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I have a linear mapping from one basis to another $ F: \mathbb R^{3} \rightarrow \mathbb R^{4}$ with the bases being: $$ \mathbf v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ \end{pmatrix}, \mathbf v_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \\ \end{pmatrix}, \mathbf v_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix} $$

and

$$ \mathbf w_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \\ \end{pmatrix}, \mathbf w_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ \end{pmatrix}, \mathbf w_3 = \begin{pmatrix} 0 \\ 0 \\ -1 \\ 0 \\ \end{pmatrix}, \mathbf w_4 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 2 \\ \end{pmatrix} $$ The transofmration is then given by $F( \mathbf v_1)=\mathbf w_4, F(\mathbf v_2) = \mathbf w_2 + \mathbf w_3, F(\mathbf v_3)=\mathbf w_1 $. I should then represent $F$ with respect to the two relevant standard basis. I started out naively by simply writing the transformation matrix with respect to the three given basis vectors. $$ D = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \\ 2 & 1 & 0 & \end{pmatrix} $$ I then recalled the relationship for transformation matrix with respect to another basis. $$ D = C^{-1}*A*C $$ Where A is the transformation with respect to the standard basis, C´s columns represent second pair basis vectors and D is then finally the transformation with respect to this second pair of the basis vectors. So far so good. I thought that I just need to go in the other direction - I know the transformation with respect to the non-standard bases, so I just need to figure out what A looks like. So the result should become simply $$ A = C*D*C^{-1} $$ However, this does not seem to be the correct approach, since upon multiplying the matrices the dimensions do not work out! Can somebody spot an error in my approach? Does the above relationship work only for matrices with the same dimensions?

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  • $\begingroup$ Where do you want to change the basis ? For determining the matrix corresponding to a linear map, both basis of domain and codomain come into picture. $\endgroup$ – Mambo Apr 24 '16 at 21:35
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Your matrix $D$ that takes in $v$ and outputs $w$ should be with respect to the ordered basis $w$, not the column vectors of $w$. I.e., if the transformation is as you have described, it should be:

given by $F( \mathbf v_1)=\mathbf w_4, F(\mathbf v_2) = \mathbf w_2 + \mathbf w_3, F(\mathbf v_3)=\mathbf w_1 $

$$D = \begin{pmatrix}0&0&1\\0&1&0\\0&1&0\\1&0&0\end{pmatrix}$$

The transformation matrix (in this case, it is coordinate, I assume) is uniquely determined coefficients of the transformation with respect to the basis and contains the coefficients of $w$ used to generate $F(v_i)$, not the column vectors of $w$ itself. Notice that: $$F(v_1) = 0w_1 + 0w_2 + 0w_3 + 1w_4$$

The rest follow easily.

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  • $\begingroup$ Ok, I guess I understand now - it is just a matter of redefining the new basis vectors as the standard basis vectors. The matrix you wrote then represents the transformation matrix with respect to the given bases vectors. The thing I wrote, on the other, seems to be then the same transformation matrix with respect to the standard basis, right? $\endgroup$ – whypi314 Apr 25 '16 at 13:45
  • $\begingroup$ Yes, that's exactly right $\endgroup$ – q.Then Apr 25 '16 at 14:02

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