- #1
Dustinsfl
- 2,281
- 5
Let A be a diagonalizable matrix whose eigenvalues are all either 1 or -1. Show that [tex]A^{-1}=A[/tex].
[tex]A=X\begin{bmatrix}
\pm 1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \pm 1
\end{bmatrix}X^{-1}
[/tex] and [tex]A^{-1}=X\begin{bmatrix}
\pm 1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \pm 1
\end{bmatrix}^{-1}X^{-1}
[/tex]
How do I show they are equal?
[tex]A=X\begin{bmatrix}
\pm 1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \pm 1
\end{bmatrix}X^{-1}
[/tex] and [tex]A^{-1}=X\begin{bmatrix}
\pm 1 & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & \pm 1
\end{bmatrix}^{-1}X^{-1}
[/tex]
How do I show they are equal?