# Show $\begin{pmatrix} 5 & 1 \\ -6 & 0 \end{pmatrix}$ and $\begin{pmatrix} 7 & -5 \\ 4 & -2 \end{pmatrix}$ are similar. [closed]

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!

• Do you know about characteristic polynomial, eigenvalues, eigenvectors...? Commented Mar 25, 2016 at 22:23
• First you should check that trace and determinant agree. If so, then just solve your $CA = BC$ thing, it is really not that hard. Commented Mar 25, 2016 at 22:31

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 $S^{-1}AS = T^{-1}BT$.