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DavidLiew
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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.
In order to prove this statement, we can use the fact that a diagonalizable matrix has a full set of linearly independent eigenvectors. Since the rank of a matrix is equal to the number of linearly independent columns, and each eigenvector corresponds to a non-zero eigenvalue, the number of non-zero eigenvalues will be equal to the rank of the matrix.
Yes, an example of such a matrix is a 2x2 matrix with eigenvalues of 3 and 0, but with only one linearly independent eigenvector. This matrix would have a rank of 1, but two non-zero eigenvalues.
The diagonalizability of a matrix does not necessarily affect the rank. There are cases where a matrix is diagonalizable but has a rank less than the number of non-zero eigenvalues, as shown in the previous example. However, if a matrix is diagonalizable and has a full set of linearly independent eigenvectors, then the rank will be equal to the number of non-zero eigenvalues.
Yes, there is a relationship between the rank and the diagonalizability of a matrix. If a matrix is diagonalizable, then the rank will be equal to the number of non-zero eigenvalues. However, if the rank is less than the number of non-zero eigenvalues, then the matrix may still be diagonalizable, but it is not guaranteed.
In order to prove that a matrix is diagonalizable, we must show that it has a full set of linearly independent eigenvectors. This can be done by finding the eigenvalues and corresponding eigenvectors of the matrix and checking if they are linearly independent. If they are, then the matrix is diagonalizable. Alternatively, we can also use the criteria that a matrix is diagonalizable if it has n distinct eigenvalues, where n is the size of the matrix.