Proof for Quadratic matrix inequality

Question

I would like to understand how we can show $(v^TA^TAv)^{1/2}≤∥A^TA∥^{1/2}(v^Tv)^{1/2}, \forall A \in \mathbb{R}^{n \times n}, v \in \mathbb{R}^{n}$.

I appreciate your answer.

Answer 1

$A^TA$ is symmetric and positive semi-definite and therefore diagonalizable with real and nonnegative eigenvalues $\lambda_1^2\ge\lambda_2^2\ge\cdots\ge\lambda_n^2$. $A^TA=U^TDU$ where $D$ is diagonal and the and $U$ is unitary. $$v^TA^TAv=v^TU^TDUv=\sum_{i=1}^n \lambda_i^2x_i^2\le \lambda_1^2x^Tx=\lambda_1^2 v^Tv$$ where $x=[x_i]_{i=1}^n=Uv$. But $\lambda_1=\|A^TA\|$. We have the required conclusion.