# How to show, that a Hermitian matrix is positive definite, if all eigenvalues are positive.

I want to show, that a Hermitian matrix is positive definite, if all eigenvalues of the matrix are positive. And the other way round.

Also I wonder, if every Hermitian, strict diagonally dominant matrix is positive definite.

Thank you guys.

If $A$ is a positive definite Hermitian matrix, so that $A^{\dagger} = A$ and $\langle x, Ax \rangle > 0$ for all $x \ne 0$, where $\langle \cdot, \cdot \rangle$ is the Hermitian inner product, then there exists a unitary matrix $U$, $U^{\dagger} = U^{-1}$, diagonalizing $A$. Thus we may write

$U^{\dagger}AU = \Lambda = \text{diag} (\lambda_1, \lambda_2, . . ., \lambda_N), \tag{1}$

where $N$ is the size of $A, U$ and the $\lambda_i$ are the (real) eigenvalues of $A$. If some $\lambda_i \le 0$, then consider the column vector $e_i$ the components of which are $e_{ij} = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta; thus $e_i$ has $1$ for its $i$-th component and zeroes everywhere else. We clearly have

$\Lambda e_i = \lambda_i e_i, \tag{2}$

whence

$\langle e_i, \Lambda e_i \rangle = \langle e_i, \lambda_i e_i \rangle = \lambda_i \le 0. \tag{3}$

If, in accord with (1), we substitute $U^{\dagger}AU = \Lambda$ into (3), we see that

$\langle e_i, U^{\dagger}AU e_i \rangle = \langle e_i, \Lambda e_i \rangle = \lambda_i \le 0, \tag{4}$

so setting $x = Ue_i$ we have

$\langle x, Ax \rangle = \langle Ue_i, AU e_i \rangle = \langle e_i, U^{\dagger}AUe_i \rangle = \lambda_i \le 0, \tag{5}$

contradicting the hypothesis that $A$ is positive definite, i.e. $\langle x, Ax \rangle > 0$ for all nonzero $x$. Thus we must have $\lambda_i > 0$ for all $\lambda_i$.

Going the other way is just about as easy; if all the $\lambda_i > 0$, and $U, \Lambda$ are chosen as above, for any vector $x \ne 0$ set $y = U^{\dagger}x$; then

$x = Ix = (UU^{\dagger})x = U(U^{\dagger}x) = Uy, \tag{6}$

and we have

$\langle x, Ax \rangle = \langle Uy, AUy \rangle = \langle y, U^{\dagger}AUy \rangle = \langle y, \Lambda y \rangle, \tag{7}$

and expanding $y$ in the basis $e_i$, so that

$y = \sum_i y_i e_i, \tag{8}$

we find

$\langle y, \Lambda y \rangle = \sum \bar y_i y_i \lambda_i > 0, \tag{9}$

since $y = Ux \ne 0$ implies some $\bar y_i y_i > 0$. Now we have, from (7) and (9),

$\langle x, Ax \rangle = \langle y, \Lambda y \rangle > 0 \tag{10}$

for all $x \ne 0$, showing $A$ is positive definite. QED.

As for the other question, I don't know an answer at present; but if I were to go deeper into it, I might well first ask if strict diagonal dominance is preserved under unitary transformations and/or similar riddles.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• Hm, showing that an HPD matrix has positive eigenvalues takes just a one line argument :) – Algebraic Pavel Nov 10 '13 at 23:38
• @Algebraic Pavel: you mean as in $0 < \langle x, Hx \rangle = \langle x, \lambda x \rangle = \lambda \langle x, x \rangle$? – Robert Lewis Nov 10 '13 at 23:50
• essentially yes, but of course going through the spectral theorem is the easiest to cover both cases. Just to say: $A$ is HPD iff $\Lambda$ is HPD and then it's all about diagonal matrices. – Algebraic Pavel Nov 11 '13 at 0:03

If $A$ is HPD and $\lambda\in\mathbb{R}$ is its eigenvalue such that $Ax=\lambda x$ for some $x$ such that $x^*x=1$, then $0<x^*Ax=\lambda x^*x=\lambda$. Hence $\lambda$ is positive.

The other way around works with the spectral theorem for Hermitian matrices. Every Hermitian matrix $A$ can be unitarily transformed to a diagonal matrix $\Lambda$. It is not hard to check that if $\Lambda$ is HPD (that is, $A$ has positive eigenvalues) then $A$ is HPD (see the other answer).

Every Hermitian and strictly diagonally dominant matrix is positive definite. This is a consequence of the Gershgorin theorem which says that the spectrum is contained in the union of closed circles in the complex plane, where each circle is centered in the diagonal element and its radius given by the sum of the absolute values of the off-diagonal elements of the row. For $A=(a_{ij})_{i,j=1}^n$, it means that $$\Lambda(A)\subset\bigcup_{i=1}^n D_i,$$ where $$D_i=\{z\in\mathbb{C}: |a_{ii}-z|\leq\sum_{j=1}^n|a_{ij}|\}.$$ Since $A$ is diagonally dominant, all $D_i$'s are centered on the positive part of the real line and do not contain $0$. That means (since Hermitian matrices have real eigenvalues), that no eigenvalue of $A$ is $0$ or negative. Hence $A$ is HPD.

This also works for irreducible weakly diagonally dominant matrices, it's just not that easy.