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Let {|u>,|v>} be a basis for a linear space, suppose that <u|v>=0, then prove that:

[tex]A|v>=<A>I|v>+\delta A|u>[/tex]

where, A is hermitian operator, and [tex]<A>=<v|A|v>,\delta A= A-<A>I[/tex]

where I is the identity operator.

my attempt at solution:

basically, from the definitions i need to prove that [tex]\delta A|u>=\delta A|v>[/tex]

now, obviously <A>=<v|A|v>=[tex]\int du <v|u><u|A|v>[/tex]=0

cause <u|v> equals zero, so we actually need to prove that A|u>=A|v>

but A|u>=u|u> where u is the eigenvalue of |u>, then if we multiply it by <u| we get:

<u|A|u>=u (i guess that |u> and |v> are normalised), but from the same assertion as above we get that <u|A|u>=0 and so u=0, and so is v=0, so we get that A|u>=A|v>=0.

is this correct or totally mamabo jambo as i think?

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# Linear algebra question (using braket notation).

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