This question asks for the dimension of a specific vector space, which is equivalent to finding the number of independent entries required to define any matrix in that space. A symmetric matrix is defined by the property \(A^T = A\), meaning its entries are symmetric across the main diagonal (\(a_{ij} = a_{ji}\)).
Step 1: Understanding the Question:
We need to determine how many elements in an \( n \times n \) matrix we can choose freely, given the constraint that the matrix must be symmetric. This number will be the dimension of the vector space of all such matrices.
Step 2: Key Formula or Approach:
The total number of independent entries can be found by summing the number of independent diagonal elements and the number of independent off-diagonal elements.
Step 3: Detailed Explanation:
Let's break down the matrix into its components:
Diagonal Elements: An \( n \times n \) matrix has \( n \) diagonal elements (\(a_{11}, a_{22}, \dots, a_{nn}\)). There are no symmetry constraints on these elements, so all \(n\) of them can be chosen independently.
Off-Diagonal Elements: The total number of elements in the matrix is \(n^2\). After accounting for the \(n\) diagonal elements, there are \(n^2 - n\) off-diagonal elements. The symmetry condition \(a_{ij} = a_{ji}\) means that once we choose an element above the main diagonal (e.g., \(a_{12}\)), the corresponding element below the diagonal (\(a_{21}\)) is automatically determined. Therefore, we only need to choose the elements in the upper triangle (or lower triangle). The number of such elements is \( \frac{n^2 - n}{2} \).
The total number of independent entries (the dimension) is the sum of these two counts:
\[ \text{Dimension} = (\text{Independent diagonal elements}) + (\text{Independent off-diagonal elements}) \]
\[ \text{Dimension} = n + \frac{n^2 - n}{2} = \frac{2n + n^2 - n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2} \]
Step 4: Final Answer:
The dimension of the vector space of all \( n \times n \) real symmetric matrices is \( \frac{n(n+1)}{2} \).