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This is the second part of a problem I've meaning to solve, here's the First part

$B = \{v_1,v_2,v_3\}\ and\\ C=\{w_1,w_2,w_3,w_4\}$

$$A= \begin{pmatrix} 1 & 1 & 2 \\ 1 & -1 & 0 \\ 1 & 2 & 3 \\ -1 & 3 & 2 \\ \end{pmatrix} $$

Being A the matrix associated to the linear transformation that takes vectors from $B$ and returns a vector's coordinates in the base $C$

$Let\ B'\ be\ an\ ordered\ basis\ such\ that\\ B' = \{v_1,2v_2+v_3,v_2+v_3\}\ and\\ C'=\{w_1,w_2,w_3+w_4,w_3-w_4\} \\ Calculate\ |f|_{B'C'}$

My confusion comes from the basis, which is composed of linear combinations of vectors. Normally if I would like to find a change of basis matrix, I would replace each vector from the first base, in my linear transformation, then find it's coordinates in the other base, and assemble the matrix. I'm not sure about how to proceed.

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  • $\begingroup$ When in doubt, go back to definitions. What do the coordinates of a vector relative to some basis represent? $\endgroup$ – amd Apr 29 at 6:32
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If $\mathcal{B}$ and $\mathcal{C}$ are bases for finite-dimensional vector spaces $V$ and $W$, and $T: V \to W$ is a linear map, one produces $[T]_{\mathcal{B},\mathcal{C}}$ by the following algorithm: take the $j$th vector in $\mathcal{B}$, apply $T$ to it, write the result as a combination of the elements in (the ordered) basis $\mathcal{C}$, and place the coefficients in the $j$th column of $[T]_{\mathcal{B},\mathcal{C}}$.

So, for example, in your case: the first column of the matrix $A = [f]_{\mathcal{B},\mathcal{C}}$ says that $f(v_1) = w_1+w_2+w_3-w_4$. Changing basis in a space amounts to composing with an extra identity function. Namely, since $f = {\rm Id}_{\Bbb R^4} \circ f \circ {\rm Id}_{\Bbb R^3}$, one can write that $$[f]_{\mathcal{B}',\mathcal{C}'} = [{\rm Id}_{\Bbb R^4}]_{\mathcal{C},\mathcal{C}'}[f]_{\mathcal{B},\mathcal{C}}[{\rm Id}_{\Bbb R^3}]_{\mathcal{B}',\mathcal{B}} = [{\rm Id}_{\Bbb R^4}]_{\mathcal{C}',\mathcal{C}}^{-1}[f]_{\mathcal{B},\mathcal{C}}[{\rm Id}_{\Bbb R^3}]_{\mathcal{B}',\mathcal{B}}. $$To find these representations of the identity transformation, one follows the same recipe. For example, to find $[{\rm Id}_{\Bbb R^3}]_{\mathcal{B}',\mathcal{B}}$, one takes the elements of $\mathcal{B}'$, applies ${\rm Id}_{\Bbb R^3}$ (that is to say, does nothing), then writes the result as a combination of the elements in $\mathcal{B}$ (this is trivial, the elements of $\mathcal{B}'$ were defined as certain combinations of the elements in $\mathcal{B}$), and finally place the coefficients in columns.

That being understood, you can compute

$$[f]_{\mathcal{B}',\mathcal{C}'} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{pmatrix}^{-1}\begin{pmatrix} 1 & 1 & 2 \\ 1 & -1 & 0 \\ 1 & 2 & 3 \\ -1 & 3 & 2 \\ \end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1\end{pmatrix}$$

yourself. Also note that whoever wrote this problem tried to be your friend here: inverting the order $4$ matrix above is easy, since it is has block-diagonal form (i.e., invert each block).

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  • $\begingroup$ Thanks a lot! The property you applied,does it have a special name? $\endgroup$ – iLu Anzu Apr 29 at 12:30
  • $\begingroup$ It didn't have a name, but I found it in a book, at it even has a nice representation look $\endgroup$ – iLu Anzu Apr 29 at 12:55
  • $\begingroup$ Yes, this is exactly the picture I have in mind when thinking about those problems. $\endgroup$ – Ivo Terek Apr 29 at 16:31

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