# Hermitian, positive definite matrices

1. Mar 26, 2016

### pyroknife

1. The problem statement, all variables and given/known data
I am trying to prove the following:
if $A\in C^{m\ \text{x}\ m}$ is hermitian with positive definite eigenvalues, then A is positive definite. This was a fairly easy proof. The next part wants me to prove if A is positive definite, then $\Delta_k$=\begin{bmatrix} a_{11} & \cdots & a_{1k} \\ \vdots & \ddots & \vdots\\ a_{k1} & \cdots & a_{kk} \end{bmatrix}

is also positive definite. k = 1...m

2. Relevant equations

3. The attempt at a solution
Since the first part I already proved if some matrix is hermitian with positive eigen values then that matrix is positive definite. I basically need to prove the $\Delta_k$ is hermitian with positive eigenvalues, therefore its positive definite.

It is obvious that $\Delta_k$ is hermitian since it is simply just a sub matrix of A, but how do I go about proving that this matrix also has positive eigenvalues? Any hints? Seems like it should be obvious, but I can't think of how to prove it.

2. Mar 26, 2016

### Orodruin

Staff Emeritus
How does $\Delta_k$ act on vectors outside of the k-dimensional subspace spanned by the first k unit vectors? (If you use it as the upper left block of an mxm matrix which is otherwise zero)

3. Mar 26, 2016

### pyroknife

Hmm. Does this involve eigenvalues?

Or are you suggesting that I check $x^T \Delta_k x$

4. Mar 26, 2016

### Ray Vickson

Well that is exactly the DEFINITION of positive-definite.

5. Mar 26, 2016

### pyroknife

Yes. I was thinking of just proving the eigenvalues are positive, based on the first part of the problem. If I think $x^T \Delta_k x$, this would not involve eigenvalues in any way, right?

6. Mar 26, 2016

### Ray Vickson

Well, the eigenvalues of a $k \times k$ submatrix may be different from any of those of the original matrix, so I cannot see how using eigenvalues will help; you no longer know they are all > 0!

7. Mar 26, 2016

### pyroknife

Hmm. I see. I thought there was some sort of relation between the eigen values of the original matrix and a sub-top-left matrix. I guess not.
If I do $x^T \Delta_k x$, is there a simple way to show this for a generic $\Delta_k$ matrix? I can write out the vector-matrix multiplication & summing them up to show its >0, but that seems tedious. I am wondering if there's a much simpler way to show this.

8. Mar 26, 2016

### Ray Vickson

There might be such eia relation, but I have never heard of one, and I doubt it would be easy to find even if it exists.

I honestly do not understand your difficulty; it all follows in one line and with almost zero work from the definition of positive-definiteness of the original matrix $A$.

9. Mar 26, 2016

### pyroknife

Hmm, I am not seeing this.
Since A is positive definite, that means:
$x^T A x > 0$ for all $x\neq 0$
But I am not sure how this would relate to $\Delta_k$. There's something I'm not visualizing. Are you looking at this from an element by element multiplication perspective?

10. Mar 27, 2016

### Nathanael

@pyroknife Have you considered Orodruins hint?

Think about the mxm matrix which is all zeros except the upper left corner is Δk ... how is this matrix related to A?

11. Mar 27, 2016

### pyroknife

The matrix with all zeros except the upper left corner is $\Delta_k$ is essentially A, when k = m. Should I visualizing the submatrices $\Delta_k$ a matrix of zeros except with the upper left corner as $\Delta_k$?

12. Mar 27, 2016

### Nathanael

Let me use $\Delta_k^{mxm}$ to represent the mxm matrix of all zeros except the upper left corner is Δk (just so I can reference it easily).
Ok, true. But in general, when k ≠ m, how do you get from A to $\Delta_k^{mxm}$?
What I mean specifically is, what matrix takes you from A to $\Delta_k^{mxm}$?

13. Mar 27, 2016

### pyroknife

Oh. hmm, let's see. if k = m, then the matrix would just be identity.
If k < m, then this gets a little complicated. Let me see if I can get a relationship for this matrix. This matrix basically needs to introduce m-k columns&rows of zeros at the end of $\Delta_k^{mxm}$

14. Mar 27, 2016

### Ray Vickson

You want to know if the quadratic form
$$Q_k(x_1, x_2, \ldots, x_k) = (x_1, x_2, \ldots, x_k) \pmatrix{a_{11} & a_{12}& \cdots & a_{ik} \\ a_{21} & a_{22} & \cdots & a_{2k}\\ \vdots & \vdots & \ddots & \vdots \\ a_{k1} & a_{k2} & \cdots & a_{kk} } \pmatrix{x_1 \\ x_2 \\ \vdots \\x_k}$$
is always positive for $(x_1, x_2, \ldots, x_k) \neq (0,0,\ldots, 0)$.

15. Mar 27, 2016

### pyroknife

Yes exactly. But it seems to show that the quadratic form is > 0, I'd need to carry out all the matrix multiplications and verify this is > 0?

16. Mar 27, 2016

### pyroknife

I have already showed that A is positive definite (not in this thread, but let's just assume I showed you guys).
Which means

$Q_m(x_1, x_2, \ldots, x_m) = (x_1, x_2, \ldots, x_m) \pmatrix{a_{11} & a_{12}& \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m}\\ \vdots & \vdots & \ddots & \vdots \\ a_{k1} & a_{k2} & \cdots & a_{mm} } \pmatrix{x_1 \\ x_2 \\ \vdots \\x_m} > 0$

Now I need to show that if I take out (m-k) elements from $x$ and m-k rows&columns from A, then this still yields positive definite matrix.

17. Mar 27, 2016

### Nathanael

You shouldn't have to go too far from the identity matrix to find it; maybe put some zeros in the right spots?

Let me try to hint at the proof in another way:
If $\vec x ^TA\vec x >0$ for all $\vec x$ then in particular it must be true for all $\vec x$ of the form $\vec x^T = [x_1, ... , x_k, 0, ..., 0]$ right?
(Sorry I don't know how to write a column vector in latex.)

18. Mar 27, 2016

### pyroknife

Yes that is true. So we are one step closer, but the next part, if we add the (m-k) columns and rows of zeros, I am not sure why that would still yield greater than zero.
Edit: Never mind, I undertand that if we modify A and add the (m-k) columns and rows of zeros, it would still yield a positive definite matrix. $x^T \Delta_k^{mxm} x = x_k^T \Delta_k x_k$ is that correct?

where $x_k = \begin{bmatrix} x_1 \\ x_2 \\ \vdots\\ x_k \end{bmatrix}$

19. Mar 27, 2016

### pyroknife

I think the above proves it, but I'm still thinking about the matrix, when multiplied to A, yields zeros in the last (m-k) columns&rows. I was thinking it should just be a slight modification of the identity matrix, but I can't get it to work.

20. Mar 27, 2016

### Nathanael

Are you convinced that $\vec x^T\Delta_k^{mxm}\vec x>0$ for all $\vec x \in C^{m}$ implies that $\vec v^T\Delta_k\vec v>0$ for all $\vec v \in C^{k}$ ?
(This was the one part of the proof that I'm not sure of the best way to prove, but it seems clear enough.)

If so, then the goal becomes to show $\vec x^T\Delta_k^{mxm}\vec x>0$. This follows nicely from the positive-definiteness of A if you think about the matrix which takes you from A to $\Delta_k^{mxm}$

Now what is that matrix? I'm not sure how to hint at it without just saying it, but like I said it's similar to the identity matrix except with some extra zeros...