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Let $$f : \Bbb R^3 \to \Bbb R^4; $$ $$ (a,b,c) \mapsto (a,b,c,a+b+c) $$ Furthermore let $A =\{v_1,v_2,v_3\},B =\{w_1,w_2,w_3,w_4\}$ be bases of $ \Bbb R^3$ and $\Bbb R^4$ respectively where

$v_1=(3,4,4),v_2=(1,1,1),v_3=(0,-7,6)$

$w_1=(2,3,2,-1),w_2=(1,-2,3,-4),w_3=(0,0,0,1),w_4=(-3,8,7,7)$

Find the matrix representation $M^B_A(f)$.

Okay, What I have to do is:

1) Find $M_A^K(f)$ Where K is the standard basis.

2) Put all the column vectors of $M_A^K(f)$ as a linear combination of the vectors in B.

3) Put the coefficients for that linear combination in a new matrix.

I am having trouble with part 2.


$M_A^K = \begin{pmatrix} 3 & 1 & 0 \\ 4 & 1 & -7 \\ 4 & 1 & 6 \\ 11 & 3 & -1 \\ \end{pmatrix}$

since $f(3,4,4) = (3,4,4,11)$ and so on. Now I must use all the column vectors here to find a linear combination for each one of the the vectors in B. Already for the first one I run into trouble given that

$$ \left[ \begin{array}{ccc|c} 3 & 1 & 0 & 2 \\ 4 & 1 & -7 & 3 \\ 4 & 1 & 6 & 2 \\ 11 & 3 & -1 & -1 \\ \end{array} \right] $$ has no solutions. I don't know what I missed. I checked and $v_i$ and $w_i$ are indeed linearly independent so it's not that. I should also add that I made the exercise up so there could be something intrinsically wrong with the question.

What did I do wrong?


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    Yes, I think there is something fundamentally wrong with your exercise. Your image maps onto a completely different vectorspace. So when you try to get $\begin{pmatrix} 2\\3\\2\\-1\end{pmatrix}$ as a linear combination of your 3 vectors, they can't "portray" it.2017-01-29
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    If you want an extended answer with visual representation (I will have to shorten your dimensions for that though) let me know2017-01-29

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