How to show two matrices are similar

Question

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.

If you know that $A=S^{-1}BS$, then you are done. Otherwise write $SA=BS$ with $S=(s_{ij})$ and compute directly. You have to say, furthermore, over which domain. — Dietrich Burde, Oct 02 '16 at 19:51
Two matrices $A$ and $B$ are said to be similar if there exists a nonsingular matrix $S$ such that $A=S^{-1} B S$. So, if you already know that... — parsiad, Oct 02 '16 at 19:51
It means there is a change of basis from one matrix to the other. So they are similar if they represent the same linear map under a different basis. This condition is presented exactly as the other comments described. — IAmNoOne, Oct 02 '16 at 19:55

score 1 · Accepted Answer · answered Oct 02 '16 at 19:58

1

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., $A\sim B$.

answered Oct 02 '16 at 19:58

Dietrich Burde

130,978

Hmm So $SA=BS$ and you multipy both side by $SAS^{-1}$= $BSS^{-1}$ – Fernando Martinez Oct 02 '16 at 20:52
I thinkI also have to solve for $2s_1-4c_2=0$ – Fernando Martinez Oct 02 '16 at 21:00
Yes, but $2s_1-4s_2=0$ is the same equation as $s_1-2s_2=0$, if $2\neq 0$ in the field. – Dietrich Burde Oct 03 '16 at 08:11

How to show two matrices are similar

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