Matrix representation relative to bases

[Robb]

Homework Statement

Please see attached file. I'm not quite sure if I'm on the right track here. I think the basis for F is throwing me off as well as T(f). Please advise. Thanks!

Homework Equations

The Attempt at a Solution

 

[GwtBc]

Ok I can see the logic that led you through this attempt but it's not quite correct. First off you have a 3x3 matrix for a transformation that goes from a 3-D space to a 2-D one. This is already troublesome. You want your number of columns to be equal to $\dim{U}$ and the number of rows to be equal to the dimension of your target space i.e. $\dim{V}$. So in this case you want a 2x3 matrix.

As for the entries of the matrix, each column is directed by letting $T$ act on a basis vector from $U$. For example, if we have the vector $\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}$ which represents a constant polynomial equal to $1$, then $T( \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix})$ gives us $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ as represented by the basis $F$ in the space $V$.

 

[Robb]

Ok I can see the logic that led you through this attempt but it's not quite correct. First off you have a 3x3 matrix for a transformation that goes from a 3-D space to a 2-D one. This is already troublesome. You want your number of columns to be equal to $\dim{U}$ and the number of rows to be equal to the dimension of your target space i.e. $\dim{V}$. So in this case you want a 2x3 matrix.

As for the entries of the matrix, each column is directed by letting $T$ act on a basis vector from $U$. For example, if we have the vector $\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}$ which represents a constant polynomial equal to $1$, then $T( \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix})$ gives us $\begin{pmatrix} 1\\ 0 \end{pmatrix}$ as represented by the basis $F$ in the space $V$.

Ok, I guess I'm unsure of
$T( \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix})$ gives us ##
\begin{pmatrix}
1\\
0

How does this go from a 3x1 to a 2x1 vector. I'm not seeing the math behind it.

 

[GwtBc]

Ok, I guess I'm unsure of
$T( \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix})$ gives us ##
\begin{pmatrix}
1\\
0

How does this go from a 3x1 to a 2x1 vector. I'm not seeing the math behind it.

That's just how $T$ is defined. It's a transformation that takes a vector from a 3-D space to a 2-D space. In terms of the actual calculation, you let $T$ operate on a basis vectors of E (which is a basis for $U$). So the first one is $\begin{pmatrix} 1\\ 0\\0 \end{pmatrix}$ which represents the polynomial $f(t) = 1$. Then $f(3) = 1$ and $f'(3)=0$, which if represented by the basis vectors of $F$ (which is a basis for $V$) is $\begin{pmatrix} 1\\ 0 \end{pmatrix}$