Positive-definite Hermitian matrix

**algorithm** · April 4th 2012, 11:23 AM

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.

**girdav** · April 4th 2012, 11:26 AM

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

**algorithm** · April 4th 2012, 11:36 AM

Thanks.

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

**girdav** · April 4th 2012, 11:41 AM

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

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