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

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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 &#955; in the Reals such that T(v) = &#955;v 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 = &#955;v and let Tw = Âµw. Want to show that &#955;=Âµ.
We also have T(av +bw) = a&#955;v + bÂµw for every a,b in the base field,F

https://brainmass.com/math/linear-algebra/linear-operators-finite-dimensional-vector-space-fields-and-mappings-157561

#### Solution Preview

Proof:
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)=&#955;, then for each x in V, we have T(x)=xT(1)=&#955;x. 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.

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