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The matrices $A=\begin{pmatrix}5 & -3 \\ 4 & -2\end{pmatrix}$ and $B=\begin{pmatrix}-1 & 1\\-6 & 4\end{pmatrix}$ are similar. By knowing that similar matrices have the same eigenvalues, find a matrix $T$ such that $A=TBT^{-1}.$

any idea or proof is welcome :) thanks .

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No, they are not similar. – Chris Eagle Dec 2 '12 at 16:55
They cannot be similar because neither their determinant nor their trace are equal...unless you're working on a field of characteristic $\,2\,$ ... – DonAntonio Dec 2 '12 at 16:55
sorry, I fix the mistake – Iuli Dec 2 '12 at 16:56
Oh, now that looks better. – DonAntonio Dec 2 '12 at 16:58
up vote 2 down vote accepted

Evaluate $\,A'$s eigenvalues:


Thus, the eigenvalues of $\,A\,$ are $\,1,2\,$. Find now one eigenvector for each eigenvalue:

$$(i)\;\;t=1:\;\;\;\;\;\;-4x+3y=0\Longleftrightarrow y=\frac{4}{3}x\Longrightarrow \binom{3}{4}$$ $${}$$

$$(i)\;\;t=2:\,\,\,\,\,\,-3x+3y=0\Longleftrightarrow x=y\Longrightarrow \binom{1}{1}$$

Well, as we know, we get that


Take it from here

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But $A \neq TBT^{-1}$ – Iuli Dec 2 '12 at 17:12
I messed up both the letter and the rows and columns of the last matrix. Now, $$\,S^{-1}AS=\begin{pmatrix}1&0\\0&2\end{pmatrix}=R^{-1}BR\,$$ ,and you can get the matrix $\,R\,$ for $\,B\,$ by the same method as we got $\,S\,$ for $\,A\,$ above, so now yes: take it from here – DonAntonio Dec 2 '12 at 17:20
Could you give me more details, please. After I found matrix $S$ which is the next step? Finding matrix $R$ and then what about $T$ ? thanks:) – Iuli Dec 2 '12 at 17:21
Read my last, edited message, @Iuli – DonAntonio Dec 2 '12 at 17:22
@DonAntonie Thanks :) it is interesting. Could you give some ideas about…. merci :) – Iuli Dec 2 '12 at 17:27

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