Question: Let $A$ be an $m$ x $n$ matrix. Show that matrix multiplication, $x \to Ax$, defines a homomorphism $\phi$ :$\mathbb R$${^n} \to$ $\mathbb R$${^m}$.
I have no notes on this because my professor wants to see if I can answer this before the next class. So far I know that scalar multiplication is given by $av$=$a(a_1,….,a_n)$=$(aa_1,….aa_n)$
The scalar multiplication $x \to Ax$ "stretches" $x$ by a factor $\vert A \vert$ and reversing its direction when $A$ is negative (we put quotes around stretches because $Ax$ is shorter than $x$ when $\vert A\vert$ $\lt 1$)
Definition of homomorphism: Let $G_1$ and $G_2$ be two groups. Then $\phi$ : $G_1 \to G_2$ is called a homomorphism iff $\forall$ $a,b \in G_1$. So $\phi$$(ab)$=$\phi (a)$$\phi (b)$
This all the knowledge or information I have about this problem. I do not know how to show the question
Best Answer
First of all for $\phi: \mathbb R^n \to \mathbb R^m$ such that $\phi(x)=Ax$ implies that $x$ isn't a scalar. $x$ is a vector. With that little tidbit of information, we can conclude that $\phi$ is doing a form of matrix multiplication on the right. Thus for your action we have, $$\phi(x)= \begin{bmatrix} p_1\\p_2\\\vdots\\p_m \end{bmatrix},\; \mathrm {s.t.}\: p_k=\sum_{i=1}^n a_{(i,k)}x_i .$$ So let $x,y\in \mathbb R^n$ such that $\phi(x)=\begin{bmatrix} p_1\\p_2\\ \vdots\\ p_m \end{bmatrix}$ and $\phi(y)=\begin{bmatrix} q_1\\q_2\\ \vdots\\ q_m \end{bmatrix}$. It will now be sufficient to prove that $\phi(x+y)=\phi(x)+\phi(y)$. So,
$$\begin{array}{ccc} \phi(x+y) = \begin{bmatrix} r_1\\r_2\\\vdots\\r_m \end{bmatrix},\; \mathrm {s.t.}\: r_k &=&\sum_{i=1}^n a_{(i,k)}(x_i+y_i)\\ &=&\sum_{i=1}^n a_{(i,k)}(x_i)+\sum_{i=1}^n a_{(i,k)}(y_i)\\ \begin{bmatrix} p_1+q_1\\p_2+q_2\\\vdots\\p_m+q_m \end{bmatrix}=\begin{bmatrix} p_1\\p_2\\ \vdots\\ p_m \end{bmatrix}+\begin{bmatrix} q_1\\q_2\\ \vdots\\ q_m \end{bmatrix}&=&\phi (x)+\phi(y) \end{array}$$ as desired.