Let's say D is the diagonalized form of A. Then the diagonal elements of D are precisely the eigenvalues of A. The rank of D is the number of linearly independent columns. Obviously this equals the number of non-zero eigenvalues. Since the rank of A and the rank of D are the same, the conclusion follows.
#3
Zorba
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If A is similar to B then \textrm{rk}(A)=\textrm{rk}(B), then consider the rational canonical form, and it follows as Landau stated above.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?