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 .
$\endgroup$ 41 Answer
$\begingroup$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
$\endgroup$ 5
