HermitianSpaces: an Adjoint property proof (help)

1. Sep 6, 2008

emlio

1. The problem statement, all variables and given/known data

$$V$$ is a linear space over $$C$$, finite n-dimensional

$$h: VxV \rightarrow C$$ is an Hermitian Product, POSITIVE DEFINED

and so

$$(V,h)$$ Hermitian Space
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$$L: V \rightarrow V$$, is a Linear Endomorphism of V

$$L^{*}$$ is the ADJOINT of $$L$$

$$h(L(v),w) = h(v,L^{*}(w)) \forall v,w\in V$$ (adjoint definition)

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Given:

$$B=\left\{v_{1},v_{2},v_{3},....................,v_{n}\right\}$$ is a h-ORTHORNORMAL basis for $$V$$

$$M\itshape^{B}_{B}(L)$$ i.e. matricial representation "from basis B to basis B" of L

$$M\itshape^{B}_{B}(L^{*})$$ i.e. matricial representation "from basis B to basis B" of L* (the adjoint of L)

I need to proof that:

$$M\itshape^{B}_{B}(L^{*})= (M\itshape^{B}_{B}(L))^{*}$$

2. Sep 6, 2008

Dick

Start by writing down the expression for the ij_th element of both of those matrices (i.e. what is a general matrix element) in terms of the {v_i}. Now remember '*' of a matrix is the conjugate transpose.

3. Sep 6, 2008

emlio

Tnx for the quick replay, I'll try as you say...

4. Sep 6, 2008

emlio

I solved it !!!!!!!!!!!!!!!!:rofl:

$$B=\left\{v_{1},v_{2},v_{3},....................,v_{n}\right\}$$ is a h-ORTHORNORMAL basis for $$V$$

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$$M\itshape^{B}_{B}(L)=(a_{x,y})$$ j-th Columns is:

$$L(v_{j})= a_{1j}v_{1} +....+a_{nj}v_{n}$$

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$$M\itshape^{B}_{B}(L^{*})=(b_{x,y})$$ k-th COLUMN IS:

$$L^{*}(v_{k})= b_{1j}v_{1}+....+b_{nk}v_{n}$$

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$$h(L(v_{j}),v_{k}) = h(v_{j},L^{*}(v_{k}))$$

$$h( a_{1j}v_{1} +....+a_{nj}v_{n} , v_{k}) = h(v_{j}, b_{1j}v_{1}+....+b_{nk}v_{n})$$

................

applying properties of "h"scalar product and properties of h-ohrtonormal bases

...............
$$a_{jk}*h(v_{k},v_{k}) = \bar{b}_{kj}*h(v_{j},v_{j})$$

$$a_{jk} = \bar{b}_{kj}$$

$${b}_{jk} = \bar{a}_{kj}$$ !

5. Sep 6, 2008

emlio

There are some typing errors, I've written "j" instead of "k" in some parts, the proof is ok

6. Sep 6, 2008

Dick

Looks like the idea is right. Good job.