self-adjoint operator

I would be grateful for some help/tips/relevant website with this question.

Let (V,<,>) be a complex inner product space with an orthonormal basis {v1,v2,.......vn}. Let L:V------>V be a linear operator. Explain what is meant by saying that L is self-adjoint.

i) L is self-adjoint if and only if <Lvi,vj>=<vi,lvj>.
ii) a_ij=<Lvj,vi>.

I know if If L is self-adjoint, then all eigenvalues of L are real but how can i use this to solve i) and ii).

Thanks in advance.

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Originally Posted by charikaar

I would be grateful for some help/tips/relevant website with this question.

Let (V,<,>) be a complex inner product space with an orthonormal basis {v1,v2,.......vn}. Let L:V------>V be a linear operator. Explain what is meant by saying that L is self-adjoint.

i) L is self-adjoint if and only if <Lvi,vj>=<vi,lvj>.
ii) a_ij=<Lvj,vi>.

I know if If L is self-adjoint, then all eigenvalues of L are real but how can i use this to solve i) and ii).

Thanks in advance.

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Having all real eigenvalues is a property of all self-adjoint linear operators but there are non-self-adjoint linear operators that also have real eigenvalues. "Explain what is meant by saying that L is self-adjoint" is asking for the definition of self-adjoint. Look up the definition.

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i)L is self-adjoint if and only if <Lvi, vj>= <vi, lvj>

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You mean "<Lvi, vj>= <vi, Lvj>".

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ii) a_ij= <Lvj,vi>.

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This makes no sense because you haven't said what the "a_ij" are. What is the question really?

yes i meant <Lvi, vj>= <vi, Lvj>.
A=a_ij is the matrix representing L with respect to the basis {v1,......v_n}.

Can you get me started please.

Quote:

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Originally Posted by charikaar

I would be grateful for some help/tips/relevant website with this question.

Let (V,<,>) be a complex inner product space with an orthonormal basis {v1,v2,.......vn}. Let L:V------>V be a linear operator. Explain what is meant by saying that L is self-adjoint.

i) L is self-adjoint if and only if <Lvi,vj>=<vi,lvj>.
ii) a_ij=<Lvj,vi>.

I know if If L is self-adjoint, then all eigenvalues of L are real but how can i use this to solve i) and ii).

Thanks in advance.

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L is self adjoint means L = L*, but we know L* is the one and unique operator for which <Lv, u> = <v, L*u> for all u,v, so (i) is the correct choice.

Tonio

Thanks for that Tonio.

How do i prove i) and ii).

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Originally Posted by charikaar

Thanks for that Tonio.

How do i prove i) and ii).

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I have no idea what you're asking: I thought your question one that if L is a slef-adjoint operator then which one of (i) or (ii) is true, and that's already been answered.
(i) and (ii) cannot be proved since (i) is the definition of self adjoint operator and (ii) makes no sense since we've no idea what a_ij are.

Tonio

A=a_ij is the matrix representing L with respect to the basis {v1,......v_n}.

this is question 4 of my homework question on this website http://www.maths.qmul.ac.uk/~cchu/MTH6122/lode091.pdf

A lot of "proving" things depends on using the precise statements of definitions. Now, I tend to define a linear tranfsormation, L, from V to V, to be "self adjoint" if and only if <Lu, v>= <u, Lv> for all vectors u, and v in V, because I would start from the definition that, if L is a linear transformation from U to V, then its "adjoint", L* is the linear transformation from V to U such that <Lu,v>= <u,L*v> where u is in U, v is in V, the first inner product is taken in V and the second in U.

But your problem asks you to "Explain what is meant by saying that L is self-adjoint" and then asks you to show that <Lu,v>= <u, Lv> so apparently, that is NOT the definition you are using.

So the ball is back in your court- what, exactly, is the definition of "self adjoint" your text is using?

We don't have text book for this course apart from short lecture notes which are not helpful to solve the above problem. You definition is correct.


for ii) a_ij=<Lvj,vi>

Lv_j= L(a_1jv_1+a_2jv_2+....a_njv_n)

then <L(a_1jv_1+a_2jv_2+....a_njv_n),v_i>=(a_i_j) Is this correct?

Yes, since your basis is orthonormal, that is correct.