Find out the characteristic and minimal polynomial of $vv^*$ where $v$ is a column vector $n \times 1$ column vector.

Question

Find the characteristic and minimal polynomial of $vv^*$ where $v \in \Bbb C^n$ is a column vector $n \times 1$ column vector.

Now considering $A=vv^*$ I have $A^2=cA$ where $c=v^*v$ Hence $x(x-c)$ is the minimal polynomial (unless $v$ is the zero vector) with eigenvalue $0,c$. Characteristic polynomial has only roots $0$ and $c$. Now what?

I am getting that it's a Hermitian matrix. Now in the search option also I am not getting any link. Please help!!

Add Okay, I think I got it. $A^n=cA^{n-1}$ so $A$ satisfies $x^{n-1}(x-c)$

Answer 1

One has to be very careful here; if $p$ is a monic polynomial of degree $n$ that annihilates $A$, then $p$ is not necessarily the characteristic polynomial of $A$.

For instance, if $$ A= \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, $$ then $p_A(t) = t^4$ yet the polynomial $q(t) = t^4 + t^2$ annihilates $A$.

For the problem above, notice that if $v \ne 0$, then $$ vv^* \sim \begin{bmatrix} \bar{v}_1 & \bar{v}_2 & \cdots & \bar{v}_n \\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix}, $$ where $\sim$ denotes row-equivalence. Since the first-row contains at least one nonzero value, it follows that zero is an eigenvalue and $\dim(E_0) = n-1$. Since $vv^*$ is Hermitian, there is a unitary matrix $U$ such that $U^*(vv^*)U = \text{diag}(v^*v,\underbrace{0,\dots,0}_{n-1})$. Thus, the characteristic polynomial is $t^{n-1}(t - c)$.