Show that $\|Ax\|_2^2 = \lambda \|x\|_2^2$

Question

Let $A \in \mathbb{R}^{n \times n}$, let $\lambda$ be an eigenvalue of $A^TA$ and $x \in \mathbb{R}^n \setminus \{0\}$ be the corresponding eigenvector, then show that $$\|Ax\|_2^2 = \lambda \|x\|_2^2 \ \text{and hence} \ \lambda \geq 0$$

Answer:

Here $||.||_2$ denote matrix $ \ 2-$norm i.e, $||A||_2=\sigma_{\max} (A)=\sqrt{\lambda},$ where $\sigma_{\max}$ is the largest singular value of matrix $A$ and $\lambda$ is largest eigenvalue of $A^TA$.

Now we have,

$(A^TA)x=\lambda x \Rightarrow ||(A^TA)x||_2=||\lambda x||_2$

How to conclude the proof?

help me.

Since

Answer 1

A simpler proof is that $$||Ax||_2^2{=x^TA^TAx\\=x^T(A^TAx)\\=x^T(\lambda x)\\=x^T\lambda x\\=\lambda||x||_2^2}$$