1. ## Transformation Matrices

"T1(v) = 7v - 7u
T1(u) = -6v + 5u
T2(v) = -3v + 5u
T2(u) = -5v - 6u

Find (T2T1)(v) and (T2T1)(u)"

I know to find the matrix T2T1 I just have to multiply them together, but I'm not sure how to find the transformation matrices for T1 or T2. Can anyone nudge me in the right direction on how to solve this?

2. Are we to assume that u and v are basis vectors for this space?

If not, then I have no idea what something like "T1(v) = 7v - 7u" could mean. If so, set the basis vectors in a specific order, say first u, then v.
Apply the linear transformation to u and v in order:
T1(u)= 5u -6v (notice that because I decided on the order "first u, then v" I wrote u first v second.) That is a linear combination of the basis vectors with coefficients 5 and -6. That is the first column of our matrix.

T1(v)= -7u+ 7v. The second column is the coeffcients -7 and 7. The matrix representing T1 in this ordered basis is
$\displaystyle \begin{bmatrix}5 & -7 \\-6 & 7\end{bmatrix}$

Do you see why that works? u itself is 1u+ 0v and so would be represented, in this ordered basis, by $\displaystyle \begin{bmatrix}1 \\ 0\end{bmatrix}$. Multiplying any two column matrix by that would give exactly the first column. On the other hand, v= 0u+ 1v and so would be represented by $\displaystyle \begin{bmatrix}0 \\ 1\end{bmatrix}$. Multiplying that by any matrix gives the second column.

3. transform basis from $\displaystyle (\vec{e}_1, \vec{e}_2) \rightarrow (\vec{v}, \vec{u}), [T_1(\vec{v})]_{\mathfrak{B}}=7\vec{e}_1-7\vec{e}_2=\begin{bmatrix}7\\-7 \end{bmatrix},[T_1(\vec{u})]_{\mathfrak{B}}=-6\vec{e}_1+5\vec{e}_2=\begin{bmatrix}-6\\5 \end{bmatrix}$
$\displaystyle [T_1(\vec{x})]_{\mathfrak{B}}=\begin{bmatrix}7&-6\\-7&5 \end{bmatrix}[\vec{x}]_{\mathfrak{B}}=A_1[\vec{x}]_{\mathfrak{B}}$
same as $\displaystyle [T_2(\vec{x})]_{\mathfrak{B}}=\begin{bmatrix}-3&-5\\5&-6 \end{bmatrix}[\vec{x}]_{\mathfrak{B}}=A_2[\vec{x}]_{\mathfrak{B}}$

$\displaystyle [T_2(T_1(\vec{x}))]_{\mathfrak{B}}=A_2A_1[\vec{x}]_{\mathfrak{B}}=\begin{bmatrix}14&-7\\77&-60 \end{bmatrix}[\vec{x}]_{\mathfrak{B}}$
$\displaystyle [T_2(T_1(\vec{v}))]_{\mathfrak{B}}=A_2A_1[\vec{v}]_{\mathfrak{B}}=A_2A_1\vec{e}_1=\begin{bmatrix}14\ \77 \end{bmatrix}$ , $\displaystyle T_2(T_1(\vec{v}))=14\vec{v}+77\vec{u}\ ,\ T_1(T_2(\vec{u}))=-7\vec{v}-60\vec{u}$