# Math Help - building a transformation from the image

1. ## building a transformation from the image

give an example for a transformation T:R^3 -> R^3
in which $(ImT)^{\perp}=\{(x,y,z)|x-3y+2z=0$

and find T(x,y,z) formula

??

i solvled this x-3y+2z=0 equation
i got two vectors then i took V=(a,b,c)
and said that V*U1=0 and V*U2=0
and that v is ImT

the result is Im T=sp{(-1,3,0)}

i got 2 thing to do here
the first one:
T(0,0,1)=(-1,3,0)
T(0,1,0)=(-1,3,0)
T(1,0,0)=(-1,3,0)
correct?

regarding the second one:
how to find T(x,y,z) formula
?

2. ## re: building a transformation from the image

$(-1,3,0)$ cannot be in the image of $T$. Consider the element $(1,1,1)\in (\mathrm{im} T)^{\perp}$, for example.

$(-1,3,0)\cdot (1,1,1)=2\neq 0\Rightarrow (-1,3,0)\notin \mathrm{im} T$.

The correct answer, I believe, is that $\mathrm{im} T=\mathrm{span}((1,-3,2))$. I got this just by writing out a few elements of $(\mathrm{im} T)^{\perp}$, noting that if an element is in the image of $T$ then its dot product with each of those elements must be 0, and solving a system of equations based on this fact.

As far as finding a formula for $T$, I'm not 100% sure what to do. I mean, we can define a lot of different linear transformations whose image is $\mathrm{span}((1,-3,2))$. I guess just pick one.

3. ## re: building a transformation from the image

The set $\{(x,y,z)|x-3y+2z=0\}$ is just the set of vectors (x,y,z) which have zero dot product with the vector (1,-3,2), in other words, the orthogonal complement of <1,-3,2>. You want a linear transformation with that set as its orthogonal complement, that is, you need the image to be <1,-3,2>. Just send one or more basis elements to nonzero scalar multiples of (1,-3,2), and send the rest to zero.

4. ## re: building a transformation from the image

ahhhh
t(0,0,1)=(1,-3,2)
t(0,1,0)=(0,0,0)
t(1,0,0)=(0,0,0)
correct?

5. ## re: building a transformation from the image

Yes, that's one correct choice.