Does the lowest diagonal element of a real symmetric matrix form an upper bound to the lowest eigenvalue?

If I have a real symmetric matrix, is it possible to look at the lowest diagonal element and then claim that the lowest eigenvalue of the matrix must be less than or equal to that diagonal element? I think this may be the case, so if it is, is this a well known theorem?

Gerschgorin's disk theorem doesn't appear useful to me, as the disk radius extends both ways in both the positive and negative direction from the diagonal.

FWIW, in quantum chemistry we have the Hyleraas-Undheim-MacDonald theorem, which bounds eigenvalues this way, but I'm not sure if this applies to any real symmetric matrix.

For a real symmetric matrix $n\times n$ matrix $A$, consider the Rayleigh quotient $$R_A(u) = \frac{u^\top A\,u}{u^\top u}\quad\text{for}\quad u\in\mathbb{R}^n\setminus\{0\}$$ Every eigenvector $v$ is a stationary point of that quotient, that is, $$\forall i=1,\ldots,n: \left.\frac{\partial R_A}{\partial u_i}\right|_{u=v} = 0$$ and all stationary points are eigenvectors. (Exercise: Prove that.) The eigenvalue corresponding to such eigenvector $v$ is $$\lambda = R_A(v)$$ In particular, the minimum quotient corresponds to the minimum eigenvalue: $$\lambda_{\text{min}} = \min_{u\in\mathbb{R}^n\setminus\{0\}} R_A(u)$$ Now suppose the main diagonal element $a_{ii}$ of $A$ is the minimum of all main diagonal elements. Let $e_i$ be the corresponding basis vector, then: $$\lambda_{\text{min}}\leq R_A(e_i) = a_{ii}$$ and there we are.
Also see Horn's theorem. It says that two real $n$ vectors $\bf{x}$, $\bf{y}$ are the eigenvalues and the diagonal of a (real) symmetric matrix if and only if $\bf{x}\succ \bf{y}$ in the Schur order, that is \begin{eqnarray}x'_1 &\le &y'_1\\ x'_1 + x'_2 &\le& y'_1 + y'_2 \\ \ldots \\ x'_1 + \ldots + x'_n &=& y'_1 + \ldots + y'_n \end{eqnarray} where $\bf{x}'$ is the increasing rearrangement of $\bf{x}$ and same for $\bf{y}$' of $\bf{y}$.
Note that the only if part is not too hard, using reduction to principal axes, which shows that $\bf{y}$ is a mixture of $\bf{x}$ ( a sequence with all entries equal is small).
• @ccorn: Great call! Oh, I think I flipped the Schur order, since $\bf{y}$ is a mixture of $\bf{x}$, I'll correct that. – orangeskid Jul 1 '15 at 23:02