Find adjoint of a linear transformation Let $T:\mathbb{R^3} \to \mathbb{R^3}$ be a linear transformation, and 


*

*$T(1,1)=(2,4)$

*$T(1,-1)=(0,-2)$
Find $T^*(x,y)$.
I "found" (I mean, I think it's wrong...) the general form of the linear transformation:
$T(x,y)=\alpha T (1,1)+\beta T(1,-1)=\alpha(2,4)+\beta(0,-2)$
So I solved $\begin{cases} 2\alpha=x \\ 4\alpha-2\beta=y \end{cases}$ 
I got $\alpha=x/2$ and $\beta=\frac{y-2}{-2}$ and then I plugged them in:
$T(x,y)=x/2T(1,1)+\frac{y-2}{-2}(1,-1)=x/2(2,4)+\frac{y-2}{-2}(0,-2)=(x, 2x+y-2)$
Now I used the definition of adjoint transformation:
$<(x,y), T^*(1,0)>=<T(1,0), (x,y)>=<\underbrace{(1,2)}_\textrm{Using the general form of the transformation},(x,y)>$
$=x+2y$
I did the same for $T(0,1)$:
$<(x,y), T^*(0,1)>=<T(0,1), (x,y)>=<(0,-1), (x,y)>=-y$
So the adjoint would be $T^*(x,y)=(x+2y, -y)$, but the answer says it's $T^*
(x, y) = (x + y, x + 3y)$, and I don't know how they got that answer :-/
 A: After a few minutes of posting the question I figured pretty much all I did was useless.
I found that $T(1,0)=\alpha T(1,1) + \beta T(1, -1)$ so I solved for $\alpha$ and $\beta$.
$\begin{cases} \alpha+\beta=1 \\ \alpha-\beta=0 \end{cases}$ so $\alpha=1/2$ and $\beta=1/2$
So I plugged in the values $T(1,0)=1/2 T(1,1) + 1/2 T(1, -1)=1/2(2,4)+1/2(0,-2)=(1,1)$
I did the same for $T(0,1)$, and I got $T(0,1)=(1,3)$
So using the definition for adjoint transformations:
$<T^*(1,0), (x,y)>=<(x,y), T(1,0)>=<(x,y), (1,1)>=(x+y)$
$<T^*(0,1), (x,y)>=<(x,y), T(0,1)>=<(x,y), (1,3)>=(x+3y)$
Finally, $T^*(x,y)=(x+y, x+3y)$
A: First of all your take of finding the matrix representation of $T$ is wrong.
The idea behind this is that we can write
$$
T(x,y)=\left(\begin{array}&\alpha&\beta\\\gamma&\delta\end{array}\right)\left(\begin{array}&x\\y\end{array}\right).
$$
So in your case the given values should lead to four equations (each row of each given vector is one equation).
From these equations you should be able to find $\alpha,\beta,\gamma,\delta$.
The adjoint of $T$ is given by the transposed matrix representation.
