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I am reading lecture notes on abstract algebra and came across a proof that doesn’t appear to be valid. The main theorem states that SpecT$\leq$DimV. It then says that this can be proven by showing that eigenvectors with distinct eigenvalues are linearly independent. This would be acceptable, however it requires that the number of eigenvalues of a given transformation(in this case an endomorphism) be finite. Is that the case or is there something I’m misunderstanding?

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  • $\begingroup$ What are $T$ and $V$? In a finite vector space, the number of eigenvalues is finite, but this isn't the case in an infinite one. $\endgroup$ – Michelle Nov 17 '20 at 16:48
  • $\begingroup$ @Michelle $T$ is an endomorphism of the finite-dimensional vector space $V$. $\endgroup$ – Adam French Nov 17 '20 at 16:50
  • $\begingroup$ The argument does not require that the number of eigenvalues be finite. $\endgroup$ – lhf Nov 17 '20 at 19:00
  • $\begingroup$ @lhf Is this because any collection of eigenvectors with distinct values is linearly independent and so the addition of other such eigenvectors will still yield a linear combination with less than Dim$V$ terms? $\endgroup$ – Adam French Nov 17 '20 at 19:05
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Suppose $V$ is a finite-dimensional vector space and let $n=\dim V$. The characteristic polynomial of $T$ is $p_T(X)=\ker(XI_n - T)$. We can show (by induction for example) that $p_T$ is indeed a polynomial of degree $n$. We can also show that $\lambda \in \text{Sp}(T) \iff p_T(\lambda) = 0$. Since the number of roots of a polynomial is inferior or equal to its degree, we have that the number of eigenvalues of $T$ is inferior or equal to $n$.

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It then says that this can be proven by showing that eigenvectors with distinct eigenvalues are linearly independent. This would be acceptable, however it requires that the number of eigenvalues of a given transformation(in this case an endomorphism) be finite.

Actually, once you know that eigenvectors with distinct eigenvalues are linearly independent, it follows immediately that there must be at most $\dim V$ eigenvalues, without any need to appeal to the existence of the characteristic polynomial. This is the content of the theorem! The argument is simple: if there are $n$ distinct eigenvalues, then we can find $n$ linearly independent eigenvectors, so $n \le \dim V$.

The existence of the characteristic polynomial proves the stronger fact that there are still at most $\dim V$ eigenvalues when counted with (algebraic) multiplicity, but you also don't need the existence of the characteristic polynomial to prove this if you define the algebraic multiplicity as the dimension of the generalized eigenspace $\dim \ker((T - \lambda)^{\dim V})$. The argument is a bit more complicated but similar: you show that the generalized eigenspaces are "linearly independent" in the sense that the natural map from their direct sum to $V$ is injective (in fact it's an isomorphism but that isn't needed here).

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