Explore BrainMass

Explore BrainMass

    Linear Operators : Finite-dimensional Vector Space, Fields and Mappings

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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

    © BrainMass Inc. brainmass.com March 4, 2021, 8:17 pm ad1c9bdddf

    Solution Preview

    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 ...

    Solution Summary

    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.