Let \(V\) be a finite dimensional vector space over an algebraically closed field \(F\). Let \(T\) be a linear transformation of \(V\) into itself, let \(\lambda \) be an eigenvalue of multiplicity \(m\) of \(T\) (i.e., \(m\) is the multiplicity of \(\lambda \) as a root of the characteristic polynomial of \(T\)), and let \(U\) be the subspace \(U=\{v\in V\mid T(v)=\lambda v\}\). Prove that the dimension of \(U\) is \(\le m\).
Question
Math 6310 • Algebra
Fall 1997 • Question 2