Which of the following Maps is Bilinear?

Question

Problem:

Show which of the following maps are bilinear ?

$\phi_1 : \left( \left(\begin{array}{c} v_1 \\ v_2 \end{array}\right) , \left(\begin{array}{c} w_1 \\w_2 \end{array}\right)\right) \mapsto v_1w_2 + v_2w_1 $

$\phi_2 : \left( \left(\begin{array}{c} v_1 \\ v_2 \\v_3 \end{array}\right) , \left(\begin{array}{c} w_1 \\w_2 \\ w_3 \end{array}\right)\right) \mapsto v_1v_2 + 2w_1w_2 $

$\phi_3 : \left( \left(\begin{array}{c} v_1 \\ v_2 \end{array}\right) , \left(\begin{array}{c} w_1 \\w_2 \end{array}\right)\right) \mapsto v_1 + v_2 + w_1 + w_2 $

Questions:

I have already shown that $\phi_1$ is bilinear, though i want to know if i was right with my assumption. Though right now what i am struggling with is to show that $\phi_2$ and $\phi_3$ are bilinear, due to lack of knowledge of transforming the following results of $\phi_2$ and $\phi_3$ to see if it is even possible for them to be bilinear. If there is a piece of advice on seeing quickly if it's bilinear, i would love to hear it. Other than that i thank you in advance for the following pieces of advise and awnsers !

Answer 1

Remark: Remember that a $\color{red}{bilinear form}$ on a vector space $V$ over a field $F$ is a $\color{blue}{map}$ $$\color{blue}{\phi: V \times V \to F}$$ such that:

1. $\forall v_{1},v_{2},w\in V$, we have that $$ \color{blue}{\boxed{\phi(v_{1}+v_{2},w)=\phi(v_{1},w)+\phi(v_{2},w)}}$$
2. $\forall v,w_{1},w_{2}\in V$, we have that $$\color{blue}{\boxed{\phi(v,w_{1}+w_{2})=\phi(v,w_{1})+\phi(v,w_{2})}}$$
3. $\forall v,w\in V, a\in F$, we have that $$\color{blue}{\boxed{ \phi(av,w)=a\phi(v,w)}}$$
4. $\forall v,w\in V, a\in F$, we have that $$\color{blue}{\boxed{\phi(v,aw)=a\phi(v,w)}}$$

So you should try to verify each of these properties (at least when you first start studying a definition in mathematics it is best not to skip steps but to be as detailed as possible).

Now, note that for $\phi_{2}$ we have \begin{eqnarray*} \phi_{2} \left( \begin{pmatrix} v_{1}\\v_{2}\\ v_{3} \end{pmatrix},\begin{pmatrix} w_{1}\\ w_{2}\\ w_{3} \end{pmatrix} +\begin{pmatrix} u_{1}\\u_{2}\\ u_{3} \end{pmatrix}\right)&=&\phi_{2}\left( \begin{pmatrix} v_{1}\\v_{2}\\ v_{3} \end{pmatrix},\begin{pmatrix} w_{1}+u_{1}\\ w_{2}+u_{2}\\ w_{3}+u_{3} \end{pmatrix} \right)\\ &=&v_{1}v_{2}+2(w_{1}+u_{1})(w_{2}+u_{2})\\ &=&v_{1}v_{2}+2(w_{1}w_{2}+w_{1}u_{2}+u_{1}w_{2}+u_{1}u_{2})\\ &=&v_{1}v_{2}+2w_{1}w_{2}+2w_{1}u_{2}+2u_{1}w_{2}+2u_{1}u_{2}\\ &\not=&\color{blue}{v_{1}v_{2}+2w_{1}w_{2}}+\color{red}{v_{1}v_{2}+2u_{1}u_{2}}\\ &=&\color{blue}{\phi_{2}\left(\begin{pmatrix} v_{1}\\ v_{2} \\ v_{3}\end{pmatrix},\begin{pmatrix} w_{1}\\ w_{2} \\ w_{3}\end{pmatrix} \right)}+\color{red}{\phi_{2}\left( \begin{pmatrix} v_{1}\\ v_{2} \\ v_{3}\end{pmatrix},\begin{pmatrix} u_{1}\\ u_{2} \\ u_{3}\end{pmatrix} \right)} \end{eqnarray*} with $\begin{pmatrix} v_{1}\\v_{2}\\ v_{3} \end{pmatrix},\begin{pmatrix} w_{1}\\ w_{2}\\ w_{3} \end{pmatrix},\begin{pmatrix} u_{1}\\u_{2}\\ u_{3} \end{pmatrix}\in \operatorname{Dom}(\phi_{2})$.

Therefore, $\phi_{2}$ it does not satisfy condition $\color{red}{2}$ and that is enough to affirm that $\phi_{2}$ it is not a bilinear form.

I will leave it to you to conclude with literal $3$. Many successes in your studies.