To find the dimension of the vector space $V = \{ A \in M_{3 \times 3}(\mathbb{R}) : A = A^T \}$, we need to examine the properties of symmetric matrices of size $3 \times 3$.

Step-by-Step Solution:

1. Definition of a Symmetric Matrix: A matrix $A \in M_{3 \times 3}(\mathbb{R})$ is symmetric if $A = A^T$. This means that the entries of $A$ satisfy:

   • $a_{ij} = a_{ji}$ for all $i, j$.

2. Form of a Symmetric Matrix: A general symmetric $3 \times 3$ matrix has the form: $A = \begin{pmatrix} a & b & c \\ b & d & e \\ c & e & f \end{pmatrix}$ Here:

   • $a, d,$ and $f$ are the diagonal elements, which can take any value in $\mathbb{R}$.
   • $b, c,$ and $e$ are the off-diagonal elements, but each pair is equal (e.g., $a_{12} = a_{21} = b$).

3. Counting the Independent Entries: Each unique entry in this symmetric matrix corresponds to a free variable:

   • $a, b, c, d, e, f$
   • There are a total of 6 independent entries.

4. Conclusion: The dimension of the vector space $V$ of all symmetric $3 \times 3$ matrices is therefore 6, as each independent entry represents a basis element in this space.

Answer:

The dimension of the vector space $V$ is $\boxed{6}$.

Would you like further details, or do you have any questions?

Here are five related questions to deepen understanding:

1. What is the dimension of the space of $n \times n$ symmetric matrices in general?
2. How does the dimension change if we consider only skew-symmetric matrices instead?
3. Can we define a basis for the space of $3 \times 3$ symmetric matrices explicitly?
4. How would you compute the dimension of the space of symmetric matrices over complex numbers?
5. What properties do symmetric matrices have in relation to eigenvalues?

Tip: Symmetric matrices are always diagonalizable over the real numbers, making them useful in applications like Principal Component Analysis (PCA).