Prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A.
http://ltcconline.net/greenl/courses/203/MatrixOnVectors/symmetricMatrices.htm
I've read the entire page and while it's on the correct topic, it doesn't prove what I'm looking to prove.
To prove that if A is a diagonalizable matrix, then the rank of A is the number of nonzero eigenvalues of A, we need to follow some steps.
Step 1: Understand the concept of diagonalizable matrices.
A matrix A is said to be diagonalizable if it can be expressed as A = PDP^(-1), where P is an invertible matrix and D is a diagonal matrix. In other words, A is similar to a diagonal matrix.
Step 2: Understand the relationship between eigenvalues and diagonalizable matrices.
For a diagonalizable matrix A, the eigenvalues of A are the elements on the diagonal of the diagonal matrix D in the expression A = PDP^(-1). These eigenvalues appear on the diagonal of D in the same order as the corresponding eigenvectors appear as columns in P. Furthermore, the non-zero eigenvalues correspond to the columns of P that are linearly independent.
Step 3: Understand the relationship between rank and linear independence.
The rank of a matrix A is defined as the maximum number of linearly independent rows or columns of A. In other words, it represents the dimension of the column or row space of A.
Step 4: Show that the rank of A is equal to the number of nonzero eigenvalues of A.
Since the non-zero eigenvalues of A correspond to the linearly independent columns of P, we can conclude that the rank of A is equal to the number of nonzero eigenvalues. This is because each non-zero eigenvalue contributes one linearly independent column to P, and hence, contributes one linearly independent column to A.
Therefore, if A is a diagonalizable matrix, the rank of A is indeed the number of nonzero eigenvalues of A.