Eigen values of $A. \operatorname{adj}(A)$ where $\operatorname{adj}(A)$ is adjoint matrix of A are real

Question

Eigen values of $A\cdot\operatorname{adj}(A)$ where $\operatorname{adj}(A)$ is the adjoint matrix of a non-singular matrix A are real.

This is a question asked in an exam. I have a doubt that info is incomplete.

Let $lambda$ be eigenvalue of A such that $Ax=\lambda x $

This gives,

$A \cdot\operatorname{adj}(A) x = |A| x$ as $\operatorname{adj}(A) = \frac{A^{-1}}{|A|}$

Let $B = A\cdot\operatorname{adj}(A)$ then $B$ is hermitian only if $B^{\theta} = B$. It means

B = $A\cdot\operatorname{adj}(A) = |A|I$, For $B^{\theta} = B$, $|A|$ has to be real. Only then $B=A\cdot\operatorname{adj}(A)$ is hermitian, which means given matrix has eigen values. But in exam I assumed this condition that $|A| $ is real. But is there any way to do this without assuming it as real.

Answer 1

I am not sure whether you mean "adjoint" or "adjugate". (1) Assume you mean "adjoint". Then if $A{\rm adj}(A)x=\lambda x$, we can write $$\lambda \left<x,x\right> =\left<x, A{\rm adj}(A)x\right>=\left<Ax, Ax \right>,$$ where $\left<\cdot,\cdot\right>$ is a Hermitian scalar product. As $\left<x,x\right>>0$ and $\left<Ax, Ax \right>\ge 0$ , $\lambda$ is real and nonnegative.

(2) Assume you mean "adjugate". Then $A{\rm adj}(A) = (\det A)I$, where $I$ is the identity matrix. Clearly the only eigenvalue is $\det A$. This has no reason to be real unless $A$ is real.

Answer 2

If adj denotes the adjugate, then $$ A\operatorname{adj}A=(\det A)I $$ (with $I$ the suitable identity matrix) is diagonal, so its only eigenvalue is $(\det A)$.

If $A$ is supposed to be a complex matrix, its determinant may be not real.

Your exercise is probably about $AA^*$, where $A^*$ is the adjoint of $A$, that is, the conjugate transpose. Since $AA^*$ is Hermitian, the eigenvalues are surely real.

Indeed, if $B=B^*$ (that is, $B$ is Hermitian) and $\lambda$ is an eigenvalue of $B$, then $Bv=\lambda v$ for some column vector $v\ne0$. Then $$ v^*Bv=v^*(\lambda v)=\lambda(v^*v) $$ On the other hand, $$ v^*Bv=v^*B^*v=(Bv)^*v=(\lambda v)^*v=\bar{\lambda}v^*v $$ Since $v^*v\ne0$, we conclude $\lambda=\bar{\lambda}$ is real.