Where n is an odd positive integer

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SUMMARY

The discussion centers on proving that an operator T: Rn → Rn, where n is an odd positive integer, has at least one eigenvector. It is established that T must be a linear operator for eigenvectors to be defined. The characteristic equation for T is an nth order polynomial with real coefficients, which implies that complex roots appear in conjugate pairs. Given that n is odd, at least one real eigenvalue must exist, ensuring the presence of at least one eigenvector in Rn.

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squenshl
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I'm studying for a test.
In doing one of the old tests and it had a question that I couldn't do.
Let T: Rn [tex]\rightarrow[/tex] Rn be an operator on Rn,
where n is an odd positive integer. How do I prove T has at least one eigenvector in Rn
 
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Well, you can't unless it is specified that T is a linear operator- otherwise "eigenvectors" cannot be defined.

I presume that you know that the "characteristic equation" for an operator on an n dimensional space is an nth order polynomial equation. Since the underlying field is the real numbers, all coefficients of that characteristic equation are real numbers. Now, you know, do you not, that, in that situation, complex roots come in complex conjugate pairs?
 

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