# 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?

• No [[1,-2,0,1],[-2,5,1,-4],[1,0,2,-3]] – JP McCarthy Mar 22 '17 at 13:30

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:

$$\begin{pmatrix} x_1-2x_2+x_4\\ -2x_1+5x_2+x_3-4x_4\\ x_1+2x_3-3x_4 \end{pmatrix}$$

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 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.

You can not ignore $x_4$ !! The right matrix is

$\begin{pmatrix} 1 & -2 & 0 & 1\\ -2 & 5 & 1 &-4\\ 1 & 0 & 2 & -3 \end{pmatrix}.$