I'd be immensely grateful for any help with this problem:
Let be a finite-dimensional vector space, and let
be a linear operator (a linear map) such that , for every linear map .
Prove that there exists a scalar such that .
Oh, and there has been a hint provided:
Hint: show that has at least one eigenvalue and observe the corresponding eigenspace.
First, how could I show that has at least one eigenvalue? Of course, if has an eigenvalue , then there exists a vector , , such that ; but how to show that there is such an eigenvalue in the first place?
And I would really appreciate if someone could show me how to proceed from that to the final solution.
Alternatively, the hypothesis given tell you that A commutes with all linear transformations B. That is, it is in the center of the nxn matrices ( ), where n is the dim(V). The center is precisely the nxn scalar diagonal matrices. But I like TPH's method way better, gets right down to it.
We are given a linear operator , and everything we know about it, is that , for every linear operator .
This means that .
Let's mark , and so we must find (the field) such that .
If I'm not mistaken, if we manage to find a vector and scalar such that , then the existence of at least one eigenvalue will have been proved.
But how to do that is a different kettle of fish...
Show that every eigenspace of an operator is invariant for every operator which commutes with (in other words, such that ).
But this can be shown like this:
let be , let and let be the eigenspace of corresponding to an eigenvalue . Let be an arbitrarily chosen vector.
We have: ,
To return to the starting problem, it seems to me that if we can show the existence of a scalar such that for one operator such that , then by the preceding statement it can be shown to hold for every , if I'm not mistaken...
If you show that the Eigenspace is the entire vector space, then you are done, as then is the zero map, and so . I'm not quite sure how you would show that about the Eigenspace though.
On another trail of thought, the matrix is constant under any base change - does that not automatically make it a scalar multiple of the identity? Or will other matrices be preserved?