I tried to prove that the 2-norm of a row matrix ($||A||_2$) is equal to the 2-norm of a vector consisting of the same entries($||a||_2$).
I tried to use: $2$-norm of a matrix &$ 2$-norm of a vector
I know that $||a||_2=\sqrt{\sum_{i=1}^{n}|a_i|^2}$, which is basically the the Euclidean norm. Furthermore, I know that $||A||_2 = \sqrt{\rho(A^TA)}$, where $\rho(A^TA)$ is the highest eigenvalue in absolute value of matrix $A^TA$. I tried to find properties for symmetric matrices regarding their maximum eigenvalue but I couldn't really find one.
I furthermore tried to use the Cauchy-Schwarz inequality as it was given as a hint, here I got: $|(x,y)| \leq ||x||_2||y||_2$, so if we take $a$ for both $x\& y$ $||a||_2 \leq ||A||_2*||A||_2=||A||_2^2$, which doesn't really seems to help me. Am I doing anything wrong with the inequality here or can I maybe use another inequality to squeeze the norms in?
Thanks
The linear mapping $A^TA$ has one dimensional image, namely $Lin(a)$, the span of the vector a. Therefore the kernel has dimension n-1, so the eigenvalue $\lambda =0$ has multiplicity n-1 as well. On the other hand a is obviously an Eigenvector of $A^TA$. Now compute the corresponding Eigenvalue.