Question:

Let \( M_n(\mathbb{R}) \) be the real vector space of all \( n \times n \) matrices with real entries, \( n \ge 2 \). Let \( A \in M_n(\mathbb{R}) \). Consider the subspace \( W \) of \( M_n(\mathbb{R}) \) spanned by \(\{I_n, A, A^2, A^3, \ldots\}\). Then the dimension of \( W \) over \(\mathbb{R}\) is necessarily

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By Cayley–Hamilton theorem, any square matrix satisfies its own characteristic polynomial, which limits the number of linearly independent powers of the matrix to at most its size \(n\).
Updated On: Dec 6, 2025
  • \(\infty\).
  • \(n^2\).
  • \(n\).
  • at most \(n\).
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The Correct Option is D

Solution and Explanation

Step 1: Using Cayley–Hamilton Theorem.
The characteristic polynomial of an \(n \times n\) matrix \(A\) has degree \(n\), and by the Cayley–Hamilton theorem, \(A\) satisfies its characteristic equation. Thus, \(A^n\) can be expressed as a linear combination of \(I, A, A^2, \ldots, A^{n-1}\).
Step 2: Dimension bound.
This means the set \(\{I, A, A^2, \ldots, A^{n-1}\}\) spans all possible powers of \(A\). Hence, the dimension of \(W\) is at most \(n\).
Step 3: Conclusion.
Therefore, the correct answer is (D) at most \(n\).
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