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I have a 4x4 matrix $$ A = \left[ \begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \end{matrix} \right] $$ and i want to check if there is any similar matrix of matrix $A$ of the form: $$ B = \left[ \begin{matrix} 0 & 0 & x & y \\ 0 & 0 & z & w \\ a & b & 0 & 0 \\ c & d & 0 & 0 \end{matrix} \right] $$ Is there any way to come in a conclusion if there such of this form similarity ?

thanks.

mpouxa
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2 Answers2

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note that two similar matrix have the same caracteristic polinomial $P(t)$, if we calcul this polynomial for each matrix we tack an have coefficient -1 at $t^3$ and othere $0$ as coefficient at $t^3$ so the similarity is impossible

m.idaya
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    In other words, they clearly have different traces! That's simpler than my roundabout argument. – hmakholm left over Monica Jun 26 '16 at 11:14
  • very good, exactely, thanks – m.idaya Jun 26 '16 at 11:15
  • But this can be a generic criterion for making a conclusion and for other matrices? For example if i have the matrix $$C = \left[ \begin{matrix} 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 \end{matrix} \right]$$ which has a characteristic Polynomial $\lambda^4 -1 $ what can i say for this case ?I suppose that the term $\lambda^2$ in char polynomial of matrix $B$ possibly might not be 0. – mpouxa Jun 26 '16 at 14:07
  • two matrix have the same caracteristic polynomial not amplique there are similar matrix. – m.idaya Jun 26 '16 at 14:14
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Working over $\mathbb C$, let's write $B$ as a block matrix $$ B = \begin{bmatrix}0 & X \\ E & 0 \end{bmatrix} $$ Then $$ B^2 = \begin{bmatrix}XE & 0 \\ 0 & EX\end{bmatrix} $$ and since $XE$ and $EX$ have the same eigenvalues, we can do a case analysis on the number of these eigenvalues:

  • If there's only one eigenvalue for $EX$ and $XE$, then $B^2$ has only one eigenvalue. However, Wolfram Alpha tells us that $A^2$ has three different eigenvalues, so $A$ cannot be similar to such a $B$.

  • On the other hand, if $EX$ has two eigenvalues, it is diagonalizable, and so is $XE$, so $B^2$ is diagonalizable. But Wolfram Alpha just showed us that $A^2$ isn't diagonalizable, so $A$ is not similar to this kind of $B$ either.

Since $A$ is not similar to any $B$-shaped matrix over $\mathbb C$, this is in particular not the case over $\mathbb R$ either.

Community
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hmakholm left over Monica
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  • Thank you very much! You were very Helpfull ! – mpouxa Jun 26 '16 at 10:58
  • @mpouxa: Consider accepting m.idaya's much simpler argument instead. – hmakholm left over Monica Jun 26 '16 at 11:19
  • if i find a diagonalizable matrix in the second case of your answer is it enough to say that the matrix are similar ? For example i take the matrix $C$ as i used in m.idaya's answer, where $$C = \left[ \begin{matrix} 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 1 & 0 & 0 & 0 \end{matrix} \right]$$. $C^2$ is diagonalizable but for the first case it has 2 different eigenvalues. – mpouxa Jun 26 '16 at 15:22
  • @mpouxa: That is similar to $$\begin{bmatrix}0&0&1&0\0&0&0&1\0&1&0&0\1&0&0&0\end{bmatrix}$$ simply by permuting indices. – hmakholm left over Monica Jun 26 '16 at 15:32
  • Yes, But first i'm looking for a generic creterion to understand for more difficult cases when i have to consider this kind of simillarity . For example an other matrix that i'm trying to check is: $$D = \left[ \begin{matrix} 0 & 0 & 1 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 1 & 1 & 0 & 0 \end{matrix} \right]$$ which also agrees with the second case of your answer but not with the first – mpouxa Jun 26 '16 at 15:44