# Linear maping, maybe?

• May 14th 2008, 07:51 AM
Linear 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.

• May 14th 2008, 08:01 AM
Moo
Hello,

Quote:

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.

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 AM
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 AM
Moo
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".

Quote:

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

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}\$
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 :D