Linear Algebra Help! Matrices with respect to given basis Hey Guys I'm new here so I dont know much exactly how to do all the fancy symbols but here it goes:
Let $V_n$ be the vector space of real polynomials of degree at most $n$ and Let $B_n$ be the usual basis  $v_i$(t) = $t^{i-1}$ Define $T: V_2\to V_3$ by
$$
T(v(t)) = \int_{0}^{t} v(x)\,dx
$$
Let $D: V_3 \to V_2$ be the differentiation operator. Find the Matrices $[T]$ and $[D]$ with respect to the bases. Find $TD$ and $DT$ and their matrices
Verify that $[DT] = [D][T]$ and $[TD] = [T][D]$
now I found $[T]$ to be 
\begin{pmatrix}
0 &  0 &  0\\
1  &0   &0\\
0 & .5   &0\\
0&   0   &\frac 13\\
\end{pmatrix}
I cant figure out the rest though... is $[D]$ supposed to be a $3\times3$? or did I to $[T]$ totally wrong?
What exactly to they mean by $TD$?
Thanks guys
 A: You did it right for $T$. I guess this a homework problem, so let me give some help - but only hints. I will go through how to get $T$ (as you did it already) and then I will hint how to get $D$.
Hints
(1) Let us first clarify some points. Using your definitions, $T$ maps from a $3$-dimensional vector space to a $4$-dimensional one, hence $[T]$ should be $3 \times 4$ matrix. Consider the basis vectors $v(0)=1$, $v(1)=t$, $v(2)=t^2$, $v(3)=t^3$. Now $\{v(0), v(1), v(2) \}$ forms a basis for $V_2$, and $\{v(0), v(1), v(2), v(3) \}$ forms a basis for $V_3$.
Since we have that $T(v(0))=v(1)$, $T(v(1))=\tfrac{1}{2}v(2)$, $T(v(2))=\tfrac{1}{3}v(3)$ 
the matrix form of $T$, i.e. $[T]$, is as you wrote
\begin{pmatrix}
0 &  0 & 0\\
1  &0  & 0\\ 
0 & \tfrac{1}{2} & 0\\
0 & 0 & \tfrac{1}{3} 
\end{pmatrix}.
(2) On the other hand, since $D$ maps from a $4$-dimensional vector space to a $3$-dimensional one, hence $[D]$ will be a  $4 \times 3$ matrix. To get the entries of $[D]$ you have to calculate $D(v(0))$, $D(v(1))$, $D(v(2))$ and $D(v(3))$. For example, since $D(v(0))=0$ this means that the first column of $[D]$ consists only of zeros; while $D(v(1))=v(0)$ implies that the second column starts with $1$ and the rest of the entries in that column are zero. You can similarly find the other columns.
Can you find now the full matrix $[D]$? 
(3) When you know the answer to (2), let us consider the maps $TD$ and $DT$. 
Since $T$ is a map from $V_2$ to $V_3$, and $D$ is a map from $V_3$ to $V_2$, 
$DT$ will be a $V_2 \to V_2$ map, and $TD$ will be a $V_3 \to V_3$ map. Hence $[DT]$ will be a $3 \times 3$ matrix, while $[TD]$ will be a $4 \times 4$ matrix. 
To find $[DT]$ you only need to calculate $DT(v(0))$, $DT(v(1))$ and $DT(v(2))$. For example, since $T(v(0))=v(1)$ and $D(v(1))=v(0)$, we have that $DT(v(0))=D(v(1))=v(0)$.
This means that the first column of the $3 \times 3$ matrix will start with $1$ and the rest of the entries in the first column are zero. You can also similarly find the other columns of $[DT]$ (and also of $[TD]$).
If you are done with finding $[DT]$, consider the matrix multiplication
$[D][T]$. Since $[D]$ is a $4 \times 3$ matrix and $[T]$ is a 
$3 \times 4$ matrix, $[D][T]$ will be a $3 \times 3$ matrix (as $[DT]$!). Perform the matrix multiplication and compare with the previously obtained $[DT]$ matrix. (And then do the same procedure once more: evaluate the matrix multiplication $[T][D]$ and compare that to matrix $[TD]$.)
