Let V be a finite-dimensional vector space. The base field F may be either R or C here. Let T, an element of the linear mapping of V to V, L(V), be an operator. Suppose that all non-zero elements of V are eigenvectors for T. Show that T is a scalar multiple of the identity map, i.e., that there is a λ in the Reals such that T(v) = λv for all v in V.
Clues that were given:
Assume we have 2 non-linear dependent vectors . . .
Means space is 1-dimensional . . .
If there are no 2 linearly independent vectors (i.e. dimV = 1) then . . .
If v and w are linearly independent, let Tv = λv and let Tw = µw. Want to show that λ=µ.
We also have T(av +bw) = aλv + bµw for every a,b in the base field,F
The trivial case is that V is a 1-dimensional space, or dimV=1, then each element in V is a number. Let T(1)=λ, then for each x in V, we have T(x)=xT(1)=λx. The statement is obviously hold.
For non-trivial case that dimV>1, then we consider any two linearly independent vector v and w in V. From the ...
Finite-dimensional Vector Space, Fields and Mappings are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.