$T(a_1,a_2,a_3) = (3a_1+a_2,a_1+a_3,a_1-a_3)$. $(a_1,a_2,a_3)^T$ is written with regards to the standard basis.
We can figure out $T$ in matrix form by calculating $T(a_1),T(a_2), T(a_3)$. That's easy and we get:
$T$ = $\left(\begin{array}{ccc}3 & 1 & 0 \\1& 0 & 1 \\1 & 0 & -1\end{array}\right)$
$\beta$ is a new basis = $\{(1,0,0), (1,1,0), (1,1,1)\}$.
Here's what I'm trying to find: $T$ relative to this new basis.
I have three questions:
First, How do I methodically set up the problem and solve it?
Second, why does $T*C = S$ work?, where
$T$ is the transformation matrix relative to $(a_1,a_2,a_3)^T$,
$C = \left(\begin{array}{ccc}1 & 1 & 1 \\0& 1 & 1 \\0 & 0 & 1\end{array}\right)$.
and $S$ is the solution = $\left(\begin{array}{ccc}3 & 4 & 4 \\1& 1 & 2 \\1 & 1 & 0\end{array}\right)$.
Third Question: how general is $T*C$ as a solution? Did I just get lucky? It seems all I'm doing is multiplying a new vector (a vector in $\beta$) by an old transformation matrix. Why would this give me a new, correct $T$? Basically, if someone asks me to find $T$ relative to a new basis can I just use $T*C$, where $C$'s columns contain the new basis vectors?
Edit: How do you do this without using inverses?
Best Answer
Let
$\mathcal{E}=\left\{e_1,e_2,e_3\right\}$
be our canonical base. With this base, transormation T has representation
$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.
Now we have got a new base:
$\mathcal{F}=\left\{e_1,e_1+e_2,e_1+e_2+e_3\right\}$.
Let
$M_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right)$
be the transition between the two bases.
Then canonical coordinates are transormed in new coordinates (with respect to base $\mathcal{F}$ ) by inverse matrix, which is
$N_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{array} \right)$.
Take
$A=\left\{a_1,a_2,a_3\right\}$
and get new coordinates
$B=N_{\mathcal{F}}.A$.
Then, with $S=T.M_{\mathcal{F}}$
we see:
$T.A=T.M_{\mathcal{F}}.N_{\mathcal{F}}.A=S.B$.
It's not a miracle, only lin. Algebra. Key is transformation of basis, which implies transformation of coordinates. That's all.
By the way: Calculating without inverses is not possible. Your transformation with bases must be regular. They must be invertible, otherwise it didn't work.
Let's see. Other basis
$\mathcal{B}=\left\{2 e_1+5 e_3,e_1+e_2+6 e_3,3 e_1+9 e_3\right\}$,
another transition:
$M_{\mathcal{B}}=\left( \begin{array}{ccc} 2 & 1 & 3 \\ 0 & 1 & 0 \\ 5 & 6 & 9 \\ \end{array} \right)$.
The inverse:
$N_{\mathcal{B}}=\left( \begin{array}{ccc} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -\frac{5}{3} & -\frac{7}{3} & \frac{2}{3} \\ \end{array} \right)$.
Old transformation T
$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.
Transformed T:
$S=T.M_{\mathcal{B}}=\left( \begin{array}{ccc} 6 & 4 & 9 \\ 7 & 7 & 12 \\ -3 & -5 & -6 \\ \end{array} \right)$
Transformed A:
$B=N_{\mathcal{B}}.A$.
$T.A=T.M_{\mathcal{B}}.N_{\mathcal{B}}.A=S.B$
Like before.