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I am trying to prove that the geometric multiplicity of an eigenvector is bounded by the algebraic multiplicity. One particular proof of this theorem that I like is contained in the answer by Mariano Suárez-Alvarez on this Mathematics Stackexchange post. But there is one aspect of this proof that I do not understand, and I would like it if somebody would please explain it to me.

In the proof that I linked to, the author states that $t I-A$ has the same determinant as $C$. It is not obvious to me why this is the case. It certainly is not true in general that row-equivalent matrices $\mathbf A$ & $\mathbf B$ have the same determinant.

Meta: Rather than answering my current question here, perhaps it is better to edit the proof in the answer provided in the link.

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  • $\begingroup$ It appears that while $tI-A$ and $C$ may not have the same determinant, their determinants differ by a constant multiple, so they are divisible by the same power of $t-\lambda$. This is all that is actually needed for the argument to work. $\endgroup$ – Brian M. Scott Jul 27 '16 at 0:52
  • $\begingroup$ Sir. You are absolutely correct. If you make your comment into an answer then I will accept it. $\endgroup$ – EricVonB Jul 27 '16 at 2:34
  • $\begingroup$ Done! (And I’ll make that slight change in Mariano’s answer.) $\endgroup$ – Brian M. Scott Jul 27 '16 at 2:37
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It’s a slight oversight in the wording of the answer. Although $tI-A$ and $C$ may not have the same determinant, their determinants can differ only by a constant multiple, so they are divisible by the same power of $t-\lambda$. This is all that is actually needed for the argument to work.

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