Answer to Question #215422 in Linear Algebra for Sabelo

Question #215422

Suppose T € L(V ) and dim range T =k. Prove that T has at most k + 1distinct eigenvalues.


1
Expert's answer
2021-08-19T12:40:03-0400

"\\text{Since dim(V ) < \u221e, T has at most a finite number of distinct eigenvalues.}\\\\ \\text{Let $\u03bb_1, . . . , \u03bb_m$ be the distinct\neigenvalues of T, and let}:\\\\ v_1, . . . , v_m \\text{be corresponding nonzero eigenvectors.}\\\\\n\n\\text{If $\u03bb_j \\neq 0,$ then $T(v_j\/\u03bb_j ) = v_j$}\\\\\n\\text{Since at most one of $\u03bb_1, . . . , \u03bb_m $ equals 0, this implies that at least m \u2212 1 of the vectors $v_1, . . . , v_m$ are in\nrange(T)}.\\\\\n\\text{ These vectors are linearly independent, which implies that $m \u22121 \u2264 dim(rangeT) = k.$}\\\\\n\\text{Thus $m \u2264 k + 1$, as desired.}"


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