For a symmetric, positive definite matrix, why does each diagonal element exceeds the small eigenvalue? In the PDE book I am reading. It took for granted, $A$ is positive and symmetric
$a_{11}\geq \lambda$ where $\lambda$ is the smallest eigenvalue. (I initially wrote $>$, but that was wrong)
I have managed to show this is indeed the case for $2\times 2$ matrix. why is this true for an $n\times n$ matrix?
Proof for $2\times2$ matrices:
Let $\Lambda$ denote the larger eigenvalue. WLOG $a_{11}\geq a_{22}$, the determinant of the matrix is $a_{11}a_{22}-a_{12}^2>0$, this forces $a_{11}a_{22}>0$. Moreover by the trace rule $\lambda+\Lambda = a_{11}+a_{22}>0$, so
$\lambda\Lambda\leq a_{11}a_{22}$ and $\lambda+\Lambda=a_{11}+a_{22}$
squaring the 2nd relation and subtracting the first relation twice, we arrive at 
$$(\Lambda-\lambda)^2\geq (a_{11}-a_{22})^2$$ which implies
$$\Lambda -\lambda \geq a_{11}-a_{22}$$
since $\Lambda +\lambda =a_{11}+a_{22}$, we must have $\Lambda\geq a_{11}\geq a_{22}\geq \lambda$.
I cannot see how to extend this result for higher dimensions, essentially because I used 2 conditions and it is not sufficient to solve this problem in higher dimensions.
 A: It's because $A - \lambda I$ is still positive semi-definite, so all its main diagonal entries are non-negative.
(Extra detail requested: I assume that you are dealing with real matrices here. A real symmetric matrix has an orthonormal basis of eigenvectors, say $\{ v_{1},v_{2}, \ldots,v_{n}\},$ say with $Av_{i} = \beta_{i}v_{i}$ for each $i$ ( the $\beta_{i}$ are real and positive, but need not be distinct). Now if we choose a vector $v,$ and write $v = \sum_{i=1}^{n}\alpha_{i}v_{i},$ then we see that $\langle Av, v \rangle = \sum_{i=1}^{n}\beta_{i}|\alpha_{i}|^{2} \geq \lambda \langle v,v \rangle$ as $\langle v,v \rangle = \sum_{i=1}^{n}|\alpha_{i}|^{2}$ and $\beta_{i} \geq \lambda$ for each $i.$ 
Hence $\langle (A-\lambda I)v,v \rangle \geq 0.$ Hence $A- \lambda I$ is positive semi-definite, as $v$ was arbitrary.
Now the entry $b_{ii}$ of a symmetric matrix $B$ may be interpreted as $\langle Bu_{i}, u_{i} \rangle$ , where $\{u_{i} : 1 \leq i \leq n \}$ is the standard (orthonormal) basis of $\mathbb{R}^{n}$, viewed as column vectors under product $\langle x, y \rangle = x^{t}y.$ If $B$ is positive semidefinite, this is always non-negative. So we apply this to $A - \lambda I).$
A: The strict inequality is wrong. Counterexample: $\begin{bmatrix} 1\end{bmatrix}$.
An overkill is to use Cauchy's interlacing theorem with $m=1$ which tells you that given an hermitian matrix $A\in \mathcal M_{n\times n}(\mathbb C)$, where
$$A=\begin{bmatrix} a_{11} & \ldots & a_{1n}\\ \vdots & \ddots & \vdots \\a_{n1 } & \ldots & a_{nn}\end{bmatrix},$$
the eigenvalues $\alpha _1, \ldots ,\alpha _n$ of $A$ (with $\alpha _1 \leq \ldots \leq \alpha _n$) are such that $\alpha _1 \leq a_{11}\leq \alpha _n$.
A more pictorial version of the theorem can be found in Meyer's Matrix Analysis and Applied Linear Algebra. I leave the relevant part of the book below.


To use this version of the theorem you need to apply it iteratively to the principal submatrices of $A$ of smaller orders until you get to the $1\times 1$ case.
An alternative proof of this result can be found here.

As Geoff Robinson says in his answer, it suffices to note that $A-\lambda I$ is positive-semidefinite. This is a consequence of the fact that $A-\lambda I$ is still hermitian and in general the eigenvalues of $M+\mu I$ are the eigenvalues of $M$ summed with $\mu$.
