Is $||A||_F ||x||_2^2 \geq x^TAx$

Question

Given a symmetric matrix $A$ and a vector $x$

Is $||A||_F ||x||_2^2 \geq x^TAx$?

If yes, how to show this?

Answer 1

The quantity $\frac{x^T A x}{||x||_2^2}$ is the Rayleigh quotient of $A$ and its maximum value is the largest eigenvalue of $A$, $\lambda_{max}$.

Noting that $||A||_F = \sqrt{tr(A A^T)} = \sqrt{tr(A A)} = \sqrt{tr(A^2)} = \sqrt{\sum_i \lambda_i^2} \geq \sqrt{\lambda_{max}^2} = |\lambda_{max}| \geq \lambda_{max}$, we get the inequality, where the steps in the middle follow by definition of frobenius norm, A being symmetric (so by diagonalization, we can see that the eigenvalues of $A^2$ are the square of the eigenvalues of $A$, and they are real by spectral theorem) and the definition of trace as sum of the eigenvalues. The second to last step is monotonicity of sqrt, and the last step is a basic property of absolute value.

Combining these two statements, we get the desired inequality.