If A is a $3$ by $2$ matrix,if we do the singular value decomposition (SVD) to A,that is
$A= \begin{bmatrix} \vec u_1 & \vec u_2 &\vec u_3 \end{bmatrix} \begin{bmatrix} s_1 & 0 \\ 0 & s_2 \\ 0 & 0 \\ \end{bmatrix} \begin{bmatrix} \vec v_1 \\ \vec v_2 \\ \end{bmatrix} $
$\vec u_1$ , $\vec u_2$ and $\vec u_3$ are both a column vector,that is,$3$ by $1$ vector
$\vec v_1$ and $\vec v_2$ are both a row vector,that is,$1$ by $2$ vector
$s_1$ and $s_2$ are both singular value,and $s_1>s_2$.
Now ,why must $|A\vec v_1|^2$ is the highest value ?and why must $|A\vec v_1|^2$ > $|A\vec v_2|^2$?,Is there any theory can prove it?
$|f|^2=f^Hf$,if the f is a $N$ by $1$ vector