I'm confused about finding the inverse of a matrix with limited information.
a) Show that the matrix $\begin{pmatrix} 9 & 16 & -4\\
-4 & -7 & 2\\
4 & 8 & -1 \end{pmatrix}$ is diagonalizable, and diagonalize it. You may use use the fact that the eigenvalues of $A$ are $-1$ and $1$.
I already did this. $P=\begin{pmatrix} 2 & -2 & 1\\
-1 & 1 & 0\\
1 & 0 & 2 \end{pmatrix}$ and $D=\begin{pmatrix} -1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1 \end{pmatrix}$
b) Without doing any further numerical calculations, show that $A$ is invertible and find its inverse. Explain your answer.
I know the matrix is invertible since none of the eigenvalues are zero. How do I find the inverse though?
I attempted the following.
I know for eigenvalues, $Ax=\lambda x$
For $\lambda=1$, we get $Ax=x$ which implies that $Ax-x=0$.
This furthermore implies that $x(A-I)=0$. Since 0 is never an eigenvector, we must have that $A=I$. If I invert both sides, I get $A^{-1}=I^{-1}$ or $A^{-1}=I$ which tells use that $A=A^{-1}$.
For $\lambda=-1$, we get $Ax=-x$ which implies that $Ax+x=0$ or $x(A+I)=0$. This implies that $A=-I$. Similarly, $A^{-1}=-I^{-1}$ or $A^{-1}=-I$ which also implies that $A^{-1}=A$.
Is my logic correct? If it's not, where did I go wrong and how can I do this? Thanks!
Best Answer
It can be much simpler: we know that $D=P^{-1}AP$, which is equivalent to $A=PDP^{-1}$.
Now, as $D$ is diagonal, it is obvious that $D^2=I$, whence $$A^2=(PDP^{-1})(PDP^{-1})=PD^2P^{-1}=PP^{-1}=I.$$