Q11. Then the dimension of W over R is necessarily?

Question

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

(A) ∞.

(B) $n^2$.

(C) $n$.

(D) at most $n$.

Answer 1

The correct answer is:

(D) at most $n$.

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Explanation:

Let $P(x) = \det(A - xI_n)$ be the characteristic polynomial of matrix $A$. By the Cayley-Hamilton theorem, we have $P(A) = 0$.

This means that the matrix $A$ satisfies its own characteristic equation, which implies that any power of $A$ can be expressed as a linear combination of ${I_n, A, A^2, \ldots, A^{n-1}}$. Specifically, $A^n$ can be expressed as a linear combination of ${I_n, A, A^2, \ldots, A^{n-1}}$.

Therefore, the set ${I_n, A, A^2, \ldots, A^{n-1}}$ spans the subspace $W$.

Now, the number of matrices in the set ${I_n, A, A^2, \ldots, A^{n-1}}$ is at most $n$, so the dimension of $W$ is at most $n$.

Hence, the correct answer is (D) at most $n$.