HermitianSpaces: an Adjoint property proof (help)

[emlio]

Homework Statement

$$V$$ 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
---------------------------------------------------------------

$$L: V \rightarrow V$$, 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)


-------------------------------------------------------------------------------
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:

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

Please help me, I'm not able to proof it. tnx

 

[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.

 

[emlio]

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.

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

 

[emlio]

I solved it !

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

------------------------------------------------------------------

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

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



-------------------------------------------------------------------

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

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



-------------------------------------------------------------------


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

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

...

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

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

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


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

 

[emlio]

I solved it !

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

------------------------------------------------------------------

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

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



-------------------------------------------------------------------

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

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



-------------------------------------------------------------------


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

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

...

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

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

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


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

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

 

[Dick]

Looks like the idea is right. Good job.