GATE DA 2025 Set 1 — Question 37
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Linear Algebra → Vector Spaces, Basis & Dimensions → Rank of a Matrix
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Question
Let be such that . Which one of the following statements is ALWAYS correct?
Correct answer
(D) A and A² have the same rank
Solution
Given , the minimal polynomial of divides .
The possible eigenvalues of are . Since the minimal polynomial has distinct roots, is diagonalizable.
Let be the eigenvalues of . Then the eigenvalues of are .
The rank of a diagonalizable matrix is equal to the number of non-zero eigenvalues.
The non-zero eigenvalues of are and .
The non-zero eigenvalues of are and .
Thus, for every non-zero eigenvalue of , the corresponding eigenvalue of is also non-zero (specifically 1). The zero eigenvalues remain zero.
Therefore, the number of non-zero eigenvalues is the same for and , implying .Why other options are incorrect:
(A) If , holds but is not invertible.
(B) If , holds but .
(C) If , trace is 0. If , trace is . Neither must be 1.
The possible eigenvalues of are . Since the minimal polynomial has distinct roots, is diagonalizable.
Let be the eigenvalues of . Then the eigenvalues of are .
The rank of a diagonalizable matrix is equal to the number of non-zero eigenvalues.
The non-zero eigenvalues of are and .
The non-zero eigenvalues of are and .
Thus, for every non-zero eigenvalue of , the corresponding eigenvalue of is also non-zero (specifically 1). The zero eigenvalues remain zero.
Therefore, the number of non-zero eigenvalues is the same for and , implying .Why other options are incorrect:
(A) If , holds but is not invertible.
(B) If , holds but .
(C) If , trace is 0. If , trace is . Neither must be 1.
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