Linear Algebra: Adjoint

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  • #1
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Homework Statement


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>##.


Homework Equations





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.
 

Answers and Replies

  • #2
Fredrik
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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
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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.

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
Fredrik
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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
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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
Fredrik
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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.
 
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