So in my courses in linear algebra we consistently use that the characteristic polynomial is of degree n. ( that is assuming $f:V \to V$ with the dimension of $V$ being $n$) However, I have not actually see the proof of it. Does it just follow from simple definitions or is there something else one needs to notice?
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Induction on the size of the matrix, plus a suitable characterization of the determinant, for instance Laplace expansion. – Git Gud Mar 22 '16 at 21:27
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1this might be useful [lemma 2] http://www.math.ucla.edu/~tao/resource/general/115a.3.02f/week8.pdf – Andres Mejia Mar 22 '16 at 21:29
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Thanks for the link Andres , definitely contains a bunch of useful material – Quality Mar 23 '16 at 00:15
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The characteristic polynomial of a matrix $A$ is defined as $$p(t)=\det(A-tI)=\sum_{\sigma \in S_{n}}\epsilon(\sigma)\prod_{i=1}^{n}(a_{i \sigma(i)}-t\delta_{i \sigma(i)})$$ and the term in the sum with $\sigma=id$ has degree $n$.
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