Linear maping, maybe?

**Deadstar** · May 14th 2008, 07:51 AM

Let $S: R^3 -> R^2$ and $T:R^2 -> R^3$ be defined by

$S(x_1, x_2, x_3) = (x_2 + x_3, x_1)$ , $T(x_1, x_2) = (x_2, x_1, x_1 + x_2)$

Find expressions for ST and TS.

Any help please?

**Moo** · May 14th 2008, 08:01 AM

Hello,

Originally Posted by Deadstar

Let $S: R^3 -> R^2$ and $T:R^2 -> R^3$ be defined by

$S(x_1, x_2, x_3) = (x_2 + x_3, x_1)$ , $T(x_1, x_2) = (x_2, x_1, x_1 + x_2)$

Find expressions for ST and TS.

Any help please?

Are you looking for the composition of S with T and T with S ? o.O

Let's see for $SoT(x_1,x_2)$ :

$T(x_1,x_2)=(x_2, x_1, x_1 + x_2)$

---> $SoT(x_1,x_2)=S(T(x_1,x_2))=S(x_2, x_1, x_1 + x_2)$

We know that $S(m,n,p)=({\color{red}n+p},{\color{blue}m})$ .
I renamed this on purpose, because it would confuse you.
Here :
$m=x_2$ (1)
$n=x_1$ (2)
$p=x_1+x_2$ (3)

Therefore ${\color{red}n+p}=(2)+(3)=\boxed{2x_1+x_2}$
And $\boxed{{\color{blue}m}=x_2}$

Hence :

$S(x_2, x_1, x_1 + x_2)=(2x_1+x_2, \ x_2)$

---> $\boxed{SoT(x_1, \ x_2)=(2x_1+x_2, \ x_2)}$

Is it what you wanted ? Does it help ?

**Deadstar** · May 14th 2008, 10:51 AM

Yeah i think i get it now.

I dont know how to explain it exactly...

Is this right...
Since $m = x_2$ (1) for example...

Any $x_1$ in the equation $(x_2 + x_3, x_1)$ must be replaced by $m = x_2$ ...

Dunno if that sounds right the way ive explained it but it works when i try it cos i got $TS = (x_1, x_2 + x_3, x_1 + x_2 + x_3)$ which was the right answer!

**Moo** · May 14th 2008, 10:58 AM

Hmm how to explain... Actually, I replace by m, n and p because it was redundant and really confusing.
It's equivalent to say "the first coordinate of the image of a triplet by T is the sum of its two last coordinates".

Originally Posted by Deadstar

Any $x_1$ in the equation $(x_2 + x_3, x_1)$ must be replaced by $m = x_2$ ...

It sounds strange to me... Because once you replace with m, n or p, you don't have to say "I'll replace it by...".

We know that $S(m,n,p)=({\color{red}n+p},{\color{blue}m})$ .
I renamed this on purpose, because it would confuse you.
Here :
$m=x_2$ (1)
$n=x_1$ (2)
$p=x_1+x_2$ (3)

Therefore ${\color{red}n+p}=(2)+(3)=\boxed{2x_1+x_2}$
And $\boxed{{\color{blue}m}=x_2}$

This part was more for explaining the thing to you than for writing the correct answer.
You have your own way to explain it. I don't think your teacher would bother that much if you showed it your way. But make it understandable

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