Why do we express linear transformation matrices as transposes?

Question

So for instance, if we let $D: V \to V$ be a map defined by the derivative, $f_1 =cos(t)$ and $f_2 = sin(t)$, and B be the basis ${f_1, f_2}$ why is it that my book expresses the matrix associated with this transformation as

\begin{bmatrix}0&1\\-1&0\end{bmatrix}

but when you multiply this matrix with, say, \begin{bmatrix}cos(t)\\sin(t)\end{bmatrix}

this is not the derivative of those elements? Why doesn't the matrix

\begin{bmatrix}0&-1\\1&0\end{bmatrix}

make more sense?

I hope this question exposes a fundamental misunderstanding on my part. It seems like it'd be far more sensible to express transformation matrices such that when applied to the basis of the departure space or domain, we see it in terms of the arrival space's basis.

Thanks.

Answer 1

In a vector space, a vector and its coordinates are two different things.

The multiplication of the representation matrix and $[\cos t, \sin t]^T$ does not make sense.

Note that $$ Df_1=-f_2,\quad Df_2=f_1 $$ and the coordinates of $Df_1$ and $Df_2$ with respect to the base $B$ are $$ [0,-1]^T,\quad [1,0]^T. $$ And this is where the matrix $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ in your book from.