How to prove that if A is a diagonalizable matrix, then the rank of A

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The discussion establishes that if matrix A is diagonalizable, its rank is equal to the number of nonzero eigenvalues. The diagonalized form D of A contains the eigenvalues of A as its diagonal elements. Since the rank of D corresponds to the number of linearly independent columns, and the ranks of similar matrices are equal, the conclusion is that the rank of A matches the count of its nonzero eigenvalues. This proof leverages the properties of diagonalization and similarity transformations.

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This discussion is beneficial for students and professionals in mathematics, particularly those studying linear algebra, as well as educators teaching concepts related to matrix theory and eigenvalues.

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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.
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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.
 


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.
 

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