2-norm of a matrix

Question

Given that $\lambda$ is an eigenvalue of $A^TA$ with eigenvector $x$ I need to show

$(1)$ $\|Ax\|_2^2=\lambda\|x\|_2^2$

(2) $\|A\|_2\le \|A^TA\|^{1\over 2}$

I knew $\|Ax\|_2=\sum_i\sum_j a_{ij}x_i$

Answer 1

The first point is proven as follows.

From the SVD of $A = UDV^T$ we can see that eigenvalues of $A^TA = VD^2V^T$ are just squared ones from $A$. At the same time the columns of $V$ are the eigenvectors of $A^TA$. So, exploiting orthogonality of eigenvectors

$$\|Ax\|_2^2 = \|UDVx\|_2^2 = \|D\left(Vx\right)\|_2^2 = \|De_{\lambda}\|x\|\|_2^2 = \|\sqrt{\lambda}\|x\|\|_2^2= \lambda\|x\|^2$$

The proof is based on the property of the second matrix norm

$${\displaystyle \|A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}A)}}}$$. $${\displaystyle \|A^{^{T}}A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}AA^{^{T}}A)}}}.$$

The same reasoning from the first point says that the eigenvalues of $A^{^{T}}AA^{^{T}}A$ are just the ones of $A^{^{T}}A$ being squared. So $${\displaystyle \|A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}A)}}}$$ $${\displaystyle \|A^{^{T}}A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}AA^{^{T}}A)}}} = {\sqrt {\lambda _{\max }^2(A^{^{T}}A)}} = \vert\lambda _{\max }(A^{^{T}}A)\vert = \lambda_{\max}(A^TA)=\max(\lambda_{\min}^2(A),\lambda_{\max}^2(A)) \ge \|A\|_{2}^2 $$

Edit

The inequality becomes after the notion that $\lambda_{\max}(A^TA)=\max(\lambda_{\min}^2(A),\lambda_{\max}^2(A))$, because eigenvalues of $A$ could be negative.