Hermitian, positive definite matrices

[pyroknife]

Well I kind of did mean that, because as it turns out $J_kA = J_k^TAJ_k= \Delta_k^{mxm}$But wait! There's one last step in my proof, which is to find a J_k such that $J_k^TAJ_k= \Delta_k^{mxm}$

I'll just say it; J_k is the mxm matrix with the kxk identity matrix in the top left and zeros everywhere else. The transpose of J_k is still J_k, and multiplying A by J_k on either the left or right (or both) gives $\Delta_k^{mxm}$.

Hmmm, I said this earlier, but I tried it and it didn't work. Let me write it out.

Let's just assume $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1 & 1\\ 1& 1 & 1\\ \end{bmatrix}$

so $\Delta_k^{mxm} = \begin{bmatrix} 1 & 1 & 0\\ 1& 1 & 0\\ 0& 0 & 0\\ \end{bmatrix}$

But
$\begin{bmatrix} 1 & 0 & 0\\ 0& 1 & 0\\ 0& 0 & 0\\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 1\\ 1& 1 & 1\\ 1& 1 & 1\\ \end{bmatrix} \neq \Delta_k^{mxm}$

But if we multiply this matrix on both sides, then we get $\Delta_k^{mxm}$. So I don't think you can just multiply on one side?
So $J_k A \neq J_k^T A J_k$

 

[Nathanael]

Hmmm, I said this earlier, but I tried it and it didn't work. Let me write it out.
...
But if we multiply this matrix on both sides, then we get $\Delta_k^{mxm}$. So I don't think you can just multiply on one side?
So $J_k A \neq J_k^T A J_k$

Oh darn! You're right! Multiplying on the left J_kA says "get rid of the last m-k rows" and multiplying on the right AJ_k says "get rid of the last m-k columns." We need to do both in order to get $\Delta_k^{mxm}$! Sorry for any confusion, I was looking at that wrongly.

Luckily it doesn't change the proof.