Show that this is not true: $\left\langle\vec{u},\vec{v}\right\rangle^2=\vec{u}^2\cdot\vec{v}^2$ 
Show that the "calculation rule" for the following scalar product is
  wrong: 
  $$\left\langle\vec{u},\vec{v}\right\rangle^2=\vec{u}^2\cdot\vec{v}^2$$

I'm not sure if I did it correctly?
Let $\vec{u}=\begin{pmatrix}
u_1\\ 
u_2\\ 
u_3
\end{pmatrix},
\vec{v}=\begin{pmatrix}
v_1\\ 
v_2\\ 
v_3
\end{pmatrix}$
Then
$$\left\langle\vec{u},\vec{v}\right\rangle^2=\vec{u}^2\cdot\vec{v}^2 \Leftrightarrow$$
$$\left[\begin{pmatrix}
u_1\\ 
u_2\\ 
u_3
\end{pmatrix} 
\begin{pmatrix}
v_1\\ 
v_2\\ 
v_3
\end{pmatrix}\right]^2 =
\begin{pmatrix}
u_1\\ 
u_2\\ 
u_3
\end{pmatrix}^2
\begin{pmatrix}
v_1\\ 
v_2\\ 
v_3
\end{pmatrix}^2$$
And at tjos point, I realized that you cannot square $\vec{u}$ and $\vec{v}$ because multiplication of $3 \times 1$ matrix with $3 \times 1$ doesn't work due to invalid size?
Is that reason enough to say that $\left\langle\vec{u},\vec{v}\right\rangle^2\neq\vec{u}^2\cdot\vec{v}^2$ ?
I'm really not sure about that and also about the correct notation. Or maybe you can do this a little more easy?
 A: Take any orthogonal unit vectors: $$0\neq 1$$
In general to disprove something all you need is one counterexample.
A: To show that a "rule" is wrong, you only need to show that it is wrong once.
Let 
$$
\vec{u} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\quad
\vec{v} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.
$$
Work out the values of
$\left\langle\vec{u},\vec{v}\right\rangle^2$
and
$\vec{u}^2\cdot\vec{v}^2.$
Compare them. If they are not the same, the rule is false.

For the calculations, make sure you understand what 
$\left\langle\vec{u},\vec{v}\right\rangle$ means and what $\vec{u}^2$ means.
In matrix notation, if 
$$
\vec{u} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix} \quad\text{and}\quad
\vec{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3  \end{pmatrix},
$$
then
$$
\left\langle\vec{u},\vec{v}\right\rangle =
\begin{pmatrix} u_1 & u_2 & u_3 \end{pmatrix}
\begin{pmatrix} v_1 \\ v_2 \\ v_3  \end{pmatrix}
$$
and
$$
\vec{u}^2 =
\begin{pmatrix} u_1 & u_2 & u_3 \end{pmatrix}
\begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}.
$$
In other words, 
$\left\langle\vec{u},\vec{v}\right\rangle = \vec u^T \vec v$
when $\vec u$ and $\vec v$ are column vectors.
