91Ó°ÊÓ

First, let us understand the matrices in the set \(\{I_{n}, A, A^{2}, \ldots\}:\) \(I_{n}\) is the \(n\times n\) identity matrix containing 1's on the diagonal and 0's everywhere else. Therefore, the identity matrix has a dimension of \(n\). \(A\) is an \(n\times n\) matrix with unknown entries. \(A^{2}\) is the square of matrix \(A\), which is also an \(n\times n\) matrix. For each integer power in the set, the resulting matrix is of order \(n\times n\).

The span of the set \(\{I_{n}, A, A^{2}, \ldots\}\) is every linear combination of these matrices. We have to prove that the dimension of this span defined as \(\operatorname{dim}(\operatorname{span}(\{I_{n}, A, A^{2}, \ldots\}))\) is less than or equal to \(n\). Suppose we express the span as a subset \(S\) in the following form: \(S = \{\alpha_0 I_{n} + \alpha_1 A + \alpha_2 A^{2} + \cdots + \alpha_k A^{k}\}\) Where \(\alpha_i\) are scalar coefficients, and \(k \le n\). We want to show that there is a finite set of vectors that are linearly independent, so that we can express the given span with a basis of size less than or equal to \(n\).

Cayley-Hamilton Theorem states that if we let \(A\) be an \(n \times n\) matrix, then \(A\) satisfies its own characteristic polynomial. The characteristic polynomial of \(A\) is given by \(p(\lambda) = \det(A-\lambda I_n)\), where \(\lambda\) is an eigenvalue of \(A\) and \(\det()\) represents the determinant of a matrix. The Cayley-Hamilton theorem asserts that \(p(A) = 0\), where 0 is the zero matrix. The characteristic polynomial of an \(n \times n\) matrix has degree \(n\). Therefore, we can express the polynomial as: \(p(A) = \beta_0 I_n + \beta_1 A + \cdots + \beta_n A^n = 0\), where \(\beta_i\) are the coefficients of the polynomial. We know that any higher power of matrix \(A\) can be expressed as a linear combination of lower powers of \(A\). This means that for any power \(A^m\), where \(m > n\), we can always express it as a linear combination of \(\{I_{n}, A, A^{2}, \ldots, A^{n}\}\).

Since any higher power of \(A\) can be expressed as a linear combination of lower powers of \(A\), we can say that in the subset \(S\), there will be at most \(n+1\) linearly independent matrices (including the identity matrix). Since the dimension of a subspace is the number of linearly independent vectors (or matrices in this case) in its basis, we can therefore conclude that: \(\operatorname{dim}(\operatorname{span}(\{I_{n}, A, A^{2}, \ldots\})) \le n\). Thus, we have proved that the dimension of the span of the given set is less than or equal to \(n\).