I have a complex square matrix with the only eigenvalues being $1$ and $-1$. Show that $A$ and the inverse of this matrix $A^{-1}$ are similar

Question

I was thinking I use the formula for Jordan blocks I have $J=P^{-1} AP$ where

$$J= \left[\begin{array}{rr} 1 & 0 \\0 &-1 \end{array}\right]$$

Then $J^{-1}=P A^{-1}P^{-1}$

I can note $J=J^{-1}$

Then could I set the two sides equal? As in

$P^{-1} AP=P A^{-1}P^{-1}$ when simplifying $A=A^{-1}$

Thus proving similarity? I'm not entirely sure if my logic here is sound however.

Assuming $A$ is $2\times 2$, then I think that's basically the idea. You could also note that since $A$ has two distinct eigenvalues it is diagonalizable. This would be pretty much the same proof, but you wouldn't have to leverage the Jordan Form (which is strong result). — Trevor Norton, Mar 29 '17 at 17:11

Nathanael Skrepek · Answer 1 · 2017-03-29T17:17:49.847

2

If you regard your basis of eigenvectors $(v_i)_{i=1}^n$ then you know \begin{align*} Av_i = \pm v_i\quad \text{and} \quad AAv_i = v_i \end{align*} Now you know every $x\in \mathbb{C}^n$ can be written as $\sum_{i=1}^n x_i v_i$ for certain $x_i \in \mathbb{C}$. By using the linearity of the matrix multiplication you receive \begin{align*} AAx = AA\sum_{i=1}^n x_i v_i = \sum_{i=1}^n x_i AAv_i = \sum_{i=1}^n x_i v_i = x \end{align*} So $A$ is its own inverse.

edited Mar 29 '17 at 17:17

answered Mar 29 '17 at 17:10

Nathanael Skrepek

2,558

1

Can you apply your result to $A=\begin{bmatrix}1 & 1\0 & 1\end{bmatrix}$? – A.Γ. Mar 29 '17 at 19:28
How do you know that $A$ is diagonalizable? – amd Mar 29 '17 at 19:34
@amd you are right, I don't. I assumed that it is and every eigenvalue is $\pm 1$. I will think about what follows when there are generalized eigenvectors. – Nathanael Skrepek Mar 30 '17 at 08:52

I have a complex square matrix with the only eigenvalues being $1$ and $-1$. Show that $A$ and the inverse of this matrix $A^{-1}$ are similar

1 Answers1