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I am considering $3\times 3 $ matrices, one of them is diagonalizable i.e

$$ A=\left[\begin{array}{ccc} 3&0&4\\ 0&-1&0\\ -2&0&-3 \end{array}\right]= \left[\begin{array}{ccc} -1&0&-2\\ 0&1&0\\ 1&0&1 \end{array}\right] \left[\begin{array}{ccc} -1&0&0\\ 0&-1&0\\ 0&0&1 \end{array}\right] \left[\begin{array}{ccc} -1&0&-2\\ 0&1&0\\ 1&0&1 \end{array}\right]^{-1}=P_1DP_1^{-1}. $$ The second one is not diagonalizable, and we have $$ B=\left[\begin{array}{ccc} 1&0&0\\ 0&-1&1\\ 0&0&-1 \end{array}\right]= \left[\begin{array}{ccc} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array}\right] \left[\begin{array}{ccc} -1&1&0\\ 0&-1&0\\ 0&0&1 \end{array}\right] \left[\begin{array}{ccc} 0&0&1\\ 1&0&0\\ 0&1&0 \end{array}\right]^{-1}=P_2JP_2^{-1}. $$ Matrices $A$ and $B$ are called similar if there exists an invertible matrix P such that $ B=P^{-1}AP$.

In my example $A$ and $B$ are not similar, because one of them is diagonalizable and the second is not diagonalizable. I am not sure, that my explaination is proper. I would be grateful for your advices, how to explain that this matrices are not similar.

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    $\begingroup$ You say that $B$ is not diagonalizable, but then write it as $B=PJP^{-1}$ and then $B\sim J$? $\endgroup$
    – A. P.
    Jan 5, 2023 at 12:56
  • $\begingroup$ $J$ is Jordan matrix $\endgroup$
    – AnnaGlo
    Jan 5, 2023 at 13:03
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    $\begingroup$ But the original matrix $B$ is already a Jordan matrix. What else do you want? $\endgroup$ Jan 5, 2023 at 13:04
  • $\begingroup$ oh.... I did not noticed it... (too much work.) But i am not sure, that my explaination about similarity is proper. $\endgroup$
    – AnnaGlo
    Jan 5, 2023 at 13:10
  • $\begingroup$ Well, $J$ is Jordan matrix, I didn't read correctly the matrix. Now, in order you see if $A$ and $B$ are similar you can see if we can find a matrix $P$ such that $AP=PB$. You can use brute force defining $P=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$. $\endgroup$
    – A. P.
    Jan 5, 2023 at 13:20

2 Answers 2

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Since $A$ is diagonalizable matrix and $B$ is a Jordan matrix which is not a diagonal matrix, $A$ and $B$ are not similar.

Or you can say that$$\dim\{v\in\Bbb R^3\mid A.v=-v\}=2,$$whereas$$\dim\{v\in\Bbb R^3\mid B.v=-v\}=1,$$which also proves that $A$ and $B$ are not similar.

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  • $\begingroup$ thank you very much for explaination $\endgroup$
    – AnnaGlo
    Jan 5, 2023 at 13:40
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By contradiction, suppose that $A\sim B$, then there exists a matrix $P$ regular such that $AP=PB$. Let $P=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$ be a matrix such that $\boxed{\det(P)\not=0}$.

A bit of algebra shows that

  • $AP=\begin{pmatrix}3&0&4\\0&-1&0\\-2&0&-3\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\begin{pmatrix}3a+4g&3b+4h&3c+4i\\-d&-e&-f\\-2a-3g&-2b-3h&-2c-3i \end{pmatrix}$.
  • $PB=\begin{pmatrix}1&0&0\\0&-1&1\\0&0&-1 \end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\begin{pmatrix}a&b&c\\g-d&h-e&i-f\\-g&-h&-i \end{pmatrix}$

But since we supposed that $AP=PB$ that is true if and only if $$a=0,\quad b=0,\quad c=0,\quad g=0,\quad h=0,\quad i=0$$ for $d,e,f\in {\bf R}$, but thus $$\det(P)=\det\begin{pmatrix}0&0&0\\d&e&f\\0&0&0\end{pmatrix} \implies \boxed{\det(P)=0}$$ and it is a contradiction.

Therefore, $A$ is not similar to the matrix $B$.

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  • $\begingroup$ We consider $AP=PB$, i think. $\endgroup$
    – AnnaGlo
    Jan 6, 2023 at 13:21
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    $\begingroup$ Of course, I fixed this just right now. $\endgroup$
    – A. P.
    Jan 6, 2023 at 14:03

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