# How is it shown that a Hermitian matrix will be positive definite?

The following Hermitian matrix is given for all $m \in \mathbb{Z}$:

$$A_m = \left( \begin{array}{cc} m & i \\ -i & m\\ \end{array} \right)$$

(i) Show that $A_1$ is not positive definite.

(ii) Show that $A_2$ is positive definite.

(iii) Determine all $m \in \mathbb{Z}$ such that $A_m$ is positive definite.

So my understanding is: when given a matrix, to check if it is positive definite, one calculates: $\overline{z}^{t}Az$. Would this always be the same as $z^{t}A\overline{z}$? (When I calculated it out, the answer seemed to be "no"). I got the former off wikipedia, but I think I've seen the later as well...

Using the first version, I get $\overline{z}^{t} A_{m}z = \left( \begin{array}{cc} m \overline{z_{1}}-i\overline{z_{2}}, & i\overline{z_{1}}+m\overline{z_{2}} \end{array} \right) \left(\begin{array}{c}z_{1} \\ z_{2} \end{array} \right) = mz_{1}\overline{z_{1}}-iz_{1}\overline{z_{2}}+i\overline{z_{1}}z_{2}+mz_{2}\overline{z_{2}}$ and since I'm examining whether they are positive definite, I think I should try to express as many terms as squared so I think I can write $=m|z_{1}|^{2}-iz_{1}\overline{z_{2}}+i\overline{z_{1}}z_{2}+m|z_{2}|^{2}$. Here is where I don't know what to do with regard to parts (ii) and (iii)... Is there some sort of way to factor or rearrange the terms to make a case for which $m\in \mathbb{Z}$ this expression must be $\geq 0$? I think I've managed part (i) since $m=1, 0 \neq z = (-1,-i) \Rightarrow 1-i(-1)(i)+i(-1)(-i)+1 = 1+i^{2}+i^{2}+1 = 0$. The problem is that I just got that through playing around and don't really know why it should be so...

Also, just to make sure, the same approach will work for a Hermitian matrix of higher order as well, right?

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"Would $\bar{z}^{t}Az$ always be the same as $z^{t}A\bar{z}$?" - nope. The conjugation and transposition cannot be separated here. – J. M. Apr 26 '11 at 18:02
Here's one way to make things more transparent: let $z$ have explicit complex components, like $(a+bi\quad c+di)^T$... – J. M. Apr 26 '11 at 18:06
To check if Siv's solution will apply to you: what definition of "positive definite" are you working with? – J. M. Apr 26 '11 at 18:22
@J.M. : To your previous comment, thank you for the tip! Does something like this look like the right direction? $z = (a+bi, c+di)^t \Rightarrow m(a^2 + b^2) -2(ad-bc) +m(c^2+d^2)$ and since $m(a^2 + b^2) -2(ad-bc) +m(c^2+d^2) \geq 0 \Leftrightarrow m \geq \frac{2(ad-bc)}{a^2 + b^2 +c^2 +d^2}$ – ghshtalt Apr 26 '11 at 18:24
@J.M: as to the definition of positive definite: I think it has so far just been $\alpha$ is positive definite $:\Leftrightarrow \forall v \in V, v\neq 0 : \alpha(v,v) \gt 0$... so pretty basic and we haven't covered how eigenvalues relate to that yet... – ghshtalt Apr 26 '11 at 18:28

You could try to prove (and use) the following criterion: $$\begin{pmatrix} a & b \\ \bar{b} & c \end{pmatrix}$$ is positive definite iff $a > 0$ and $ac - |b|^2 > 0$ (here $a,c$ are real and $b$ is complex).

Hint: apply to the vector $\begin{pmatrix} b \\ -a \end{pmatrix}$

This criterion can be generalized to higher dimensions: if $(a_{i,j})_{1 \leq i,j \leq n}$ is hermitian, it is positive definite iff for any $1 \leq k \leq n$, $\det \left( (a_{i,j})_{1 \leq i,j \leq k} \right) > 0$. To prove it, you need to show that the maximal subspaces on which a hermitian form is positive definite have the same dimension, and proceed by induction on $n$.

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Thank you very much, this is the sort of thing I was looking for. So a matrix is positive definite if it is hermitian, finite dimensional(is that what all that $k\leq n$ stuff means?), and the determinant is positive? Other question was about the hint, I tried it and got: $z = \left( \begin{array}{c} |i|^2 \\ -(1)(-i) \end{array}\right) \Rightarrow \overline{z}^{t}A_{1}z = 4$, did I misunderstand the inputs? – ghshtalt Apr 26 '11 at 21:49
No, you have to compute $n$ determinants of $n$ submatrices of $A$ (taking only the first $k$ rows and columns). – Plop Apr 26 '11 at 22:17
In your case $a=c=m$, so you can take $z = \begin{pmatrix} 1 \\ im \end{pmatrix}$. – Plop Apr 26 '11 at 22:19
A related generalization: if $M$ is the block matrix $\begin{pmatrix} A & B \\ -B & A \end{pmatrix}$, since the block matrix $U = \frac{1}{\sqrt{2}} \begin{pmatrix} I & -iI \\ I & iI \end{pmatrix}$ is unitary and $U M U^* = \begin{pmatrix} A + iB & 0 \\ 0 & A - iB \end{pmatrix}$, you see that $M$ is positive definite if and only if $A+iB$ and $A-iB$ both are. (When $A = m$ and $B = i$ are $1 \times 1$ scalar matrices the condition is that $m \pm 1$ must both be positive, ie, that $m > 1$. This is nonelementary enough that I'm not seriously trying to help the OP here; just having some fun.) – anon Apr 26 '11 at 23:23

In this case, you can compute the eigen values and find out for what $m$'s the eigen values are positive. (Caution: In general, if the matrix is positive definite the eigen values are positive. However it is not always the case that if all the eigen values are positive the matrix is positive definite. You need the geometric multiplicity of the eigen values should be equal to the algebraic multiplicity to conclude the matrix is positive definite (which happens to be true this case for $m >1$))

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I've the feeling this is a "mosquito-nuking" solution for his situation... – J. M. Apr 26 '11 at 18:12
thank you for this answer, it is good to know that this is a possibility, unfortunately I think this method is still a little too advanced. – ghshtalt Apr 26 '11 at 18:43
For a hermitian matrix $A$, if all its eigenvalues are positive it follows that $A$ is positive definite, right? Because you can always find an orthogonal basis consisting of eigenvectors of A, and therefore $A=P^{-1}D P = P^t D P$, with D diagonal and positive. Therefore, $x^t A x=(Px)^t D (Px)>0$ for $x\neq 0$. – Ronaldo Apr 26 '11 at 21:59
Sorry, the "transpose" has to be changed by "hermitian conjugate" in my comment. – Ronaldo Apr 26 '11 at 22:01

The trace of you matrix is 2m and its determinant is m^2-1 thus you just need you m to be 1)strictly positive 2)m^2>1 thus Am is definite positive for any m integer greater than or equal 2

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