Matrix of a linear transformation of derivative type function Find the matrix A of the linear transformation $T(f(t)) =  4f'(t)+7f(t)$ from $P_2 \to P_2$ with respect to the standad basis for $P_2$ , $\{1,t,t^2\}$
So for the problem statement above, I dont fully understand how I would do this without knowing the definition of $f(t)$.
How would one go about this without knowing $f(t)$?
 A: In general we have the following fact:

If $T: V\to W$ is a linear transformation and $\beta_{1}=\{v_{1},v_{2},\ldots,v_{n}\}$ is a ordered basis for $\mathbb{R}$-vector space $V$ and $\beta_{2}=\{w_{1},w_{2},\ldots,w_{m}\}$ is a ordered basis for $\mathbb{R}$-vector space $W$  so the matrix representation for $T$ respect to basis $\beta_{1}$ and $\beta_{2}$ is given by
$$[T]_{\beta_{1}\to \beta_{2}}:=\begin{bmatrix} \uparrow & \uparrow &\ldots &\uparrow \\ [T(v_{1})]_{\beta_{2}} & [T(v_{2})]_{\beta_{2}} & \ldots &  [T(v_{n})]_{\beta_{2}}\\ \downarrow & \downarrow & \ldots & \downarrow \end{bmatrix}\in M_{m\times n}(\mathbb{R})$$where $[T(v_{i})]_{\beta_{2}}$ denote the coordinates for the vector $T(v_{i})$ respect to basis $\beta_{2}$ and it's defined as $$[T(v_{i})]_{\beta_{2}}=\begin{bmatrix} \alpha_{1}\\\alpha_{2}\\\vdots\\\alpha_{m}\end{bmatrix}\in \mathbb{R}^{m}$$ if and only if, $$T(v_{i})=\alpha_{1}w_{1}+\alpha_{2}w_{2}+\cdots+\alpha_{m}w_{m}$$for some $\alpha_{1},\alpha_{2},\ldots,\alpha_{m}\in\mathbb{R}$ and some $T(v_{i})\in W$ and $i\in \{1,2,\ldots,n\}$.

The above fact is a general result in the construction of matrix representation for a linear transformation. Now, let's to answer the question using that fact and of course it's similar way to Mr. Gandalf Sauron's answer. However I will write it for future readers.
Setting the problem in the way the above fact, we have

*

*$V=P_{2}(\mathbb{R})$ and $\beta_{1}=\{1,t,t^{2}\}=\{v_{1},v_{2},v_{3}\}$.

*$W=P_{2}(\mathbb{R})$ and $\beta_{2}=\{1,t,t^{2}\}=\{w_{1},w_{2},w_{3}\}$.

*$T: V\to W$ is defined by $T(f(t))=4f'(t)+7f(t)$ and $T$ is clearly a linear transformation is follows directly of the definition of linear transformation.

Now, in the way for to use the above fact we need to find the coordinates $[T(v_{1})]_{\beta_{2}},[T(v_{2})]_{\beta_{2}}$ and $[T(v_{3})]_{\beta_{2}}$. Let's to do it.

*

*$T(v_{1})=4(1)'+7(1)=7+0t+0t^{2}$ implies $[T(v_{1})]_{\beta_{2}}=\begin{bmatrix}7\\0\\0\\\end{bmatrix}$.


*$T(v_{2})=4(t)'+7(t)=4+7t+0t^{2}$ implies $[T(v_{2})]_{\beta_{2}}=\begin{bmatrix}4\\7\\0\\\end{bmatrix}$.


*$T(v_{3})=4(t^{2})'+7(t^{2})=0+8t+7t^{2}$ implies $[T(v_{3})]_{\beta_{2}}=\begin{bmatrix}0\\8\\7\\\end{bmatrix}$.
Now, notice that each step above we are solving a linear system for to find the $\alpha_{1},\alpha_{2}$ and $\alpha_{3}$ and of course we can save computational time by always solving all linear systems in a single row reduction.
So using the above fact we have $$\color{red}{[T]_{\beta_{1}\to \beta_{2}}=\begin{bmatrix}7&4&0\\0&7&8\\0&0&7\end{bmatrix}}$$
Well, the natural question is the following

*

*Is the presented solution overcomplicated for this problem? If the answer is yes, what would be the easiest alternative?

The answer is "yes" and the way for to answer the second question we note that the problem has two important elements that appear in many linear algebra problems.

*

*$V=W$ so $T$ is a linear operator.

*$\beta_{1}=\beta_{2}$ is the standard basis for the vectors space.

Using that and a bit linear algebra theory for see that $P_{n}(\mathbb{R})\cong \mathbb{R}^{n+1}$, where $\cong$ denote isomorphic spaces we get a linear transformation $T: \mathbb{R}^{3}\to\mathbb{R}^{3}$ defined by
$$T\begin{bmatrix}a\\b\\c\end{bmatrix}=\color{red}{\begin{bmatrix}7 & 4 & 0\\  0 & 7 & 8\\ 0&0&7\end{bmatrix}}\begin{bmatrix} a\\b\\c\end{bmatrix}, \quad \begin{bmatrix}a\\b\\c\end{bmatrix}\in \mathbb{R}^{3}$$
and that matrix representation also works for our original problem. Why? in terms of algebra these two spaces $P_{2}(\mathbb{R})$ and $\mathbb{R}^{2+1}$ are algebraically indistinguishable due to the isomorphism that exists between them. In a certain sense we are deforming the space of the polynomials of degree less than or equal to two to treat these polynomial-vectors as vectors of three-dimensional real space.
Finally, another important detail to highlight is that in the mentioned result there is the term "ordered basis" note that this part is important because if you change the order of the basis vectors, you change the nature of the matrix, be careful with that part.
A: How would you have done it if it was a linear transformation from any vector space $V$ to itself?
You take a basis $\mathcal{B}=\{v_{1},v_{2},v_{3}\}$ and compute $T(v_{1}),T(v_{2}),T(v_{3})$ .
Then you express $T(v_{i})=c_{1i}v_{1}+c_{2i}v_{2}+c_{3i}v_{3}\,\,,i\in \{1,2,3\}$.
And then you write $[T]_{\mathcal{B}}=\begin{bmatrix}
c_{11} & c_{12} & c_{13}\\
c_{21} & c_{22} & c_{23}\\
c_{31} & c_{32} & c_{33}
\end{bmatrix}$.
So you do exactly the same in this case:-
$T(1)= 7\cdot 1 + 0\cdot t + 0\cdot t^{2}$
$T(t)= 4\cdot 1 + 7\cdot t + 0\cdot t^{2}$
$T(t^{2})= 0\cdot 1 + 8\cdot t+ 7\cdot t^{2}$
So $T_{\{1,t,t^{2}\}}=\begin{bmatrix}
7 & 4 & 0\\
0 & 7 & 8\\
0 & 0 & 7
\end{bmatrix}$
