trace inequality for positive semidefinite matrix

Question

Let $A$ be a square real matrix $n \times n$ positive semidefinite. Let $v \in \mathbb{R}^n$. Is it true that $$ v^TAv \le trace(A) \|v\|^2 $$ ?

Any help will be very appreciated! Thanks!

Answer 1

By the spectral theorem, if $A$ is symmetric it can be written as $Q^T\Lambda Q$ with $\Lambda $ the diagonal matrix whose entries are the eigenvalues of $A$ and $Q$ an orthogonal matrix. Then we have $$v^TA v=v^TQ^T\Lambda Qv=w^T\Lambda w,$$ where $w=Qv$. Since $Q$ is orthogonal, we have $$\|w\|=\langle w,w\rangle =\langle Qv,Qv\rangle =\langle v,Q^TQv\rangle =\langle v,v\rangle = \|v\|,$$ and thus $$w^T\Lambda w=\sum_{i=1}^n\lambda_iw_i^2\leq \left(\sum_{i=1}^n\lambda_i\right)\|w\|^2=\operatorname{tr}(A)\|w\|^2=\operatorname{tr}(A)\|v\|^2,$$ because all the eigenvalues are non-negative, and $w_i^2\leq \sum_i w_i^2=\|w\|^2$.