Show a matrix is a linear transformations

Question

So I know I need to multiply my transformations by (1,0) and (0,1) but I’m unsure how to get the transformations as if I multiply just by the given matrix I get the same answer please help.

Answer 1

A linear transformation is a function, L, from a vector space to a vector space such that, for any numbers, a and b, and vectors, u and v, L(au+ bv)= aL(u)+ bL(v).

Here L is the matrix multiplication, $\begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$

Let $u=\begin{bmatrix}m \\ n\end{bmatrix}$ and $v= \begin{bmatrix}p \\ q \end{bmatrix}$.

$L(au+ bv)= \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}\begin{bmatrix}am+ bp \\ an+ bq\end{bmatrix}$ $= \begin{bmatrix}\frac{am+ bp}{2}- \frac{\sqrt{3}(an+ bq)}{3} \\ \frac{\sqrt{3}(am+ bq)}{2}+ \frac{am+ bp}{2}\end{bmatrix}$

While $aL(u)+ bL(v)= a\begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}\begin{bmatrix}m \\ n\end{bmatrix}+ b\begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}\begin{bmatrix}p\\ q\end{bmatrix}$ $= a\begin{bmatrix}\frac{m}{2}- \frac{\sqrt{2}m}{2} \\ \frac{\sqrt{3}m}{2}+ \frac{n}{2}\end{bmatrix}$+ b\begin{bmatrix}\frac{p}{2}-\frac{\sqrt{3}q}{2} \\ \frac{\sqrt{3}p}{2}+ \frac{q}{2}\end{bmatrix}$

Do you see that those are the same?