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

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

Find expressions for ST and TS.

Any help please?

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- May 14th 2008, 07:51 AMDeadstarLinear maping, maybe?
Let $\displaystyle S: R^3 -> R^2$ and $\displaystyle T:R^2 -> R^3$ be defined by

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

Find expressions for ST and TS.

Any help please? - May 14th 2008, 08:01 AMMoo
Hello,

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

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

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

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

We know that $\displaystyle S(m,n,p)=({\color{red}n+p},{\color{blue}m})$.

I renamed this on purpose, because it would confuse you.

Here :

$\displaystyle m=x_2$ (1)

$\displaystyle n=x_1$ (2)

$\displaystyle p=x_1+x_2$ (3)

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

And $\displaystyle \boxed{{\color{blue}m}=x_2}$

Hence :

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

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

Is it what you wanted ? Does it help ? - May 14th 2008, 10:51 AMDeadstar
Yeah i think i get it now.

I dont know how to explain it exactly...

Is this right...

Since $\displaystyle m = x_2$**(1)**for example...

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

Dunno if that sounds right the way ive explained it but it works when i try it cos i got $\displaystyle TS = (x_1, x_2 + x_3, x_1 + x_2 + x_3)$ which was the right answer! - May 14th 2008, 10:58 AMMoo
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".

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...".

Quote:

We know that $\displaystyle S(m,n,p)=({\color{red}n+p},{\color{blue}m})$.

I renamed this on purpose, because it would confuse you.

Here :

$\displaystyle m=x_2$ (1)

$\displaystyle n=x_1$ (2)

$\displaystyle p=x_1+x_2$ (3)

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

And $\displaystyle \boxed{{\color{blue}m}=x_2}$

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 :D