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


It is not entirely clear from your question, but it looks like you're trying to prove that $$ \max_{x \in \Bbb C^2, |x| = 1} |Ax| = s_1. $$ To that end: we have $A = U \Sigma V^T$. Note the following:

  • For $x \in \Bbb C^2$, $|Ax| = |U (\Sigma V^Tx)| = |\Sigma V^Tx|$.
  • For any vector $x$, $y = V^Tx$ is a vector of the same magnitude.

We can therefore state that $$ \max_{x \in \Bbb C^2, |x| = 1} |Ax| = \max_{x \in \Bbb C^2, |x| = 1} |\Sigma (V^Tx)| = \max_{y \in \Bbb C^2, |y| = 1} |\Sigma y|. $$ From here, it's easy to prove the desired result by noting that if $y = (y_1,y_2)^T$, we have $$ |\Sigma y|^2 = s_1 |y_1|^2 + s_2 |y_2|^2 . $$

  • $\begingroup$ Why is $ |U (\Sigma V^Tx)| = |\Sigma V^Tx|$? $\endgroup$ – electronic component Nov 3 at 0:39
  • $\begingroup$ what does your objected function mean? $\endgroup$ – electronic component Nov 3 at 5:58
  • $\begingroup$ @electroniccomponent the columns of $U$ are orthonormal. So, for any vector $z \in \Bbb C^3$, we have $$ |Uz|^2 = |z_1 u_1 + z_2 u_2 + z_3 u_3|^2 = |z_1|^2 + |z_2|^2 + |z_3|^2 = |z|^2 $$ $\endgroup$ – Omnomnomnom Nov 3 at 7:38
  • $\begingroup$ $\max_{x \in \Bbb R^2, |x| = 1} |Ax|$ means "maximize $|Ax|$ over all $x \in \Bbb C^2$ subject to the constraint that $|x| = 1$" $\endgroup$ – Omnomnomnom Nov 3 at 7:38

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