# Linear Algebra: Adjoint

1. Apr 1, 2014

### nateHI

1. The problem statement, all variables and given/known data
Let $V=\mathbb{C}_2$ with the standard inner product. Let T be the linear transformation deined by $T<1,0>=<1,-2>$, $T<0,1>=<i,-1>$. Find $T^*<x_1,x_2>$.

2. Relevant equations

3. The attempt at a solution
Find the matrix of T and then take the conjugate transpose....

This seems like an uncharacteristically easy problem and I'm wondering if I'm missing something.

2. Apr 1, 2014

### Fredrik

Staff Emeritus
I don't think you have. This looks like a problem that's only meant to test if you understand the concept of "the matrix of T". You'd be surprised how many people don't.

3. Apr 1, 2014

### nateHI

Cool thanks!

What happened to the post from Halls of Ivy? It was an interesting read. Oh well, maybe I'll come back after I turn in my assignment and solve it the "hard" way myself.

4. Apr 1, 2014

### Fredrik

Staff Emeritus
It was deleted because he gave you too much information. Here in the homework forum, we can't really show you a complete solution unless you have already posted another complete solution.

5. Apr 1, 2014

### nateHI

My complete answer. Does it look OK?

The method I will use is to calculate the matrix of $T$ and take the conjugate transpose to get $T^*<x_1,x_2>$ in matrix form. Let $M_T$ be the matirx of $T$ and $M_{T^*}$ be the matrix of $T^*$. Since
$T<1,0>=<1,-2>=1e_1-2e_2$
$T<0,1>=<i,-1>=ie_1-1e_2$
then
$M_T=\left (\begin{matrix} 1&i\\-2&-1\end{matrix}\right )$
and
$M_{T^*}=\left (\begin{matrix} 1&-2\\-i&-1\end{matrix}\right )$

I'll still come back later to do it the hard way because I can see something like that popping up on a test.

6. Apr 2, 2014

### Fredrik

Staff Emeritus
Yes, it looks fine. You didn't post the formula $(M_T)_{ij}=(Te_j)_i$, but since your $M_T$ is consistent with it, I assume that you were using it.

One of your options to calculating $M_{T^*}$ as the conjugate transpose of $M_T$, is to start with $(M_{T^*})_{ij}=(T^*e_j)_i$, and then rewrite the right-hand side as an inner product and use the definition of the adjoint operation.

Since the problem is asking you to find $T^*x$ for an arbitrary $x\in\mathbb R^2$, you don't have to explicitly write down a matrix. You can just start like this:
$$T^*x=T^*\left(\sum_j x_j e_j\right)=\sum_j x_j T^*e_j=\sum_j x_j\sum_i(T^*e_j)_i e_i,$$ were $(T^*e_j)_i$ is defined as the ith component of the vector $T^*e_j$ with respect to the ordered basis $(e_1,e_2)$. This is of course the ij-component of the matrix of $T^*$, but you don't have to know that to continue this calculation.

Last edited: Apr 2, 2014
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