Find adjoint of a linear transformation

Question

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 :-/

Answer 1

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, $T^*(x,y)=(x+y, x+3y)$

Answer 2

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.