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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.

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  • $\begingroup$ @amd corrected. Pls check now $\endgroup$ – Magneto Aug 22 '18 at 20:03
  • $\begingroup$ Are you sure adj means adjugate and not adjoint (i.e. conjugate transpose)? $\endgroup$ – Ian Aug 22 '18 at 20:14
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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.

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  • $\begingroup$ madam in my text book, adjoint matrix is given as adjugate matrix in ur explanation. If it is complex conjugate everything goes straightforward. Wiki also says adjoint or some times adjugate. Can u pls elaborate? $\endgroup$ – Magneto Aug 22 '18 at 21:26
  • $\begingroup$ @kusuma pls see here -- nptel.ac.in/courses/122104018/node29.html $\endgroup$ – Magneto Aug 23 '18 at 5:37
  • $\begingroup$ Oh, I see. Your text is using slightly older terminology. The word "adjoint" also sometimes means what I called the "adjugate", but this is not common in recent publications. See also en.m.wikipedia.org/wiki/Adjugate_matrix (the introduction mentions different terminology). $\endgroup$ – Kusma Aug 23 '18 at 5:52
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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.

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  • $\begingroup$ Sir in my text book, adjoint matrix is given as adjugate matrix in ur explanation. If it is complex conjugate everything goes straightforward. Wiki also says adjoint or some times adjugate. Can u pls elaborate? $\endgroup$ – Magneto Aug 22 '18 at 21:26
  • $\begingroup$ @Magneto What language is the book written in? $\endgroup$ – egreg Aug 22 '18 at 21:37
  • $\begingroup$ english......... $\endgroup$ – Magneto Aug 23 '18 at 4:31
  • $\begingroup$ nptel.ac.in/courses/122104018/node29.html Pls see this here. $\endgroup$ – Magneto Aug 23 '18 at 5:36
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    $\begingroup$ @Magneto The page call adjoint what's commonly called adjugate. The product $A\operatorname{adj}(A)$ is always a diagonal matrix, where the diagonal has entries equal to $\det A$; so its only eigenvalue is $\det A$. The exercise makes little sense if this adjoint/adjugate is used; more significant, as I said, is to consider the adjoint also called Hermitian transpose. $\endgroup$ – egreg Aug 23 '18 at 8:06

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