Let T : R3 -> R2 be given by

T((x,y,z)) = (x+2y+z, x-y+z).

For part(a),Can i say something like because no matter what value of x,y,z I choose, there is always an unique solution for (x+2y+z, x-y+z).

For part(b), I am not too sure how to prove for each b in R2 is the image of at most one a in R3.

  • $\begingroup$ Keep in mind that your function may not be onto or one-one. $\endgroup$ – Brad Aug 14 '14 at 19:41

For part (a), no, you need to say "for every $(u,v) \in \mathbb R^2$, there is $(x,y,z)$ so that $(x+2y+z, x-y+z) = (u,v)$", and then prove it (if it is indeed onto). If it is not onto, you need to find a $(u,v)$ that can't be written $(x+2y+z, x-y+z)$.

(If it is onto, one way to prove this is to give an explicit formula for $x,y,z$ in terms of $u,v$)

For part (b), "one-to-one" is equivalent to "$T(x,y,z) = (0,0)$ only if $(x,y,z) = (0,0,0)$". (because this map is linear)

You may find it helpful to write the matrix of this linear transformation; the problem may look more familiar to you.

  • $\begingroup$ So for part a, it is not onto, as I am only able to write y in form of u,v not x and z. for part b, it is one - one, as 0 +2 * 0 + 0 = 0 and 0 - 0 + 0 = 0. Am i correct this time? $\endgroup$ – user164945 Aug 14 '14 at 20:05
  • $\begingroup$ No: You can explicitly solve for $x,y,z$ in terms of $u,v$ (though it is not unique). Consider $x = \dfrac13 (u + 2v), y = \dfrac13 (u -v), z = 0$ (there are other solutions). $\endgroup$ – BaronVT Aug 14 '14 at 20:11
  • $\begingroup$ For part (b), you have shown that $T(0,0,0) = (0,0)$ which is to be expected one way or the other (this is true of every linear transformation). You have to find a non-zero $(x,y,z)$ that does this (hint, start by taking $y=0$, then you just need to choose $x,z$ so that $x + z = 0$.) $\endgroup$ – BaronVT Aug 14 '14 at 20:12

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