# 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?

• The eigenvalues of $A^2$ are 1 and 4. If $v$ is an eigenvactor of $A$ corresponding to $-2$, then it is an eigenvector of $A^2$ corresponding to $4$, and you can also check that it is an eigenvector of $A-A^2$ corresponding to $-6$. Do something similar for the second eigemvalue to verify $A-A^2$ does not have $0$ as an eigenvalue. Dec 29, 2022 at 20:14

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.

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 $$\begin{bmatrix}-2,\,0\\0,\,-6\end{bmatrix}^{-1}=\begin{bmatrix}-1/2,\,0\\0,\,-1/6\end{bmatrix}$$.

• You mean like when $\lambda = 0$ in $p(0) = (0 + 1) (0 + 2) = 2$ this tells us that the matrix has the determinant 2 aka nonzero, right since we don't take the eigenvalue into account since it is equal to zero? I don't understand why we need to check if determinant is nonzero for $\lambda = 1$. Also I don't understand how we can conclude that $det (A - A^2) = det A$ Dec 30, 2022 at 11:19
• $\lambda$ is an eigenvalue of $A$ if and only if $\det(A-\lambda I)=0$. Since, $1$ and $0$ are not eigenvalues then this determinant is nonzero for $\lambda=0$ $(\det(A) \neq 0)$ and $\lambda=1$ $(\det(A-I) \neq 0)$. Then $A$ and $A-I$ are both invertible. For any two matrix $A,B$, $\det(AB)=\det(A)\det(B)$. So, we can see $\det(A-A^2) = \det(A(I-A)) = \det(A)\det(I-A)$ which is nonzero because as we show both matrices are invertible. Dec 30, 2022 at 16:11

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.

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$$B^{-1}=-\frac1{12}(B+8I).$$

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.