# A clarification to a proof that $x^{\top}Ax$ is maximized at $\max \lambda(A)$ for symmetric $A$

I'm looking for a clarification of an answer to

Showing that $x^{\top}Ax$ is maximized at $\max \lambda(A)$ for symmetric $A$

(This is a clarified formulation of my original question which I deleted)

Given a $$n \times n$$ symmetric matrix $$A$$,

$$\max_{x : ||x||_2 = 1} x^{\top}Ax = \max \lambda(A),$$

where $$\max \lambda(A)$$ is the maximum eigenvalue of $$A$$.

I've approached solving the problem in exactly the same way as @Ryan at the above page. The main point is in the upper bound introduced as: $$x^{\top}Ax = \sum_{i=1}^n \lambda_i \tilde{x}_i^2 \le \max \lambda(A)\sum_{i=1}^n \tilde{x}_i^2$$

and this is correct.

However, this proves only that $$x^{\top}Ax \le \max \lambda(A),$$ but not the original statement $$\max_{x : ||x||_2 = 1} x^{\top}Ax = \max \lambda(A)$$

• The usual Lagrange multipliers proof (on the unit sphere) shows that at a maximum point $x$ (which must occur by compactness and continuity), $Ax=\lambda x$ for some constant $\lambda$. This, of course, means that $x$ is an eigenvector with eigenvalue $\lambda$. – Ted Shifrin Mar 6 '19 at 22:50
• This exactly clarifies it! Many thanks. Only a true mathematician could answer this, and I'm not the one :) – dnqxt Mar 6 '19 at 22:58
• You're welcome. Feel free to check out my various lectures on YouTube (linked in my profile) on multivariable calculus and linear algebra. – Ted Shifrin Mar 6 '19 at 23:24

Mazimize $$x^TAx$$, s.t. $$x^Tx=1$$. Using Lagrange multiplier $$\lambda$$ we have $$\arg\max_{x}( x^TAx - \lambda( x^Tx-1))$$ so taking derivative w.r.t. vector $$x$$ and equating to zero we have $$2Ax-2\lambda x=0 \to Ax=\lambda x.$$ Namely, the set of solutions of this maximization problem should satisfy $$Ax = \lambda x$$, i.e., such vectors $$x$$s are vectors that multiplying $$A$$ by $$x$$ is the same as multiplying $$x$$ by a scalar. That is, by definition $$x$$ have to be the eigenvectors and thus $$\lambda$$ its corresponding eigenvalue. Now, you have a set of solutions $$\{(\lambda_i, x_i)\}_{i=1}^n$$ and you have to choose the pair the maximize the original function, so clearly $$\max x^TAx = x ^T \lambda_{\max} x = \lambda_{\max} x^T x = \lambda_{\max}.$$
Take for example $$A=I$$
If $$A$$ is symmetric, then its eigenvalues are real. Moreover, we have (Rayleigh coefficient) $$\frac{x^TAx}{||x||^2} = \lambda$$ and since $$||x||^2=1$$, we get $$x^T A x = \lambda$$. So, $$\max_{||x||^2 = 1} x^T A x = \max \lambda.$$