Prove diagonal entries of positive definite matrices cannot be smaller than the eigenvalues 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?

Proof by contradiction:
We know that, if $\lambda$ is an eigenvalue of $A$, then $\lambda - p$ is an eigenvalue of $A-pI$.
So if $\lambda$ is an eigenvalue of $A$, then $\lambda - a{_i}{_i}$ is an eigenvalue of $A-a{_i}{_i}I$. Now, if $a{_i}{_i}$ is smaller than all the eigenvalues of $A$, then each $\lambda - a{_i}{_i}$ is positive and that makes $A-a{_i}{_i}I$ a positive definite matrix.
But $A-a{_i}{_i}I$ contains $0$ as its diagonal element on row $i$. So $A-a{_i}{_i}I$ cannot be positive definite and so $a{_i}{_i}$ cannot be smaller than all the eigenvalues of $A$
