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Let $f$$ \begin{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} \end{pmatrix} $= $ \begin{pmatrix} x_1+2x_2+x_3 \\ x_1-2x_2 \\ x_1+x_3 \\ 3x_1-4x_2 \\ \end{pmatrix} $ be a linear map from $\Bbb{R}^3 \to \Bbb{R}^4$

Let $v_1= \begin{pmatrix} 1 \\ 0 \\ 1 \\ \end{pmatrix} $ $v_2= \begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} $ $v_3= \begin{pmatrix} 1 \\ 0 \\ 0 \\ \end{pmatrix} $ and $w_1= \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0\\ \end{pmatrix} $ $w_2= \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1\\ \end{pmatrix} $ $w_3= \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0\\ \end{pmatrix} $ $w_4= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 2\\ \end{pmatrix} $ Let $B=(v_1,v_2,v_3)$ and $ C=(w_1,w_2,w_3,w_4)$

Find the matrix $M_{B,C}(f)$

Hi so this is the question . What i did was i took a vector $v_1$ i applied f to it and got a vector like $(2,1,2,3)^T$ then i tried to find the coefficeients of the linear combination from $f(v_1)=aw_1+bw_2+cw_3+dw_4$ and solved the system of equation. I put the solutions as coloumns in my matrix. I repeated this 3 times and got a matrix like

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

My question is how can i check if what i did was correct ?

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    $\begingroup$ It is the right approach, but you might need to check you work. I have $(2,1,0,1)^T$ in the first column. $\endgroup$ – Doug M Feb 2 '17 at 23:14
  • $\begingroup$ ok thanks, how can i check if the matrix is the right one ? $\endgroup$ – asddf Feb 2 '17 at 23:15
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Here is another way to do it:

$B$ is the representation of the principle vectors of B in the standard basis.

$B^{-1}$ is then the representation of the principle vectors in the standard basis as represented in the basis of $B.$

We can do a similar exercise for $C.$

$fB$ would take the principle vectors in $B$ to a matrix in the standard basis of $\mathbb R^4.$

Then we have to convert a matrix in the standard basis into the basis for $C.$

$C^{-1}fB$

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