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Suppose V is a real vector space. Suppose an operator on V, T, has no eigenvalues.

Prove that det T $\gt 0$

I know that every operator on an odd dimensional real vector space has an eigenvalue and therefore the V in this problem must be even. The determinant has the formula $(-1)^n$ (det T) where n is dim V. Thus the determinant is $(-1)^{2n}$ (det T). I can't seem to figure out how to get the sign of the det T part of the equation. Any helpful hints are appreciated!

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1 Answer 1

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Since $n$ is even, the characteristic polynomial of $T$ is given by $$ p(x) = \det(T - xI) = x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $$ and $$ \lim_{x \to \infty}p(x) = \lim_{x \to -\infty}p(x) = \infty $$ We note that $p(0) = \det(T)$. If $p(0) = 0$, then $0$ is an eigenvalue. If $p(0) < 0$, then the intermediate value theorem tells us that $p$ has at least $1$ (in fact, at least $2$) real roots.

If $T$ has no eigenvalues, then $p$ has no real roots, which means that $\det T > 0$.

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