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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, 2017 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, 2017 at 11:19
  • $\begingroup$ Matrics dimensions are 3x3,both od them $\endgroup$
    – Veky
    Jun 13, 2017 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, 2017 at 11:58
  • $\begingroup$ Find $P$ such that $P^{-1}AP = B$ .maths.manchester.ac.uk/~peter/MATH10212/notes9 $\endgroup$
    – strider
    Jun 13, 2017 at 12:09

3 Answers 3

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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, 2017 at 13:23
  • $\begingroup$ @Widawensen Yes. Note that $p(SAS^{-1}) = S\,p(A)\,S^{-1}$. $\endgroup$ Jun 13, 2017 at 13:26
  • $\begingroup$ I see, and because similarity preserves rank we have conclusion from your (and mine) answer .. $\endgroup$
    – Widawensen
    Jun 13, 2017 at 13:29
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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, 2017 at 12:59
  • $\begingroup$ Why the quotation marks around the word “similar”? $\endgroup$ Jun 14, 2017 at 8:04
  • $\begingroup$ @JoséCarlosSantos Maybe it is just a form of emphasizing the word.. $\endgroup$
    – Widawensen
    Jun 14, 2017 at 8:09
  • $\begingroup$ @Widawensen Using italics would be a much better choice then. $\endgroup$ Jun 14, 2017 at 8:10
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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, 2017 at 14:27

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