This figure shows the vectors $\vec{a}, \vec{b}, \vec{u}, \vec{v}, \vec{w}$ and $\vec{z}$
if $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a linear transformation and the images of $\vec{a}$ and $\vec{b}$ are shown below, draw the images of $\vec{u},\vec{v},\vec{w}$ and $\vec{z}$
I'm not sure exactly how to do this, well honestly, i have no clue whatsoever, if someone can show me the first couple – surely i can mimic the rest. Also, it doesn't even look like $a$ and $b$ are shown below but it looks as if $a$ and $w$ images are shown below?
Best Answer
wright $w$ and $v$ in function of $a$ and $b$ so you have : $$ w=\alpha a+\beta b \iff \left(\begin{array}{c}1\\4 \end{array}\right)=\alpha \left(\begin{array}{c}1\\1 \end{array}\right)+\beta \left(\begin{array}{c}-1\\2 \end{array}\right)\iff \left(\begin{array}{c}1\\4 \end{array}\right)=\left(\begin{array}{c}\alpha-\beta \\\alpha +2\beta\end{array}\right) $$ so $\alpha=2 $ and $\beta=1 $ and then $$ w=2a+b \implies T(w)=2T(a)+T(b)=2\left(\begin{array}{c}1\\-1 \end{array}\right)+\left(\begin{array}{c}1\\3 \end{array}\right)=\left(\begin{array}{c}3\\1 \end{array}\right) $$ same argument for $v$ we verify that $v=b-a$ and then : $$ v=b-a \implies T(v)=T(b)-T(a)=\left(\begin{array}{c}0\\4 \end{array}\right) $$