Eigenvalues of $\mathrm{adj}(A)$

Question

If A is a singular square matrix , what are the eigenvalue of $\mathrm{adj}(A)$ ?

My attempt:

I know that if $\mathrm{rank}(A) < n- 1$ then $\mathrm{adj}(A) = 0$. I think that the eigenvalue of $\mathrm{adj}(A)$ will $0$.

Is its correct ?

Any hints/solution will be aappreciated ..

Answer 1

As this blog post points out, we have $\operatorname{adj}{(U^{-1}AU)} = U^{-1}(\operatorname{adj}(A)) U$ for any invertible $U$. Since any matrix $A$ can be reduced to a triangular one $B$ by such a transformation, $B=U^{-1}AU$, we have $$ \operatorname{adj}(B) = U^{-1}(\operatorname{adj}(A)) U. $$ The diagonal elements of $B$ are $\lambda_i$, the eigenvalues of $A$, and a calculation shows that the adjugate of an upper triangular matrix is also upper triangular, and the diagonal elements are $\prod_{j \neq i} \lambda_j$.

So the eigenvalues of $\operatorname{adj}(A)$ are the same as those of $\operatorname{adj}(B)$, which are the diagonal elements, i.e. $\prod_{j \neq i} \lambda_j$. Of course, this means that if $\dim\ker{A}>1$, $\operatorname{adj}(A) = 0$, while if $\dim\ker{A} = 1$, $\operatorname{adj}(A)$ has exactly one nonzero eigenvalue, the product of the others.

Answer 2

The following is just a collection of direct properties:

If $\mathrm{rank}(A)<n-1$, then the only eigenvalue of $\mathrm{adj}(A)$ is $0$ as $\mathrm{adj}(A)=\mathbf0$.

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If on the other hand $\mathrm{rank}(A)=n-1$, then $0$ is again an eigenvalue of $\mathrm{adj}(A)$ as $\mathrm{rank}(\mathrm{adj}(A))=1$, i.e. $\mathrm{dim}(\mathrm{ker}(\mathrm{adj}(A)))=n-1$ by the rank-nullity theorem. Note, that $0$ thus has geometric multiplicity $n-1$ instead of $n$ as before.

Note that this does not mean that there is not more to the $\mathrm{rank}(A)=n-1$-case.