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?
 A: The proposition as stated is false.  Consider the positive definite matrix
$$
A=\left[ \begin{array}{cc}
1 & 0 \\
0 & 2 \\
\end{array}\right]
$$
It has eigenvalues $1$ and $2$.  It has diagonal elements $1$ and $2$.  Certainly,
$a_{11} = 1 < 2 = \lambda_2$,
the eigenvalue corresponding to the eigenvector $[0~1]^T$.  
One can prove that it's impossible for every eigenvalue to be smaller than every diagonal element.
A: The most general statement I know in this direction is that the diagonal elements of a symmetric matrix are majorized by its eigenvalues. We say that a vector $x \in \mathbb{R}^n$ is majorized by a vector $y \in \mathbb{R}^n$ if, for every $i$, $1 \le i \le n$, 
$$
x_{(1)} + \ldots + x_{(i)} \le y_{(1)} + \ldots + y_{(i)}, 
$$
where $x_{(i)}$ is the $i$-th largest entry of $x$, and similarly for $y_{(i)}$, and $x_1 + \ldots + x_n = y_1 + \ldots + y_n$. This is written as $x \prec y$. An equivalent characterization of majorization is that $x \prec y$ if and only if $x$ is a convex combination of vectors obtained by permuting coordinates of $y$.
Let $a$ be the vector of diagonal entries of the matrix $A$, and $\lambda$ the vector of its eigenvalues. Then $a \prec \lambda$, and, in particular, $a$ is a convex combination of permutations of $\lambda$. This strengthens the following facts:


*

*the largest diagonal entry is at most the largest eigenvalues

*the smallest diagonal entry is at least the smallest eigenvalue

*the sum of diagonal entries is equal to the sum of the eigenvalues


To prove that $a \prec \lambda$, you can use the following formula, valid for any $1 \le k \le n$:
$$
\lambda_{(1)} + \ldots + \lambda_{(k)} = \max\mathrm{Tr}(U^T A U),
$$
where the maximum is over $n \times k$ matrices $U$ such that $U^T U = I_k$, the $k\times k$ identity matrix. 
In fact, the Schur-Horn theorem shows that $a$ is the vector of diagonal entries of a matrix with eigenvalues $\lambda$ if and only if $a \prec \lambda$. So, in this sense, there is nothing more you can say than this.
A: 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.
A: 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 $A$
