Question

Suppose A\(\in L(V)\) such that rank (A)=1. Prove that there exists \(\vec u_0 \in V\) and y\(\in V'\) such that \(A\vec v=[\vec v, y]\vec u_0\) for every \(\vec v \in V\). Find all eigenvalues of A and indicate their multiplicities. What is Tr(A)?
Appreciate any help. Please

Answer 1

@zzr0ck3r

Answer 2

its been a while since ive done real linear algebra

Answer 3

what is L?

Answer 4

it's just linear transformation set

Answer 5

Is V' V transpose?

Answer 6

V is field, and V' is dual of V. V' is set of operator on V

Answer 7

yeah I cant help you:( I have only taken one linear algebra class, so I really only know basic stuff. This is why I can never help your posts

Answer 8

@wio

Answer 9

Since rank(A)=1, we see that the dimension of the linear transformation l_A with matrix A is equal to dim(V)-1. That means the dimension of the image of l_A is equal to 1.

Let L_A: V->V be the linear transformation with matrix A.
Since rank(A)=1, we have that dim(Im(l_A))=1, where Im(l_A) is the image of l_A.
Then Im(l_A) is spanned by a vector u_0 in V. That is, Im(l_A)=<u_0>.
For any v in V, l_A(v) is in Im(l_A). So we have that l_A(v)=k*u_0 for some scalar k.
It is clear that l_A(v)=Av by definition. Therefore, l_A(v)=Av=[v,y]u_0 for all v in V, where [v,y]=k

According to Riesz representation theorem, we have that l_A(v)=Av=[v,y]u_0 for some y.

Answer 10

That is the proof for the first part.
All of the eigenvalues of A are 0 and 1.
the multiplicity of 0 is dim(V)-1, and the multiplicity of 1 is 1.

Answer 11

The eigenvalues of A are 0 and [u_0, y].
The multiplicity of 0 is dim(V)-1.
The multiplicity of [u_0, y] is 1.
Trace(A)=n*0+1*[u_0,y]=[u_0,y].
(n-1)*0
instead of n*0.

Tr(A)=Trace(A).