How form a linear mapping $f:\mathbb{R}^{4} \rightarrow \mathbb{R}^{3}$ to a matrix correctly?(solved) Given is $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3$
$f(x)= \begin{pmatrix}
x_1-2x_2+x_4\\ 
-2x_1+5x_2+x_3-4x_4\\ 
x_1+2x_3-3x_4
\end{pmatrix}$
How can I form this to a matrix correctly?
We have $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3$, and we have $x_1,x_2,x_3,x_4$
I think because we go to $\mathbb{R}^3$, we will only have $x_1,x_2,x_3$
So when I form a matrix, I will ignore $x_4$:
\begin{pmatrix}
 1 & -2 &  0\\ 
-2 &  5 &  1\\ 
 1 &  0 &  2
\end{pmatrix}
Is it fine like that?
 A: Given $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3$ defined below:
$$f(x)= \begin{pmatrix}
x_1-2x_2+x_4\\ 
-2x_1+5x_2+x_3-4x_4\\ 
x_1+2x_3-3x_4
\end{pmatrix}$$
is simply described just by left multiplication of the matrix:
$$
\begin{pmatrix}
1 & -2 & 0 & 1 \\
-2 & 5 & 1 & -4 \\
1 & 0 & 2 & -3 
\end{pmatrix}
$$
which was accomplished by taking the coefficients. Your input in the domain is a vector in $\mathbb{R}^{4}$, which you left multiply by the above matrix, and your output in the range space is a vector in $\mathbb{R}^{3}$ (the vector we used to make the matrix in the first place), namely:
$$\begin{pmatrix}
x_1-2x_2+x_4\\ 
-2x_1+5x_2+x_3-4x_4\\ 
x_1+2x_3-3x_4
\end{pmatrix}$$
A: Once your function is from $\Bbb R^4$ to $\Bbb R^3$ your matrix has to able to accepty a vector in $\Bbb R^4$ and give as an answer a vector in $\Bbb R^3$. It means that your matrix must have a dimension $3\times4$.
$A\begin{pmatrix}
 x_1\\ 
 x_2\\ 
 x_3\\
 x_4 
\end{pmatrix}=\begin{pmatrix}
x_1-2x_2+x_4\\ 
-2x_1+5x_2+x_3-4x_4\\ 
x_1+2x_3-3x_4
\end{pmatrix}=\begin{pmatrix}
 1 & -2 &  0 & 1\\ 
-2 &  5 &  1 &-4\\ 
 1 &  0 &  2 & -3
\end{pmatrix}.\begin{pmatrix}
 x_1\\ 
 x_2\\ 
 x_3\\
 x_4 
\end{pmatrix}$  
so
$$A=\begin{pmatrix}
 1 & -2 &  0 & 1\\ 
-2 &  5 &  1 &-4\\ 
 1 &  0 &  2 & -3\end{pmatrix}$$
A: A linear map from $\mathbb{R}^m \rightarrow \mathbb{R}^n$ can be represented by an $n\times m$ matrix. Lets look at a specific example, if $m = 4$ and $n = 3$, then we can represent $f(x)$ by $Ax$ in the following way:
\begin{align}
Ax =
\begin{bmatrix}
a_{1,1} & a_{1,2} &a_{1,3} & a_{1,4}\\
a_{2,1} & a_{2,2} &a_{2,3} & a_{1,4}\\
a_{3,1} & a_{3,2} &a_{3,3} & a_{1,4}
\end{bmatrix}
\begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix}
=
\begin{bmatrix} 
\sum_{j=1}^4a_{1,j}x_j \\
\sum_{j=1}^4a_{2,j}x_j \\
\sum_{j=1}^4a_{3,j}x_j \\
\end{bmatrix} = f(x).
\end{align}
Can you figure out what the entries of $A$ are? 
Hint: you know what the right-most vector is.
A: You can not ignore $x_4$ !! The right matrix is
$\begin{pmatrix}
 1 & -2 &  0 & 1\\ 
-2 &  5 &  1 &-4\\ 
 1 &  0 &  2 & -3
\end{pmatrix}.$
