1. Aug 3, 2009

### evilpostingmong

Suppose U is a finite-dimensional real vector space and T ∈
L(U). Prove that U has a basis consisting of eigenvectors of T if
and only if there is an inner product on U that makes T into a

The question is, what exactly do they mean by "makes T into a self adjoint
operator?" Is it that there exists an inner product of eigenvectors of T
say <v, v> that allows T to be self adjoint?

2. Aug 3, 2009

### HallsofIvy

Staff Emeritus
A linear transformation, T, on an inner product space, V, is "self adjoint" if and only if <Tu, V>= <u, TV> for all u and v in V and "< , >" is the inner product defined for V. In other words, the concept of "self adjoint" is only defined for a specific inner product.

An inner product space is, of course, a vector space, V, for which there is a specific inner product defined. In any finite dimensional vector space, every choice of basis defines an inner product: with basis $\{e_1, e_2, \cdot\cdot\cdot, e_n\}$, given vectors u and v, we can write $u= a_1e_1+ a_2e_2+ \cdot\cdot\cdot+ a_ne_n$ and $v= b_1e_1+ b_2e_2+ \cdot\cdot\cdot+ b_ne_n$ and then define $<u, v>= a_1b_1+ a_2b_2+ \cdot\cdot\cdot+ a_nb_n$.

Since this is an "if and only if" statement, you need to do two things:
1) Show that if T is self adjoint under inner product < , >, then there exist basis for the vector space consisting entirely of eigenvectors of T.

2) Show that if there exist a basis for the vector space, consisting entirely of eigenvectors of T, then T is self adjoint using the inner product defined, as above, by that basis.

Last edited: Aug 4, 2009
3. Aug 3, 2009

### evilpostingmong

Ok I will use everything you have defined for a vector space, vectors, and whatnot.
Let dimV=n.
Now if T is self adjoint, we know that <Tv, u>=<v, Tu>. If T is self adjoint,
then T's matrix must be nxn. Therefore T is diagonizable. This would allow T to have j-distinct eigenvalues along
its diagonal. So there is some basis for T which contains eigenvectors, one for each eigenvalue.

Last edited: Aug 3, 2009
4. Aug 3, 2009

### evilpostingmong

Now for the other direction. If there exists a basis for U consisting of eigenvectors of some T,
then Te_i=c_ie_i. Choosing two eigenvectors from the basis we have <e_i, e_k>.
So <Te_i, e_k>=<c_ie_i, e_k>=c_i<e_i, e_k>=<e_i, c_ie_k>=<e_i, Te_k>.

5. Aug 3, 2009

### Office_Shredder

Staff Emeritus
You haven't stated what <,> is here! And ciek is not Tek in general.

As always, it's better if you can avoid describing T as a matrix for your proof if at all possible. If it's necessary, you still have to justify everything. Skipping from "If T is self adjoint, then T's matrix must be nxn" which is a meaningless conclusion since we already know T maps a vector space to itself, to "Therefore T is diagonalizable". Why?

6. Aug 3, 2009

### evilpostingmong

I don't know why I said ciek=Tek, that was pretty dumb. Just got back from a long night at work.
This will prove 2.
Set u=a1e1+...+anen and v=b1e1+...+bnen where ek is an eigenvector.
For the inner product <u, v> we have a1b1+...+anbn using the example from above.
Now T has eigenvalues where Tek=ckek. Now <Tu, v>=<T(a1e1), v>+...+<T(anen), v> =c1a1b1+...+cnanbn <u, Tv>=<u, c1b1e1>+...+<u, cnbnen>=a1c1b1+...+ancnbn=c1a1b1+...+cnanbn=<Tu, v>

Last edited: Aug 4, 2009
7. Aug 4, 2009

### evilpostingmong

For "1" we know T is self adjoint so <Tv, u>=<u, Tv>. But should
I just assume that U is invariant under T, and claim that T(v)=c1b1e1+...+cnbnen
and that bkek is in some nullspace of T-ck?. I mean we know that U is invariant
under T and we know that T(bkek)=ck(bkek) so shouldn't that lead to ( T-ck)ek=0?

8. Aug 4, 2009

### HallsofIvy

Staff Emeritus
Did you really mean to say that every "nxn" matrix is diagonalizable? That is certainly NOT true!

Nothing you have said here makes any sense!

9. Aug 4, 2009

### evilpostingmong

I know. Office Shredder told me that. I fixed both parts after his post. Did you
see the two posts before yours?

Last edited: Aug 4, 2009