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Let $$A=\begin{pmatrix} 1 & 1 & 5\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{pmatrix},\qquad B=\begin{pmatrix} 1 & 7 & 0\\ 0 & 2 & 7\\ 0 & 0 & 2 \end{pmatrix}.$$ How to prove that these two matrices are similar?

Well, I tried to prove that above characteristic polynomial of both matrices ..., but it isn't correct. Because they have same elements on main diagonal,I thought that is easier to calculate, but it's not enough condition to prove that.

Can you give me some idea? Thank you

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    $\begingroup$ What are the dimensions of the matrix? You definitely need a MathJax tutorial (math.meta.stackexchange.com/questions/5020/…) to make your post more clear and informative $\endgroup$ – Toby Mak Jun 13 '17 at 11:16
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    $\begingroup$ For $1\times 1$, $2\times 2$ and $3\times 3$ matrices, characteristic polynomial + minimal polynomial are a complete similarity invariant. $\endgroup$ – user228113 Jun 13 '17 at 11:19
  • $\begingroup$ Matrics dimensions are 3x3,both od them $\endgroup$ – Veky Jun 13 '17 at 11:26
  • $\begingroup$ You have two upper-triangular matrices with the same entries on the main diagonal, these entries are the eigenvalues at the same time, however their eigenvectors are not the same.. – check what they are... it seems that the eigenspaces for both matrices are different.. $\endgroup$ – Widawensen Jun 13 '17 at 11:58
  • $\begingroup$ Find $P$ such that $P^{-1}AP = B$ .maths.manchester.ac.uk/~peter/MATH10212/notes9 $\endgroup$ – strider Jun 13 '17 at 12:09
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Two matrices, A and B, are "similar" if and only if there exist an invertible matrix, P, such that $A= PBP^{-1}$ or, equivalently, $AP= PB$. We can determine whether or not this particular A and B are similar directly from that. Write the presumed matrix, P, as $P= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$

Then $AP= \begin{bmatrix}1 & 1 & 5 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}= \begin{bmatrix}a+ d+ 5g & b+ e+ 5f & c+ f+ i \\ 2d & 2e & 2h \\ 2c & 2f & 2i\end{bmatrix}$

And $PB= \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 & 7 & 0 \\ 0 & 2 & 7 \\ 0 & 0 & 2\end{bmatrix}= \begin{bmatrix} a & 7a+ 2b & 7b+ 2c \\ d & 7d+ 2e & 7e+ 2f \\ g & 7g+ 2h & 7h+ 2i \end{bmatrix}$

Thus, A and B are similar if and only if there exist a, b, c, d, e, f, g, h, and i satisfying a+ d+ 5g= a, b+ e+ 5f= 7a+ 2b, c+ f+ i= 7b+ 2c, 2d= d, 2e= 7d+ 2e, 2h= 7e+ 2f, 2c= g, 2f= 7g+ 2h, and 2i= 7h+ 2i.

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  • $\begingroup$ General approach but it seems that in the light of two other answers the search for $a,b,c,...$ is in vain for this case. $\endgroup$ – Widawensen Jun 13 '17 at 12:59
  • $\begingroup$ Why the quotation marks around the word “similar”? $\endgroup$ – José Carlos Santos Jun 14 '17 at 8:04
  • $\begingroup$ @JoséCarlosSantos Maybe it is just a form of emphasizing the word.. $\endgroup$ – Widawensen Jun 14 '17 at 8:09
  • $\begingroup$ @Widawensen Using italics would be a much better choice then. $\endgroup$ – José Carlos Santos Jun 14 '17 at 8:10
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$A - 2I$ and $B - 2I$ (where $I$ denotes the identity matrix) do not have the same rank. So, the matrices can't be similar.

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  • $\begingroup$ Can we say more generally that if two matrices $A$ and $B$ are similar then their any polynomials $p(A)$ and $p(B)$ should be also similar ? $\endgroup$ – Widawensen Jun 13 '17 at 13:23
  • $\begingroup$ @Widawensen Yes. Note that $p(SAS^{-1}) = S\,p(A)\,S^{-1}$. $\endgroup$ – Omnomnomnom Jun 13 '17 at 13:26
  • $\begingroup$ I see, and because similarity preserves rank we have conclusion from your (and mine) answer .. $\endgroup$ – Widawensen Jun 13 '17 at 13:29
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It seems that according to Matrix Calculator

for the matrix $A$ we have eigenvectors (columns here are eigenvectors)

$ \begin{bmatrix} 1.000 & 0.707 & 0.981 \\ 0.000 & 0.707 & 0.000 \\ 0.000 & 0.000 & 0.196 \\ \end{bmatrix}$

For the matrix B

$ \begin{bmatrix} 1.000 & 0.990 & -0.990 \\ 0.000 & 0.141 & -0.141 \\ 0.000 & 0.000 & 0.000 \\ \end{bmatrix}$

The matrix made from the eigenvectors of $A$ is invertible, but from eigenvectors of $B$ not.

So $A$ is diagonalizable, $B$ is not. Hence $A$ and $B$ are not similar.

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  • $\begingroup$ For the final conclusion I could add that if it happened $A=PDP^{-1}$ and $B=TAT^{-1}$ then it would be $B=TPDP^{-1}T^{-1}=(TP)D(TP)^{-1}$ so $B$ would be diagonalizable, but it isn't due to the lack of full set of needed eigenvectors.. $\endgroup$ – Widawensen Jun 13 '17 at 14:27

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