Linear Mathematics | Part IB, 2003

Let α\alpha be an endomorphism of a finite-dimensional real vector space UU and let β\beta be another endomorphism of UU that commutes with α\alpha. If λ\lambda is an eigenvalue of α\alpha, show that β\beta maps the kernel of αλι\alpha-\lambda \iota into itself, where ι\iota is the identity map. Suppose now that α\alpha is diagonalizable with nn distinct real eigenvalues where n=dimUn=\operatorname{dim} U. Prove that if there exists an endomorphism β\beta of UU such that α=β2\alpha=\beta^{2}, then λ0\lambda \geqslant 0 for all eigenvalues λ\lambda of α\alpha.

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