The discussion focuses on the correct matrix representation for a linear transformation T from a 3-dimensional space U to a 2-dimensional space V. A 3x3 matrix is incorrectly proposed for this transformation, while the correct representation requires a 2x3 matrix, aligning the number of columns with the dimension of U and the number of rows with the dimension of V. The transformation T is defined such that it acts on basis vectors from U, producing corresponding outputs in V, exemplified by the transformation of the vector (1, 0, 0) to (1, 0).
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GwtBc said: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##.
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}##Robb said: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.