Explain why the inertia of a Hermitian matrix is uniquely determined by its rank and signature

Question

Explain why the inertia of a Hermitian matrix is uniquely determined by its rank and signature

Inertia of hermitian matrix $H$ denoted as $ i(A) = (i_+, i_{-}, i_0)$

Signature $sig(A)=i_+ - i_-$

The book doesn't metions explicity but I am guessing we are counting multiplicities as well, that means that if a $9\times 9$ matrix is given having eigenvalues $2,2,0,0,0,-1,-3,-3,-3$

then $i(A)= (2,4,3)$ (please correct me if I am wrong)

I know that for hermitian $A$, $i_0 =0$

Also $A$ is unitarily diagonalisable, so the rank of $A$,

$$r(A) = i_+ + i_- $$

$$ sig(A) = i_+ - i_- $$

so I get

$$i_+ =\frac{r(A)+sig(A)}{2}$$

$$i_- = \frac{r(A)-sig(A)}{2}$$