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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 .

Iuli
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1 Answers1

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Evaluate $\,A'$s eigenvalues:

$$p_A(t):=\det(tI-A)=\left|\begin{array}{}t-5&\;\;\;3\\-4&t+2\end{array}\right|=t^2-3t+2=(t-2)(t-1)$$

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

$$S=\left(\begin{array}{}3&1\\4&1\end{array}\right)$$

Take it from here

DonAntonio
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  • But $A \neq TBT^{-1}$ – Iuli Dec 02 '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 02 '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 02 '12 at 17:21
  • Read my last, edited message, @Iuli – DonAntonio Dec 02 '12 at 17:22
  • @DonAntonie Thanks :) it is interesting. Could you give some ideas about http://math.stackexchange.com/questions/247720/constructing-matrix-with-nullspace-containing-particular-vector. merci :) – Iuli Dec 02 '12 at 17:27