I have the following exercise to solve:
$V$ is an $\mathbb{R}$-vector space with basis $\{u_1, u_2\}$ and $f\colon V\to V$ is the linear transformation defined by:
- $f(u_1)= -3u_1 + 2u_2$
- $f(u_2)= 5u_1-u_2$
And I must calculate $f(x_1u_1+x_2u_2)$ if $x_1, x_2 \in \mathbb{R}$.
I've done the following:
$-3(x_1u_1 + x_2u_2) +2(x_1u_1 + x_2u_2)+5(x_1u_1 + x_2u_2) -(x_1u_1 + x_2u_2)= 3(x_1u_1 + x_2u_2)$
I've watched a lot of online tutorials where it was shown that $f(x) = m*c+a$ and $f(q*v+w) = m(qv+w) +a$, but what happens in my case where I have $u_1$ and $u_2$ inside my $f()$?
Saying $$f(x_1u_1+x_2u_2) = -3(x_1u_1 + x_2u_2) +2(x_1u_1 + x_2u_2)+5(x_1u_1 + x_2u_2) -(x_1u_1 + x_2u_2)$$ is just not the way linearity works. It seems you in some way applied both given rules of $f$ to all of the input at the same time. Instead, you should've used the definition of linearity to get $$ f(x_1u_1+x_2u_2)= f(x_1u_1) + f(x_2u_2) = x_1f(u_1) + x_2f(u_2) $$ Then calculate the result from there using the two rules you were given.
It is this possibility to massively simplify the input to a function (from the linear combination $x_1u_1+x_2u_2$ as a single input to just the basis vectors $u_1$ and $u_2$ separately) that makes linear functions so nice to work with.