Solved: Prove that if A is a diagonalizable matrix, then the rank of A is the number of

Chapter 5, Problem 41

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Prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A. (b) Let P be the matrix having the vectors in B as columns. Prove that the product AP can be expressed as AP = P 0Ik X 0 Y [Hint: Compare the first k column vectors on both sides.] (c) Use the result in part (b) to prove that A is similar to C = 0Ik X 0 Y and hence that A and C have the same characteristic polynomial. (d) By considering det(I C), prove that the characteristic polynomial of C (and hence A) contains the factor ( 0) at least k times, thereby proving that the algebraic multiplicity of 0 is greater than or equal to the geometric multiplicity k.