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Let $V=\mathbb{R}^4$, let $W=\mathbb{R}^3$ and let $\phi$ be the linear map $$\left(x,y,z,t\right)\rightarrow \left(x-2z+t,2y+z,x+4y+t\right)$$

Write down the matrix of A of $\phi$ with respect to the standard bases of $\mathbb{R}^4$ and $\mathbb{R}^3$.

Could someone please help me with the method for finding the matrix A.

I understand we use the standard bases of $\mathbb{R}^4$: $$e_1=\left(1,0,0,0\right), e_2=\left(0,1,0,0\right), e_4=\left(0,0,1,0\right), e_4=\left(0,0,0,1\right)$$ and the standard bases of $\mathbb{R}^3$: $$f_1=\left(1,0,0\right), f_2=\left(0,1,0\right), f_3=\left(0,0,1\right)$$

But I'm not sure how.

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You apply $\phi$ to the base vectors $e_i$ and express it in terms of the base factors $f_i$: $$ \phi(e_1) = (1, 0, 1) = f_1 + f_3 \\ $$ etc. In matrix form
$$ (\phi(e_j))_i = \sum_{j=1}^3 a_{ij} f_j $$ this gives $$ A = \begin{pmatrix} 1 & 0 & -2 & 1 \\ 0 & 2 & 1 & 0 \\ 1 & 4 & 0 & 1 \end{pmatrix} $$

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First, we need to check where our basis for $\mathbb{R}^4$ is sent. $$e_1\mapsto (1,0,1)=f_1+f_3$$ $$e_2\mapsto (0,2,0)=2f_2+4f_3$$ $$e_3\mapsto (-2,1,0)=-2f_1+f_2$$ $$e_4\mapsto (1,0,1)=f_1+f_3$$ So, using the matrix "algorithm" $$ A=\begin{bmatrix} 1&0&-2&1\\ 0&2&1&0\\ 1&4&0&1\\ \end{bmatrix}.$$

More precisely: given a linear transformation $T:V\to W$, where $V$ and $W$ are abstract vector spaces (over $\mathbb{F}$) with bases $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$ respectively, the $i^{th}$ column of the matrix representing $T$ with respect to these bases is

$$ \begin{pmatrix} a_{1,i}\\ \vdots\\ a_{m,i} \end{pmatrix}.$$ Here the $a_{j,i}\in\mathbb{F}$ are the coefficients satisfying $$ T(v_i)=\sum_{k=1}^m a_{k,i}w_k.$$

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Hint: Look at the images of the basis elements of $\mathbb{R}^4$. The images will span the image of the map. That is, you can write any element in the image as a linear combination of the images of the basis elements. Thus your matrix will have the images of the basis vectors of $\mathbb{R}^4$ as its columns.

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