TITLE: Transformation matrices in a basis
QUESTION [0 upvotes]: Let $F=\mathbb{R}$,$X=\mathbb{R^3}$ and $Y=\mathbb{R^2}$. Further $B_X$ and $B_Y$ are given by:
$B_X:=\{(1,0,0),(1,0,-1),(1,-1,-1)\}$
$B_Y:=\{(1,0),(1,-1)\}$
Let $f:X\rightarrow X$ and [...] be given as:
$f(x,y,z)=(y,x-z,x+z)$ $x,y,z\in F$
Calculate the transformation matrix (ger: "Darstellungsmatrix") $A=M(f,B_X,B_X)$ and [...]
so i just plug in all vectors of $B_X$ in $f$ and see what i get?
$f(1,0,0)=(0,1,1)\\
f(1,0,-1)=(0,2,0)\\
f(1,-1,-1)=(-1,2,0)\\$
Giving me $A=\{(0,1,1),(0,2,0),(-1,2,0)\}$ (in Matrix-form)
Is this it already?
And what do i have to do when calculating $B=M(g,B_X,B_Y)$, while
$g(x,y,z)=(2x-2y,x+y)$ ? Do i just plugin $B_X$ in g, and see what i get?

REPLY [1 votes]: Correcting your method: You're almost there.  But you found the components of the image vectors with respect to the standard basis -- not $B_X$.  So now you just need to find the components of those three vectors with respect to (wrt) $B_X$.  For instance $(0,2,0) = 0(1,0,0)+2(1,0,-1)-2(1,-1,-1)$.  So the components of this vector wrt $B_X$ are $(0,2,-2)$.  This will then be the second column of your matrix.

Transition matrix method: Let $\mathcal E$ be the standard basis for $X$.  Then we know that $$f(x,y,z)=(y,x-z,x+z) \iff [f]_{\mathcal E\leftarrow \mathcal E} = \pmatrix{0 & 1 & 0 \\ 1 & 0 & -1 \\ 1 & 0 & 1}$$
Now we construct the transition matrix $I_{\mathcal E \leftarrow B_X}$.
Luckily we're already given the components of the elements of $B_X$ in terms of $\mathcal E$.  So 
$$I_{\mathcal E \leftarrow B_X} = \pmatrix{1 & 1 & 1 \\ 0 & 0 & -1 \\ 0 & -1 & -1}$$
Then to find $I_{B_X \leftarrow \mathcal E}$, we simply find the inverse of this matrix:
$$I_{B_X \leftarrow \mathcal E} = {I_{\mathcal E \leftarrow B_X}}^{-1} = \pmatrix{1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & -1 & 0}$$
Then to find $[f]_{B_X \leftarrow B_X}$ we just multiply:
$$[f]_{B_X \leftarrow B_X} = I_{B_X \leftarrow \mathcal E}[f]_{\mathcal E \leftarrow \mathcal E}I_{\mathcal E \leftarrow B_X} = \pmatrix{1 & 0 & -1 \\ 0 & 2 & 2 \\ -1 & -2 & -2}$$

You can use a similar approach to find $[g]_{B_Y\leftarrow B_X}$ by multiplying out:
$$[g]_{B_Y\leftarrow B_X} = I_{B_Y \leftarrow \mathcal E_2}[g]_{\mathcal E_2\leftarrow \mathcal E_3}I_{\mathcal E_3 \leftarrow B_X}$$
where $\mathcal E_2$ is the standard basis for $\Bbb R^2$ and $\mathcal E_3$ is th standard basis for $\Bbb R^3$.
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