GATE 2025 DA Q.37: solution and answer

GATE 2025 · DAQ.37 · 2 marks · MCQ

Let $A \in R^{n \times n}$ be such that $A^{3} = A$ . Which one of the following statements is ALWAYS correct?

(A)
$A$ is invertible
(B)
Determinant of $A$ is $0$
(C)
The sum of the diagonal elements of $A$ is $1$
(D)
$A$ and $A^{2}$ have the same rank

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Answer and solution

ANSWER

The answer is (D): $A$ and $A^{2}$ always have the same rank.

Official answer: DSource: Official answer key, GATE 2025 DA (IIT Roorkee)

Both $A = I$ and $A = 0$ satisfy $A^{3} = A$ , so they are valid test matrices for every option.
Option (A): for $A = 0$ we have $0^{3} = 0$ , but $0$ is not invertible. So (A) is not always correct.
Option (B): for $A = I$ we have $I^{3} = I$ and $det I = 1 \neq = 0$ . So (B) is not always correct.
Option (C): for $A = 0$ the sum of the diagonal elements is $0 \neq = 1$ , and for $A = I_{2}$ it is $2 \neq = 1$ . So (C) is not always correct.
Option (D), first inequality: since $A = A^{3} = A \cdot A^{2}$ , the column space of $A$ lies inside the column space of $A^{2}$ is not needed. We only need $rank (X Y) \leq min (rank X, rank Y)$ . Applied to $A = A \cdot A^{2}$ it gives $rank (A) \leq rank (A^{2})$ .
Option (D), second inequality: applied to $A^{2} = A \cdot A$ the same rule gives $rank (A^{2}) \leq rank (A)$ .
Together, $rank (A) = rank (A^{2})$ for every real $A$ with $A^{3} = A$ .
Cross-check by eigenvalues: $A$ is annihilated by $x^{3} - x = x (x - 1) (x + 1)$ , which has distinct roots. So $A$ is diagonalizable with eigenvalues in ${0, 1, - 1}$ . Its rank is the number of nonzero eigenvalues. $A^{2}$ has eigenvalues $λ_{i}^{2}$ , and $λ_{i} \neq = 0 ⟺ λ_{i}^{2} \neq = 0$ , so the rank is unchanged.
Only (D) holds for every such $A$ .

KEY IDEA

For an ALWAYS-correct statement, test the extreme cases $A = 0$ and $A = I$ to knock out the false options. Then prove the survivor in general. Here $A = A \cdot A^{2}$ and $A^{2} = A \cdot A$ squeeze the ranks together, because a product never has larger rank than either factor.

WATCH OUT FOR

Assuming $A^{3} = A$ forces $A$ to be invertible (cancelling $A$ to get $A^{2} = I$ ) or to have a particular determinant or trace. Cancelling $A$ is illegal when $A$ is singular. The matrix $0$ and the matrix $I$ both satisfy the condition and give different invertibility, determinant and trace.

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GATE 2025 DA Q.37: solution and answer

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