Show that $ A − A^2$ is invertible given $A$'s eigenvalues? Suppose that the $2 \times 2$ matrix $A$ has the characteristic polynomial $p(\lambda) = (\lambda + 1)(\lambda + 2).$ Show that $ A − A^2$ is invertible and determine the eigenvalues to the inverse.
So this is how I tried.
$p(\lambda)=0$ gives me that $\lambda_1 = -1$ and $\lambda_2 = -2$
This means that we have at least two linearly independent vectors which means that the matrix $A$ is diagonalizable.
So we have: $A = PDP^{-1}$
$A - A^2 = PDP^{-1} - PDP^{-1} PDP^{-1} = PDP^{-1} - PD^2 P^{-1} = P(D - D^2) P^{-1} $
$D = ([-1, 0]^T , [0, -2]^T)$
$D - D^2 = ([-2, 0]^T [0, -6]^T)$
But this all feels unnecessary and I feel lost. Am I even thinking right?
 A: The eigenvalues of $A$ are $-1$ and $-2$. Hence, the eigenvalues of $A^2$ are $1$ and $4$. Hence, the eigenvalues of $A - A^2$ are $-1 - 1 = -2$ and $-2 - 4 = -6$. Since the eigenvalues are non-zero, it is invertible.
A: Since $\det(A-\lambda I) \neq 0$ for $\lambda=1$ and $\lambda=0$ we can conclude $\det(A-A^2)=\det(A)\det(I-A)\neq 0$ so invertible. $A$ is similar to diagonal matrix $\begin{bmatrix}-1,\,0\\0,\,-2\end{bmatrix}$ then $A-A^2$ is similar to $\begin{bmatrix}-1,\,0\\0,\,-2\end{bmatrix}-\begin{bmatrix}1,\,0\\0,\,4\end{bmatrix}=\begin{bmatrix}-2,\,0\\0,\,-6\end{bmatrix}$. Then the inverse is similar to $\begin{bmatrix}-2,\,0\\0,\,-6\end{bmatrix}^{-1}=\begin{bmatrix}-1/2,\,0\\0,\,-1/6\end{bmatrix}$.
A: If $A$ is diagonalisable then $A= U D U^{-1}$ for some diagonal $D= \operatorname{diag}(\lambda_1,...,\lambda_n)$. If $p$ is a polynomial, it is straightforward to check that $p(A) = U p(D) U^{-1}$ and $A^{-1} = U D^{-1} U^{-1}$.
The point being that you just need to look at the eigenvalues of $A$ under the mappings $x \mapsto x-x^2$ and $x \mapsto {1 \over x-x^2}$ to answer the question.
A: By Cayley-Hamilton,$$\begin{align}B&:=A-A^2\\&=A-(-3A-2I)\\&=4A+2I.\end{align}$$
$$\begin{align}
B^2&=16A^2+16A+4I\\&=16(-3A-2I)+16A+4I\\&=-32A-28I\\&=-8B-12I
\end{align}$$
hence $B$ is invertible and$$B^{-1}=-\frac1{12}(B+8I).$$
A: Because $p(\lambda)=(\lambda+1)(\lambda+2)=\lambda^2+3\lambda+2$ is the characteristic polynomial of $A$, then $p(A)=0$ by the Cayley-Hamilton Theorem. Therefore,
\begin{align}
    p(\lambda)I&=p(\lambda)I-p(A) \\
            &= (\lambda^2+3\lambda+2)I-(A^2+3A+2I) \\
            &= (\lambda^2I-A^2)+3(\lambda I-A) \\
            &= (\lambda I-A)\{(\lambda I+A)+3I\}
\end{align}
Therefore, if $p(\lambda)\ne 0$ for a given $\lambda$, then $(\lambda I-A)$ is invertible with inverse
$$
      (\lambda I-A)^{-1}=\frac{1}{p(\lambda)}((\lambda+3)I+A)
$$
You want to invert $A-A^2=A(I-A)$, which is equivalent to inverting $A$ and $I-A$, and multiplying their inverses:
$$
    A^{-1}=-\frac{1}{p(0)}(A+3I)=-\frac{1}{2}(A+3I) \\
   (I-A)^{-1}=\frac{1}{p(1)}(A+4I)=\frac{1}{6}(A+4I)
$$
Therefore,
$$
  (A-A^2)^{-1}=-\frac{1}{12}(A+3I)(A+4I).
$$
I'll let you multiply that out in order to obtain the inverse.
