Q:

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

A:

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
$$
|\Sigma y|^2 = s_1 |y_1|^2 + s_2 |y_2|^2 .
$$

