# Find matrix corresponding to linear transformation mapping $\mathbb{R}^3$ onto $\mathbb{R}^2$

Trying to review for a quiz and I'm woefully underprepared.

Question is:

For each of the following linear transformations $L$ mapping $\mathbb{R}^3$ into $\mathbb{R}^2$, find a matrix $A$ such that $L(x) = Ax$ for every $x$ in $\mathbb{R}^3$.

a) $L((x_1, x_2, x_3)^T) = (x_1 + x_2, 0)^T$

I don't quite know how to approach this problem. It's probably very simple, but the change from $\mathbb{R}^3 \to \mathbb{R}^2$ or from $\mathbb{R}^2 \to \mathbb{R}^3$ confuses me, I'm not sure how to treat it.

I'm looking for a slow demonstration of steps rather than an actual answer. Calculations are nice, but I'd rather learn the steps and reasoning behind them so I can complete these types of questions on my own later.

• Think about what the matrix $A$ needs to do to a $3$D vector like $(x_1, x_2, x_3)^T$. To begin with, it needs to send it to a $2$D vector $(x_1+x_2, 0)$. So, the matrix needs to be $2 \times 3$. The $2$D resultant vector has it's second part $0$. So, the entire second row of the matrix needs to be made of $0$'s also. Can you proceed reasoning like this? Commented Oct 13, 2015 at 5:51

Since the function takes in a triple and outputs a double, you are looking for a $2$x$3$ matrix such that $$\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{bmatrix}\cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} x_1+x_2 \\ 0 \end{bmatrix}$$

So we have $$a_1\cdot x_1 + b_1 \cdot x_2 + c_1 \cdot x_3 = x_1+x_2$$ $$a_2\cdot x_1 + b_2 \cdot x_2 + c_2 \cdot x_3 =0$$

Therefore, $a_1=1, b_1=1$ and the rest equal $0$.

Since $T: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ you are supposed to get a $2*3$ matrix i.e. 2 rows and 3 columns, so that the matrix acts on vectors from $\mathbb{R}^3$ and gives you vectors in $\mathbb{R}^2$.

First see how the map $T$ acts on the basis elements of $\mathbb{R}^3$, using this we construct the matrix $A$.

Basis for $\mathbb{R}^3$ is $\{(1,0,0)^T,(0,1,0)^T,(0,0,1)^T\}$.

$T((1,0,0)^T)=(1,0)^T$

$T((0,1,0)^T)=(1,0)^T$

$T((0,0,1)^T)=(0,0)^T$

Now in the matrix $A$ place the result of $T$ acting on the basis elements, as the columns of $A$ i.e.

$$A=\begin{bmatrix} 1& 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$