Show that the matrices $A =\left(\begin{matrix} 5 & 1 \\ -6 & 0 \end{matrix}\right)$ and $B = \left(\begin{matrix} 7 & -5 \\ 4 & -2 \end{matrix}\right) $

are similar over $\mathbb{R}$. In other words, show that there is an invertible matrix $C$ with real coefficients such that $A = C^{-1}BC$.

For this problem I don't know how to show they are similar. I tried to solve for the matrix $C$, by setting up $C = \left(\begin{matrix} a & b \\ c & d\end{matrix}\right)$, such that $CA = BC$. But I feel it's not the easiest way to do this. Please help me figure this out, thank you!

closed as off-topic by colormegone, Shailesh, John B, Antonios-Alexandros Robotis, Stefan Mesken Mar 26 '16 at 1:00

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  • 1
    Do you know about characteristic polynomial, eigenvalues, eigenvectors...? – DonAntonio Mar 25 '16 at 22:23
  • 1
    First you should check that trace and determinant agree. If so, then just solve your $CA = BC$ thing, it is really not that hard. – Will Jagy Mar 25 '16 at 22:31
up vote 6 down vote accepted

Hint: The characteristic polynomial of both matrices is $x^2-5x+6 = (x-3) \cdot (x-2)$. Thus both matrices are diagonalizable with same eigenvalues, so there exist invertible matrices $S, T$ with $S^{-1}AS = T^{-1}BT$.

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