Suppose T belongs to L(V,V) where L(A,W) denotes the set of linear mappings from Vector spaces A to W, is such that every subspace of V with dimension dim V - 1 is invariant under T. Prove that T is a scalar multiple of the identity operator.(adsbygoogle = window.adsbygoogle || []).push({});

My attempt : Let U be one of the sub spaces of V which has dimension = dim V -1.

Let a basis to U be defined as (V1, ....,Vj-1,Vj+1,.......Vn) where j ranges from 1 to n

Now, this basis can be extended to form a basis of V as (V1,......,Vn)

=> v = a1V1+....+aj-1Vj-1+aj+1Vj+1.......+anVn+ajVj

=>Tv =a1T(V1) + ....... + an T(Vn) + ajT(Vj)

===================

A

now since T belongs to L(V,V), => T(Vi) belongs to V

or A belongs to V -------------(1)

However, by the problem statement:

subspace of dimension dim V - 1 is invariant under T

=>A belongs to U

=>T(Vi) (i≠j) belongs to span ( Basis of U )

Since, j ranges from 1 to n

=> T(Vi) = λiVi where λ is a scalar.

Now, having proved that every basis vector is an eigen vector, I am stuck at trying to prove that T is a scalar multiple of the identity operator for any vector V

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Linear Algebra : Proving that Every map is an identity operator

**Physics Forums | Science Articles, Homework Help, Discussion**