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?
Best Answer
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 $|A-\lambda I|$.