# Find the Matrix of a Linear Transformation.

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.

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The first column of the matrix is the vector $T(1)$, the second column is the vector $T(t)$, and the third column is the vector $T(t^2)$. Try it out. All these vectors should be written with their coordinates in terms of the basis $\{1, t, t^2\}$, but luckily this is a easy basis to work with in $P_2$. – Henry T. Horton Dec 8 '12 at 22:43
@HenryT.Horton as in evaluate the equation for each? – Jared Terrell Dec 8 '12 at 22:46
Yes. For example, $T(3 + 2t) = 3(3 + 2t)' + 7(3 + 2t) = 3 \cdot 2 + 21 + 14t = 27 + 14t + 0t^2$, so $T(3 + 2t)$ would correspond to the vector $$\begin{pmatrix} 27 \\ 14 \\ 0 \end{pmatrix}.$$ – Henry T. Horton Dec 8 '12 at 22:49
Ahh! So then I express my resulting answer after evaluation in terms of being a combination of the three parts of my basis? That makes perfect sense...Thanks alot @HenryT.Horton – Jared Terrell Dec 8 '12 at 22:51
Aside: linear algebra is the study of linear transformations. If linear transformations are your weak spot, you should be seriously worried. – Chris Eagle Dec 8 '12 at 23:02

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?)

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