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

I am trying to solve an exercise to prove if $A$ is semi-positive definite matrix, then its adjugate matrix $A^{*}$ is also semi-positive definite.

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

• In my experience, positive matrices are generally assumed to be self-adjoint... – Eric Auld Aug 2 '13 at 16:14

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 $|A-\lambda I|$.

• sorry, I was still typing and did not notice your post. – giorgi Aug 2 '13 at 16:51

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.