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Question

Math 6310 • Algebra

Fall 1997 • Question 2

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 \(\le m\).

Raw LaTeX

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 \(\le m\).