How would I show that the following two matrices are similar
$A=
\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}\quad$ and $\quad B =
\begin{bmatrix}
-1 & 4 \\
-1 & 3
\end{bmatrix}$
I know they have same eigenvalue
But I am not sure how to show it. I am not sure how to find S so that $B=S^{-1}AS$ I have not learned the jordan canoical form or whatever it is called.
Best Answer
You do not need Jordan normal form. You only need linear equations and the determinant. Let $S=\begin{pmatrix} s_1 & s_3 \cr s_2 & s_4\end{pmatrix}$. Then $SA=BS$ is equivalent to the set of linear equations $$ s_1 - 2s_2=0,\; s_1 + 2s_3 - 4s_4=0,\; s_2 + s_3 - 2s_4=0. $$ Solve this over $K$, and do not forget to check $\det(S)=s_1s_4-s_2s_3\neq 0$. If there exists a solution, then $SA=BS$, i.e., $B=S^{-1}AS$, i.e., $A\sim B$.