# why must $|A\vec v_1|^2$ is the highest value ?if the $\vec v_1$ is the vector which is corresponding to the biggest singular value of A

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 $$|\Sigma y|^2 = s_1 |y_1|^2 + s_2 |y_2|^2 .$$
• Why is $|U (\Sigma V^Tx)| = |\Sigma V^Tx|$? – electronic component Nov 3 '19 at 0:39
• @electroniccomponent the columns of $U$ are orthonormal. So, for any vector $z \in \Bbb C^3$, we have $$|Uz|^2 = |z_1 u_1 + z_2 u_2 + z_3 u_3|^2 = |z_1|^2 + |z_2|^2 + |z_3|^2 = |z|^2$$ – Ben Grossmann Nov 3 '19 at 7:38
• $\max_{x \in \Bbb R^2, |x| = 1} |Ax|$ means "maximize $|Ax|$ over all $x \in \Bbb C^2$ subject to the constraint that $|x| = 1$" – Ben Grossmann Nov 3 '19 at 7:38