- #1

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## Main Question or Discussion Point

How to prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A.

Thanks and regard.

Thanks and regard.

- Thread starter DavidLiew
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- #1

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How to prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A.

Thanks and regard.

Thanks and regard.

- #2

Landau

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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

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If [tex]A[/tex] is similar to [tex]B[/tex] then [tex]\textrm{rk}(A)=\textrm{rk}(B)[/tex], then consider the rational canonical form, and it follows as Landau stated above.

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