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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, $r(x)\in\mathbb R$. Also, for a Hermitian matrix, 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.

Thanks.

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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 $a_{jj}=r(e_j)\geq \min_{x\neq 0}r(x)\geq \lambda_{\min}$.

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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. –  rookieRailer Apr 20 '12 at 21:47
    
@rookieRailer Done now. –  Davide Giraudo Apr 20 '12 at 21:49
    
(Px)∗Px=x∗x is OK, but how does it make (Px)∗DPx = x*Dx ? –  rookieRailer Apr 20 '12 at 22:20
    
I make the "change of variable" in the $\inf$ $y\leftarrow Px$. –  Davide Giraudo Apr 21 '12 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 $\min\{a_{11},\ldots,a_{nn}\}\geq\min\{\lambda_1,\ldots,\lambda_n\}$.

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