I have a following question:

Let $A \in C^{n\times n}$ be Hermitian and $\lambda_\min$ be the smallest eigenvalue of $A$, i.e., $\lambda_\min = \min\{\lambda_1, \ldots, \lambda_n\}$. Show that $\lambda_{\min} \leq \min_j a_{j}$. Hint: use the properties of the Rayleigh quotient.

I got started with the problem, as follows:

Rayleigh quotient, for any vector $x$, is given as,

$$r(x) = \frac{\langle Ax,x\rangle}{\langle x,x\rangle}.$$

If $x$ is an eigenvector, $r(x) = \lambda$.

For a Hermitian matrix, $r(x)\in\mathbb R$, the eigenvalues are real.

Now, a Hermitian matrix can be diagonalized as:

$A = UDU^*$, where U is a unitary matrix of eigenvectors, and $D$ is a diagonal matrix of eigenvalues of $A$.

I am missing the final link, that I need to solve my problem.



2 Answers 2


We show that $\lambda_{\min}=\min_{x\neq 0}\frac{x^*Ax}{x^*x}$. Indeed, let $P$ an unitary matrix and $D$ a diagonal matrix such that $P^*DP=A$. We have $$x^*Ax=x^*P^*DPx=(Px)^*DPx$$ and $(Px)^*Px=x^*x$ hence $$\min_{x\neq 0}r(x)=\min_{x\neq 0}\frac{\langle Dx,x\rangle}{\langle x,x\rangle}.$$ Now it's easier to see that $\min_{x\neq 0}r(x)\geq \lambda_{\min}$. Thanks to that, note that $a_{jj}=r(e_j)$, where $e_j$ is the $j$-th element of the canonical basis of $\mathbb C^n$. So $a_{jj}=r(e_j)\geq \min_{x\neq 0}r(x)\geq \lambda_{\min}$.

  • $\begingroup$ hmm, could you elaborate "Thanks to that, note that ajj=r(ej), where ej is the j-th element of the canonical basis of Cn, and we are done". Thanks. $\endgroup$ Commented Apr 20, 2012 at 21:47
  • $\begingroup$ @rookieRailer Done now. $\endgroup$ Commented Apr 20, 2012 at 21:49
  • $\begingroup$ (Px)∗Px=x∗x is OK, but how does it make (Px)∗DPx = x*Dx ? $\endgroup$ Commented Apr 20, 2012 at 22:20
  • $\begingroup$ I make the "change of variable" in the $\inf$ $y\leftarrow Px$. $\endgroup$ Commented Apr 21, 2012 at 8:46

A different perspective can be obtained from the Schur part of the Schur-Horn Theorem, which says that if $A$ is Hermitian, then $$ (a_{11},\ldots,a_{nn})\prec (\lambda_1,\ldots,\lambda_n), $$ where $\prec$ indicates majorization. Elementary characterizations of majorization then give that $\min\{a_{11},\ldots,a_{nn}\}\geq\min\{\lambda_1,\ldots,\lambda_n\}$.

  • $\begingroup$ Correct. Indeed, in this notation, $\lambda_n$ is the smallest eigenvalue of $A$. Majorization implies that $a_{11} + \cdots + a_{n-1,n-1} \leq \lambda_1 + \cdots + \lambda_{n-1}$. Moreover, since the trace is both the sum of the diagonal elements and the sum of the eigenvalues, the conclusion follows. Nice. $\endgroup$
    – Malkoun
    Commented Jun 8, 2019 at 11:14

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