# Thread: show that T has an inverse

1. ## show that T has an inverse

let T: R^2 -> R^2 be a linear transformation define by

$T \left[ \begin{array}{c}x\\y\end{array} \right] =\left[ \begin{array}{c}x+y\\x-y \end{array} \right]$

they are matrix they are both 2x1 incase you were wondering wth i did xD

show that T has an inverse

alright so im pretty sure i have to show that the LS has an inverse for T to have an inverse but i never really did any any question with a 1x2 matrix so i dont know how to do the inverse.. can someone help me out please

2. Originally Posted by treetheta
let T: R^2 -> R^2 be a linear transformation define by

$T \left[ \begin{array}{c}x\\y\end{array} \right] =\left[ \begin{array}{c}x+y\\x-y \end{array} \right]$

they are matrix they are both 2x1 incase you were wondering wth i did xD

show that T has an inverse

alright so im pretty sure i have to show that the LS has an inverse for T to have an inverse but i never really did any any question with a 1x2 matrix so i dont know how to do the inverse.. can someone help me out please
Consider:

$T \left[ \begin{array}{c}x+y\\x-y\end{array} \right]$

CB

3. i wish i knew where to go from there, but it's still not clear to me

T is a linear trasformation that means it's composed of many elementary matrix's

so E1E2 time the matrix $x1 = (x+y)^T$ and $x2 = (x - y)^T$

4. Originally Posted by treetheta
let T: R^2 -> R^2 be a linear transformation define by

$T \left[ \begin{array}{c}x\\y\end{array} \right] =\left[ \begin{array}{c}x+y\\x-y \end{array} \right]$

they are matrix they are both 2x1 incase you were wondering wth i did xD

show that T has an inverse

alright so im pretty sure i have to show that the LS has an inverse for T to have an inverse but i never really did any any question with a 1x2 matrix so i dont know how to do the inverse.. can someone help me out please
Do you remember way back in "precalculus" where you learned to find the inverse of a function? To find the inverse of y= f(x), swap x and y to get x= f(y) and solve for y. That gives y= $f^{-1}(x)$
You can do a similar thing here: Since T(x,y)= (x+y, x-y)= (a, b), swapping (x,y) and (a,b) gives you (a+b, a- b)= (x,y) or a+ b= x, a- b= y. Solve those two equations for a and b.

5. but that is saying it has an inverse for sure right, i'm trying to prove that T has an inverse so does this qualify for a valid proof?

6. Originally Posted by treetheta
but that is saying it has an inverse for sure right, i'm trying to prove that T has an inverse so does this qualify for a valid proof?
Show that it is 1-1 and onto.

7. Originally Posted by CaptainBlack
Consider:

$T \left[ \begin{array}{c}x+y\\x-y\end{array} \right]$

CB
Ok so you have not looked at this, lets make it more explicit:

$T \left[ T \left[ \begin{array}{c}x\\y\end{array} \right]\right] = T \left[ \begin{array}{c}x+y\\x-y\end{array} \right] =2\;\left[ \begin{array}{c}x\\y\end{array} \right]$

CB

8. Originally Posted by treetheta
but that is saying it has an inverse for sure right, i'm trying to prove that T has an inverse so does this qualify for a valid proof?
Finding its inverse is definitely a valid way of showing that it has an inverse!