Matrix $A^2+I_4$ is singular if and only if there exists a nonzero matrix $B$ in $M_{4}(\mathbb{R})$ so that $AB+BA=0$. Let $A$ be an invertible matrix in $M_{4}(\mathbb{R})$ such that $\mbox{ Tr }(A)=\mbox{ Tr }(\mbox{ adj }A)\neq0$. Prove that the matrix $A^2+I_4$ is singular if and only if there exists a nonzero matrix $B$ in $M_{4}(\mathbb{R})$ so that $AB+BA=0$.
How to proceed with this problem. I am trying to write $\det\,(A+ix)$ as a polynomial in $x$. But how to do this.
 A: Suppose $AB+BA=0$. Then the linear map $L:X\mapsto AX+XA$ is singular. Since the eigenvalues of $L$ are $\left\{\lambda_i(A)+\lambda_j(A):\ i,j\in\{1,2,3,4\}\right\}$, we must have $\lambda_i(A)+\lambda_j(A)=0$ for some $i$ and $j$. As $A$ is supposed to be invertible, such $i$ and $j$ must be unequal and $\lambda_i(A)=-\lambda_j(A)\ne0$.
In other words, we may assume that the spectrum of $A$ over $\mathbb C$ is in the form of $\{a,-a,b,c\}$. The traces of $A$ and $\operatorname{adj}(A)$ are therefore $b+c$ and $\det(A)\left(\frac1{a}+\frac1{-a}+\frac1b+\frac1c\right)=-a^2(b+c)$ respectively. By assumption, they are equal and nonzero. Hence $-a^2=1$. As a result, $A^2+I$ is singular.
Conversely, suppose $A^2+I_4$ is singular. Then $\pm i$ are two complex eigenvalues of $A$. Therefore the real Jordan form of $A$ must be equal to either $\pmatrix{R&0\\ 0&\ast}$ or $\pmatrix{R&I_2\\ 0&R}$, where $R=\pmatrix{0&-1\\ 1&0}$. However, the latter case is impossible because $A$ is assumed to have nonzero trace. Thus up to similarity we may assume that $A=\pmatrix{R&0\\ 0&\ast}$ and in this case, $AB+BA=0$ when $B=\operatorname{diag}(1,-1,0,0)$.
