Question about $P^{-1}AP=B$ equation

Question

$P^{-1}AP=\pmatrix{1&-\tfrac{1}{a}\\-\tfrac{1+a^2}{ab}&0}^{-1}\pmatrix{a&b\\-\tfrac{1+a^2}{b}&-a}\pmatrix{1&-\tfrac{1}{a}\\-\tfrac{1+a^2}{ab}&0}=\pmatrix{0&-1\\1&0}$

This statement is true. How would I find P if I would know only A and $\pmatrix{0&-1\\1&0}$? I need an easy way.

Comes from this Proving that every $2\times 2$ matrix $A$ with $A^2 = -I$ is similar to a given matrix.

I think you might be reading this incorrectly. The product of the three matrices on the LHS are equal to the RHS $\forall a, b$. In fact, the matrix on the RHS is a $90 \deg$ rotation matrix. — Amzoti, May 06 '13 at 21:07

score 1 · Accepted Answer · answered May 06 '13 at 20:30

1

Below is the most elementary way possible, but (arguably) not the easiest.

Let $P=\begin{pmatrix} x &y\\ z & w\end{pmatrix}$.

Then $P^{-1}AP=B\iff A\begin{pmatrix} x &y\\ z & w\end{pmatrix}=\begin{pmatrix} x &y\\ z & w\end{pmatrix}B$.

Now you just need to solve a linear system.

answered May 06 '13 at 20:30

Git Gud

31,356

Is there a faster method to get P? – May 06 '13 at 21:55
@David i thought I saw a faster way. But I don't anymore. – Git Gud May 06 '13 at 21:59

kafman · Answer 2 · 2013-05-06T20:41:53.967

1

May be this helps: the inverse of $P := \begin{pmatrix} u & x \\ v & y\end{pmatrix}$ is (if $\det(P) \neq 0$):

$$ P^{-1} = \frac{1}{uy-xv} \begin{pmatrix} y & -x \\ -v & u\end{pmatrix}$$.

Then multiply $P^{-1}AP$ and compare it element-wise with the given matrix on the right hand side of your equation. This yields a linear system of equations.

edited May 06 '13 at 20:41

answered May 06 '13 at 20:35

kafman

410
1
5
14

I've just seen the post of Git Gud, which is probably easier than mine as you don't have the fraction. – kafman May 06 '13 at 20:37
@GitGud You're right - thx! – kafman May 06 '13 at 20:42

Question about $P^{-1}AP=B$ equation

2 Answers2