The aim is to prove that the diagonal entries of a positive definite matrix cannot be smaller than any of the eigenvalues.
I know a positive definite matrix must have eigenvalues that are > 0, and that just because a matrix has all positive values, does not make it a positive definite matrix.
I've also looked at the wikipedia for Positive-definite matrices and understand the definition given there, but am having a hard time convincing myself that the diagonal entries have to be greater than the eigenvalues.
The starting point of the proof should be to consider $A−a_{ii}I$, where $A=A^T$, and A is the positive definite matrix.
Can anyone help push me in the right direction to complete the proof?
Best Answer
Hint:
It suffices to prove that $\min_{\|x\|=1} x^TAx = \lambda_1$ where $\lambda_1$ is the smallest eigenvalue.
Once you proved that, can you think of a $y$ of unit length such that $y^TAy=a_{ii}$? If so, then $$a_{ii}=y^TAy \geq \min_{\|x\|=1} x^TAx = \lambda_1$$
Remark: Eric Fisher is right that the current proposition is false. My current hint intend for you to prove that the diagonal entries cannot be smaller than the smallest eigenvalue.