Show that this is not true: $\left\langle\vec{u},\vec{v}\right\rangle^2=\vec{u}^2\cdot\vec{v}^2$

Question

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?

Answer 1

Take any orthogonal unit vectors: $$0\neq 1$$ In general to disprove something all you need is one counterexample.

Answer 2

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.

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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.