4.4.7

Let T be a linear operator on a finite-dimensional vector space V.

a) Suppose T has two linearly independent eigenvectors with the same eigenvalue $$ \lambda$$. Then we claim that $$ \lambda $$ is a multiple root of the characteristic polynomial of T.

Given two linearly independent eigenvectors$$(v_1,v_2) $$ of T with the same eigenvalue, we can extend this set to form a basis $$(v_1,v_2,...,v_n) $$ of V. Let A be the matrix of T with respect to this basis: $$ T(v_1,v_2,...,v_n) = (v_1,v_2,...,v_n)A$$. Since $$Tv_1 = \lambda v_1 $$ and $$Tv_2 = \lambda v_2 $$, A has the form $$ \begin{bmatrix} \lambda I_2 & A' \\ 0_M & A'' \\ \end{bmatrix} $$, where $$I_2 $$ is the 2x2 identity matrix, $$0_M $$ is an (n-2)x2 zero matrix, and $$A',A'' $$ are some 2x(n-2) and (n-2)x(n-2) matrices.

The characteristic polynomial for T is $$ det(tI-A) = det\begin{bmatrix} (t - \lambda)I_2 & -A' \\ 0_M & tI_{n-2} - A'' \\ \end{bmatrix} $$. Using expansion by cofactors twice (down the first column), we find $$ det(tI-A) = (t-\lambda)^2q(t)$$, where q(t) is some polynomial in t of degree n-2, showing that $$\lambda $$ is a multiple root of the characteristic polynomial of T.

b) The converse of the above is not true: if $$ \lambda$$ is a multiple root of the characteristic polynomial of T, T need not have two linearly independent eigenvectors with eigenvalue $$ \lambda$$. As a counterexample, let V = $$\mathbb{R}^2 $$, and let T be the linear operator defined by $$Tx = Ax $$, where $$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} $$. $$ \lambda = 1$$ is a multiple root of the characteristic polynomial $$ \lambda^2 - 2\lambda + 1 = 0$$; however, T has only one eigenvector $$ v = \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$ with eigenvalue 1.