Given are two ordered bases $B$ and $C$ of $\mathbb{R}^3$: $B = \left\{\begin{pmatrix} 8\\ -6\\ 7 \end{pmatrix}, \begin{pmatrix} -16\\ 7\\ -13 \end{pmatrix}, \begin{pmatrix} 9\\ -3\\ 7 \end{pmatrix}\right\}, C = \left\{ \begin{pmatrix} 1\\ -2\\ 1 \end{pmatrix}\begin{pmatrix} 3\\ -1\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ 1\\ 2 \end{pmatrix} \right\}$ and a linear mapping $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ with following matrix in terms of basis $B$: $\,\,\,\, A^{f}_{BB}= \begin{pmatrix} 1 & -18 & 15\\ -1 & -22 & 15\\ 1 & -25 & 22 \end{pmatrix}$

Determine the transformation matrices $A^{f}_{BC}$ and $A^{f}_{CB}$ each in two different ways (by using $A_{BB}^{f}$ and $A_{CC}^f$).

So I don't want do this / see it too much in detail with every single step calculated because it would be too long.

I know we need the change of basis $T^C_B$ and $T^B_C$ but let's not calculate them and just keep them like that.

Then for the first transformation matrix we have

$A^f_{BC} = T^C_B \cdot A^f_{BB} \cdot T^B_C, \,\,\,\,\,\,\,$ for the other possibility we have

$A^f_{BC} = T^C_B \cdot A^f_{CC} \cdot T^B_C$

and for

$A^f_{CB} = T^C_B \cdot A^f_{BB} \cdot T^B_C$

but I think that's not correct like that because I get weird results? : / as example, for the first one $A^f_{BC}$ I get the same result as for $A^f_{CC}$..


Your first formula, $$ A^f_{BC} = T^C_B \cdot A^f_{BB} \cdot T^B_C, \,\,\,\,\,\,\, $$ can be read, left to right, as "Take a vector in C-coordinates; convert to $B$-coordinate. Apply the matrix that represents $f$ in $B$ coordinate system, producing a vector of $B$-coordinates. Now finally transform those to $C$ coordinates."

That's a perfect description of $A_{CC}$, which is why it matches. If you want $A_{BC}$ (which I'm guessing transforms vectors in $C$ coords by $f$, presenting the result in $B$ coords, that'd be just $$ A^f_{BC} = A^f_{BB} \cdot T^B_C, \,\,\,\,\,\,\, $$

As usual with these things, you can kind of "cancel" adjacent similar things, so when you write "$A_{BB} T^B_C$, you can cancel the second $B$ with the third (upper) $B$ to get a $B$ and a $C$ left over, hence you've probably got the matrix $A_{BC}$. Practice with this (and a few years' experience) make this seem as natural as the chain rule, but at first it's just weird.


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