1. ## Positive-definite Hermitian matrix

Hi,

For any Hermitian matrix T,

z* T z is real and is >= 0. The * denotes conjugate transpose.

I understand that if any of the eigenvalues of T are equal to zero, then T is positive semi-definite (z* T z >= 0).

What I don't get is why T is positive definite (z* T z > 0) iff all of it's eigen values are > 0; even if all eigenvalues are > 0, couldn't there still be a z which leads to semi-definiteness?

Thanks.

2. ## Re: Positive-definite Hermitian matrix

No, since $\displaystyle T$ is Hermitian it is diagonalizable, and if $\displaystyle z^*Tz=0$ then $\displaystyle z=0$.

3. ## Re: Positive-definite Hermitian matrix

Thanks.

Could you please relate the diagonalizability to the part that if z* T z = 0 then z = 0?

4. ## Re: Positive-definite Hermitian matrix

We can write $\displaystyle T=P^*DP$ where $\displaystyle P$ is unitary and $\displaystyle D$ diagonal, containing the eigenvalues of $\displaystyle T$. If $\displaystyle z^*Tz=0$ then $\displaystyle (Pz)^*T(Pz)=0$, but writing $\displaystyle v=Pz$ we have $\displaystyle v^*Tv\geq \min_{1\leq j\leq d}\lambda_j\lVert v\rVert$ so $\displaystyle v=0$ and $\displaystyle z=0$.