Linear transform and matrix. i need to represent a transormation that is linear in matrix form. But i am a little confused with the interpretation of the application from P5 to P5:
$T(p(x))= p(x) + x p'(x)$, where p(x) can be $1,x,x²,...,x^5$, these being the basis of P5.
Now, i am not sure how to interpret this because, tecnicaly aren't we summing a vector with a number? That is, $p(x)$ is just a vector, but $x p'(x)$ isn't some inner product of the basis x and p'(x)? I think the problem is not well proposed, it would be better to write $yp'(x)$. BUT, if we ignore that the powers are basis, the problem now makes sense, we are just dealing with incognites.
Also, if we ignore the matricial point first, we can say that $T(p(x)) = (n+1)p(x)$, so in the end the matricial rerpesentation will be $TP = (n+1)P$, where n is the power of the x. BUT, if we consider right at the initial the matricial problem, and ignoring (why can we ignore? actually, it is better to ask, can we ignore it?) the fact that x apparenttly is basis, $TP= P + xP'$.
Even worst, if we consider x as the basis it should be, $T(p(x))= p(x) + x p'(x)$ is equivalent to $TP = P + X^T P'$ which does not make any sense.
Am i wrong? Or the exercised is not well written?
 A: The map $T$ is well-defined. Let $p(x)\in \mathcal{P}_5$ be a polynomial (with real coefficients) of degree at most $5$ in a variable $x$. Hence we can write $p(x)=\sum_{i=0}^5a_ix^i$. Note that the formal derivative $p'(x)=\sum_{i=0}^5ia_ix^{i-1}$ is again an element of $\mathcal{P}_5$ as well as $x\in \mathcal{P}_5$. Note that the product $xp'(x)=\sum_{i=0}^5ia_ix^{i}\in \mathcal{P}_5$ and thus $p(x)-xp'(x)=\sum_{i=0}^5(a_i-ia_i)x^i\in \mathcal{P}_5$ as well.
Explicitly, $$T(p(x))=T(\sum_{i=0}^5a_ix^i)=\sum_{i=0}^5(a_i-ia_i)x^i.$$
Now, forget about the map $T$ for a while and consider any linear map $L$ between finite-dimensional vector spaces $$L\colon V\to W$$ where $\alpha=\{v_1,\dots,v_n\}$ is a basis of $V$ and $\beta=\{w_1,\dots,w_m\}$ is a basis of $W$. Then $L$ admits a matrix representation $[L]_{\alpha}^{\beta}$ with respect to the bases $\alpha$ and $\beta$ determined as follows:
For any basis vector $v_i$ of $\alpha$, $L(v_i)\in W$ and thus can be written uniquely as $$T(v_i)=\sum_{j=1}^ma_{j,i}w_j.$$ This yields a matrix $A\in \mathbb{R}^{m\times n}$ which we call $[L]_{\alpha}^{\beta}$.
What is the purpose of this matrix? Well, take any $v\in V$ and write $v=\sum_{i=1}^nx_iv_i$. Let $Y=AX$ where $X$ is the column vector containing the $x_i$'s. Verify explicitly that $$T(v)=\sum_{j=1}^my_jw_j.$$
This argument shows that very explicitly, we can think of the linear map $L$ as a map $L_A\colon \mathbb{R}^n\to \mathbb{R}^m:X\mapsto AX$ where we think of the input $X$ as the coordinates of a vector $v$ w.r.t. the basis $\alpha$ and $AX$ as the coordinates of $T(v)$ w.r.t. the basis $\beta$.
Great, so let's return to the problem at hand, namely finding $[T]_{\alpha}^{\alpha}$ where $\alpha=\{1,x,x^2,x^3,x^4,x^5\}$.
Note that
\begin{eqnarray}
T(1) &=& 1\cdot 1+ 0\cdot x+0\cdot x^2+0\cdot x^3+0\cdot x^4+0\cdot x^5\\
T(x) &=& 0\cdot 1+ 2\cdot x+0\cdot x^2+0\cdot x^3+0\cdot x^4+0\cdot x^5\\
T(x^2) &=& 0\cdot 1+ 0\cdot x+3\cdot x^2+0\cdot x^3+0\cdot x^4+0\cdot x^5\\
T(x^3) &=& 0\cdot 1+ 0\cdot x+0\cdot x^2+4\cdot x^3+0\cdot x^4+0\cdot x^5\\
T(x^4) &=& 0\cdot 1+ 0\cdot x+0\cdot x^2+0\cdot x^3+5\cdot x^4+0\cdot x^5\\
T(x^5) &=& 0\cdot 1+ 0\cdot x+0\cdot x^2+0\cdot x^3+0\cdot x^4+6\cdot x^5\\
\end{eqnarray}
Following carefully the explanation above, we see that the coordinate of $T(1)$ w.r.t. $\alpha$ is $(1,0,0,0,0,0)$ and this yields the first column of $[T]_{\alpha}^{\alpha}$. In this way we computed $[T]_{\alpha}^{\alpha}$ as required.
A: You have a linear map $T:\mathbb K_{\le5}[x]\to \mathbb K_{\le5}[x]$ such that $p(x)\mapsto p(x)+xp'(x)$ and we want its matrix representation.
Let's take the canonical basis $\mathcal C=\{x^5,x^4,x^3,x^2,x,1\}$ and compute their images
$$x^5\mapsto x^5+5x^5=\textbf 6 x^5\\x^4\mapsto x^4+4x^4=\textbf5x^4\\x^3\mapsto x^3+3x^3=\textbf 4x^3\\x^2\mapsto x^2+2x^2=\textbf 3x^2\\x\mapsto x+x=\textbf 2x\\1\mapsto 1=\textbf 1\cdot 1\text{, hence}\\A=\mathcal M_{\mathcal C}^{\mathcal C}(T)=\begin{pmatrix}6&0&0&0&0\\0&5&0&0&0\\0&0&4&0&0\\0&0&0&2&0\\0&0&0&0&1 \end{pmatrix}.$$
For definition of matrix of representation, given $\mathbb K_{\le 5}[x]$ with its basis $\mathcal C=\{x^5,x^4,x^3,x^2,x,1\}=\{\textbf c_i\}_{i=1}^6$ we can define the isomporphism $\varphi_{\mathcal C}:\mathbb K_{\le 5}[x]\to\mathbb R^6$ such that, given a polynomial $p(x)\in\mathbb K_{\le 5}[x]$, we have
$$p(x)=\sum_{i=1}^6a_i\textbf c_i\overset{\varphi_{\mathcal C}}{\mapsto}\begin{pmatrix}a_1\\a_2\\a_3\\a_4\\a_5\\a_6\end{pmatrix}\in\mathbb R^6.$$
So the $i$-th column of $A$, $A^{(i)}$ is $\varphi_{\mathcal C}(\textbf c_i)$.
