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Let X be a column vector of dimension n > 1 with at least one non-zero entry. The number of non-zero eigenvalues of the matrix M = XXT is
  • a)
    0
  • b)
    n
  • c)
    1
  • d)
    n - 1
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Let X be a column vector of dimension n > 1 with at least one non-z...
Concept:
Eigenvalues are the special set of scalar values that is associated with the set of linear equations most probably in the matrix equations
Explanation:

Here, X is an n × 1 column vector with the entry in the ith row equal to a.
XT is a row vector having an entry in the ith column equal to a. Then, XXT is an n × 1 matrix having the entry in the ith row and ith column equal to a2 .

Since this matrix is diagonal, its eigenvalues are a2, 0, 0.......0. Hence, the number of non-zero eigenvalues of the matrix XXT is 1.
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Community Answer
Let X be a column vector of dimension n > 1 with at least one non-z...
Let X be a column vector of dimension n. In matrix notation, X can be represented as:

X = [x1, x2, x3, ..., xn]T

where xi represents the i-th element of X. The superscript T denotes the transpose operation, which converts a row vector into a column vector.

The dimension of X indicates the number of elements it contains. Therefore, the column vector X has n elements.
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Let X be a column vector of dimension n > 1 with at least one non-zero entry. The number of non-zero eigenvalues of the matrix M = XXTisa)0b)nc)1d)n - 1Correct answer is option 'C'. Can you explain this answer?
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