1
$\begingroup$

$T:R^3 \to R^2$, where $T(x,y,z)=(x+y+z, x-y-z)$

So I ask myself can every point in $R^2$ be found with the inputs from $R^3$. So the result in $R^2$ could be seen as this system of linear equations:
$$\begin{array}{c} x+y+z = b_1 \\ x-y-z = b_2 \end{array} $$

and this system has more unknowns than equations so it has infinite solutions I believe, so this is surjective. But i'm also thinking that it cannot be injective because $\text{dim}(R^3) \gt \text{dim}(R^2)$.
Would this be correct?

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes, since the system always has a solution: $x = \dfrac{b_1+b_2}{2}$. And you can let $z = 0$, then $y = \dfrac{b_1-b_2}{2}$.

$\endgroup$
1
  • $\begingroup$ Thankyou @DeepSea $\endgroup$
    – Bucephalus
    Commented Oct 1, 2017 at 2:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .