It's been a few weeks since the subject was covered in my Linear Algebra class, and unfortunately linear transformations are my weak spot, so could anyone explain the steps to solve this problem?
Find the matrix $A$ of the linear transformation $T(f(t)) = 3f'(t)+7f(t)$ from $P_2$ to $P_2$ with respect to the standard basis for $P_2$, $\{1,t,t^2\}$.
The resulting answer should be a $3 \times 3$ matrix, but I'm unsure of where to start when it comes to solving this problem.
Best Answer
NOTE Given a finite dimensional vector space $\Bbb V$ and a basis $B=\{v_1,\dots,v_n\}$ of $\Bbb V$, the coordinates of $v$ in base $B$ are the unique $n$ scalars $\{a_1,\dots,a_n\}$ such that $v=\sum_{k=1}^n a_kv_k$, and we note this by writing $(v)_B=(a_1,\dots,a_n)$.
All you need is to find what $T$ maps the basis elements to. Why? Because any vector in $P_2$ can be written as a linear combination of $1$, $t$ and $t^2$, whence if you know what $T(1)$, $T(t)$ and $T(t^2)$ are, you will find what any $T(a_0 +a_1 t +a_2 t^2)=a_0T(1)+a_1 T(t)+a_2 T(t^2)$ is. So, let us call $B=\{1,t,t^2\}$. Then
$$T(1)=3\cdot 0 +7\cdot 1=7=(7,0,0)$$
$$T(t)=3\cdot 1 +7\cdot t=(3,7,0)$$
$$T(t^2)=6\cdot t +7\cdot t^2=(0,6,7)$$
(Here I'm abusing the notation a bit. Formally, we should enclose the two first terms of the equations with $(-)_B$ )
Now note that our transformation matrix simply takes a vector in coordinates of base $B$, and maps it to another vector in coordinates of base $B$. Thus, if $|T|_{B,B}$ is our matrix from base $B$ to base $B$, we must have
$$|T|_{B,B} (P)_B=(T(P))_B$$
where we wrote $P=P(t)$ to avoid too much parenthesis.
But let's observe that if $(P)_B=(a_0,a_1,a_2)$ then $a_0T(1)+a_1 T(t)+a_2 T(t^2)=a_0(7,0,0)+a_1 (3,7,0)+a_2(0,6,7)$ is the matrix product
$$\left(\begin{matrix}7&3&0\\0&7&6\\0&0&7\end{matrix}\right)\left(\begin{matrix}a_0 \\a_1 \\a_2 \end{matrix}\right)$$
And $|T|_{B,B}=\left(\begin{matrix}7&3&0\\0&7&6\\0&0&7\end{matrix}\right)$ is precisely the matrix we're after. It has the property that for each vector of $P_2$
$$|T|_{B, B}(P)_B=(T(P))_B$$
(well, actually
$$(|T|_{B,B} (P)_B^t)^t=(T(P))_B$$
but that looks just clumsy, doesn't it?)