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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?

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  • $\begingroup$ 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. $\endgroup$ Dec 29, 2022 at 20:14

5 Answers 5

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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.

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

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  • $\begingroup$ 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$ $\endgroup$ Dec 30, 2022 at 11:19
  • $\begingroup$ $\lambda$ 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. $\endgroup$
    – Sunay
    Dec 30, 2022 at 16:11
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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.

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

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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.

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