$. $\det(\lambda I-A)=0$ and similarly for $B,C$. Denote $\sigma(X)=\operatorname{spectrum of matrix }X$. With a slight abuse of notation, to denote that $\lambda_2=2$ has algebraic multiplicity of $2$, I found $$\sigma(A)=\sigma(B)=\sigma(C)=\{\lambda_1=1,\lambda_2=2,\lambda_2=2\}$$ This means that all of them have the same Jordan Canonical Form which is (up to permutations of the diagonal Jordan blocks) $$ J_A=J_B=J_C=\begin{bmatrix} \lambda_1&0&0\\ 0&\lambda_2&1\\ 0&0&\lambda_2 \end{bmatrix} = \begin{bmatrix} 1&0&0\\ 0&2&1\\ 0&0&2 \end{bmatrix} $$ Hence, $A,B,C$ are similar. Is this justification correct? If yes, is there any faster way/trick to get to this result? Finding the eigenvalues by solving a 3-rd degree equation took me some time.

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  • $\begingroup$ well, $B$ diagonalizes the minimal polynomial is not the same as the characteristic polynomial. $x^2 - 3 x + 2$ Just check $B^2 - 3 B + 2I$ $\endgroup$ – Will Jagy Feb 23 '18 at 20:30
  • $\begingroup$ No, it is not: a n eigenvalue can have multiplicity $2$, yet be diagonalisable. This is the case if and only if the geometric multiplicity (i.e. the dimension of the corresponding eigenspace) is equal to the algebraic multiplicity. $\endgroup$ – Bernard Feb 23 '18 at 20:32
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$B$ diagonalizes

$$ \frac{1}{2} \left( \begin{array}{rrr} 4 & -2 & 2 \\ -4 & 3 & -2 \\ -2 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & -1 \\ 2 & 2 & 0 \\ 1 & 0 & 2 \\ \end{array} \right) = I $$ $$ \frac{1}{2} \left( \begin{array}{rrr} 4 & -2 & 2 \\ -4 & 3 & -2 \\ -2 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & -1 \\ -4 & 4 & -2 \\ -2 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & -1 \\ 2 & 2 & 0 \\ 1 & 0 & 2 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{array} \right) $$

Regardless, this was going to require a fair amount of work by hand. I am not so much a fan of doing things quickly as of doing things well and including alternate methods that may indicate errors. In this case, one may confirm that $B^2 - 3 B + 2 I = 0.$ For $A$ and $C$ the minimal polynomial agrees with the characteristic polynomial, which means that each eigenvalue occurs in a single Jordan block, and since $2$ has higher multiplicity this means they do not diagonalize. In contrast, for $B$ we see that $2$ must occur in more than one Jordan block. In fact, there are two, each dimension one, meaning there is a basis of two eigenvectors for eigenvalue $2.$ I have typed in one way to diagonalize $B.$ The matrix on the right has columns that are all eigenvectors

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