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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.

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?

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    $\begingroup$ Try $x^TAx$ with $x$ having zeros at the "tail". $\endgroup$ – Friedrich Philipp Mar 26 '16 at 3:39
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The relation between the eigenvalues of $A$ and the eigenvalues of $\Delta_k$ is not trivial. You can prove that the eigenvalues of $\Delta_k$ are positive by using Cauchy's Interlacing Theorem.

But showing that $\Delta_k$ is positive definite is straightforward, as Friedrich mentioned: for nonzero $x\in \mathbb C^k$, you have $$x^t\Delta_kx=y^tAy>0,$$ where $y\in\mathbb C^m$ is the vector $(x,0)$ (that is, $x$ "completed with zeroes").

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