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I am trying to prove this problem:

Let M be the space of all 2 × 2 complex matrices, satisfying 〖(X)bar〗^t = -X (skew-hermitian).

Consider M as a vector space over R.

Define a bilinear form B on M by B(X,Y) = -tr(XY)

How should I show that B takes real values, is symmetric and positive definite?

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I'm not too sure why you rolled back the edit into a more unreadable form. Is there something which I did not interpret correctly? –  EuYu Nov 8 '12 at 3:02
Jack, votes are anonymous, and requests for explanation are rarely fruitful. –  The Chaz 2.0 Nov 8 '12 at 4:08

1 Answer 1

$$X\in M\Longleftrightarrow X^t=-\overline X\Longrightarrow X=\begin{pmatrix}\,ia&x+iy\\-x+iy&ib\end{pmatrix}\;\;,\;\;a,b,x,y,\in\Bbb R$$

Thus, for example, taking

$$Y:=\begin{pmatrix}\,i\alpha&\xi+i\upsilon\\-\xi+i\upsilon&i\beta\end{pmatrix}\;\;,\;\;\alpha,\beta,\xi,\upsilon,\in\Bbb R$$

we get


Can you take it from here?

$$=a\alpha+b\beta-2(x\xi+y\upsilon)\in\Bbb R$$

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Isn't what I wrote the same? (if I understand properly your symbols...) –  DonAntonio Nov 8 '12 at 3:19
How to prove that B is symmetric and positive definite? –  Jack Nov 8 '12 at 3:19
Well, this is for you to do: $\,B(X,Y)=B(Y,X)\,$ (take into account that you're interested only on the main diagonal entries. This make calculations much less messy) –  DonAntonio Nov 8 '12 at 3:23
is the definition of positive definite given in the problem correct? –  Jack Nov 8 '12 at 3:29
I think it is, yet $\,(,)\,$ usually denotes some inner product. I don't know what they mean by this within the context of your question (not even an inner product is given!) –  DonAntonio Nov 8 '12 at 3:55

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