How is the rank of $A$ related to the characteristic polynomial of the adjugate matrix of $A$?

Question

I am trying to solve the following exercise.

Prove that if the matrix $A$ is positive semidefinite, then its adjugate matrix $A^{*}$ is also positive semidefinite.

The proof comes to that, if $A$ is not full rank (the full rank case is trivial), then rank of $A^{*}$ is less or equal to 1, and then the characteristic polynomial is

$$ | \lambda I - A^{*} | = \lambda^{n} - (A_{11} + A_{22} + \cdots + A_{nn}) \lambda^{n-1} $$

and the eigenvalue would be zero or positive, and thus finish the proof. I wonder why this is true? How can I get the formula for the characteristic polynomial of the adjugate matrix?

Answer 1

Any $n\times n$ matrix $A$ with rank $1$ or less must have a kernel of dimension equal to $n-1$ or $n$. This means that the dimension of the eigenspace of the eigenvalue $0$ must be at least $n-1$. Since the geometric multiplicity is bounded by the algebraic multiplicity, we must have that $0$ has an algebraic multiplicity of at least $n-1$ in the characteristic polynomial.

This gives us the form $\lambda^n+a_{n-1}\lambda^{n-1}$ for the characteristic polynomial. The fact that $a_{n-1}=-Tr(A)$ follows from the expansion of $|A-\lambda I|$.

Answer 2

Once A is square, $$\sum_iA_{ii}=tr(A)=\sum_i\mu_i$$ where $\mu$ are eigenvalues of A.

Once A is positive semi-definite, $\sum_i\mu_i \ge 0$

Now we have $$\lambda^n=tr(A)\lambda^{n-1}$$

from here it follows that both $\lambda^n$ and $\lambda^{n-1}$ are non-negative.