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?

  • 2
    $\begingroup$ Try $x^TAx$ with $x$ having zeros at the "tail". $\endgroup$ – Friedrich Philipp Mar 26 '16 at 3:39

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").


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.