Find the linear transformation of a polynomial Let L: P$_1$ --> P$_1$ be a linear transformation for which it is known that $L(t+1) = 2t+3$ and $L(t-1) = 3t-2$. I am supposed to find $L(at+b)$.
How do I do this? I am new to linear algebra in general and linear transformations in particular? My textbook doesn't provide an example of how to do problems like this involving polynomials.
Here's what I've been doing so far:
$L(t+1)-L(t-1) = L(2) = 2L(1) = (2t+3)-(3t-2)=-t+7$
$L(t+1)+L(t-1) = L(2) = 2L(1) = (2t+3)+(3t-2)=5t+1$
 A: You should check that $t+1$ and $t-1$ form a basis for your vector space. Then you know where $L$ sends your basis vectors, so $L$ is completely determined, since every vector in the space is of the form $c(t+1)+d(t-1)$ for some scalars $c,d$.  The reason this determines the map $L$ completely is that by linearity $L(c(t+1)+d(t-1))=cL(t+1)+dL(t-1)$ and we've already specified where to send the vectors $t+1,t-1$.  You can write coordinates for $L(t+1)$ and $L(t-1)$ in terms of this basis, but it may be simpler or illustrative to convert to the standard basis $[1,t]$ first:
$\frac{1}{2}(t+1)-\frac{1}{2}(t-1)=1$
$\frac{1}{2}(t+1)+\frac{1}{2}(t-1)=t$
Now the vector $at+b$ can be written as $\frac{a}{2}(t+1)+\frac{a}{2}(t-1)+\frac{b}{2}(t+1)-\frac{b}{2}(t-1)$ where I've used the above equalities.  If you're comfortable with coordinate matrices this can be done more neatly.
This simplifies to $\frac{a+b}{2}(t+1)+\frac{a-b}{2}(t-1)$ so the coordinates of the vector $at+b$ in the basis $[t+1,t-1]$ are $(\frac{a+b}{2},\frac{a-b}{2})$.
Now we can find $L(at+b)$:
$L(at+b)=L(\frac{a+b}{2}(t+1)+\frac{a-b}{2}(t-1))=\frac{a+b}{2}L(t+1)+\frac{a-b}{2}L(t-1)=\frac{a+b}{2}(2t+3)+\frac{a-b}{2}(3t-2)=\frac{5a-b}{2}t+\frac{a+5b}{2}$
A: View $P_1$ as $\mathbb R^2$, let $e_1=(1,0)^t$ and $e_2=(0,1)^t$. You'll easily derive that
$$L(e_1)=\frac 12\begin{pmatrix}5\\1\end{pmatrix}\quad\text{and}\quad L(e_2)=\frac 12\begin{pmatrix}-1\\5\end{pmatrix}.$$
So the matrix representation in respect of the standard base is $\frac12\begin{pmatrix}5&-1\\1&5\end{pmatrix}$.  Hence the coefficients of $L(at+b)$ may be computed as
$$\frac12\begin{pmatrix}5&-1\\1&5\end{pmatrix}\cdot\begin{pmatrix}a\\ b\end{pmatrix}=\frac12\begin{pmatrix}5a-b\\ a+5b\end{pmatrix},$$
thus $L(at+b)=\frac12(5a-b)t+\frac12(a+5b)$.
