What is the dimension and the number of basis vectors for a subspace of 3×3 symmetric matrices?
Earlier my professor told us that the dimension and the number of basis vectors for a subspace are the same. And I did get that as they are the vectors that can span the whole subspace. But when describing the subspace of symmetric matrices he says when counting the basis vectors we only consider the diagonal elements and the entries either above or below. I get that if you specify the entries above the diagonal the entries below it will be the same and so if you take a 3×3 matrix, you only need to specify 6 entries. But aren't the matrices still containing all nine entries then how is it that the subspace is 6D and not 9D?
The dimension of a vector space is not the amount of numbers somehow "involved" in representing one of its members but the least amount of numbers needed to do so.
As an example, take the subspace of $\mathbb{R}^3$ given by all the vectors perpendicular to the $xy$ plane. Every vector in this subspace can be represented like $(0,0,z)$ for some real $z$ right? This representation still involves three numbers. But the subspace is one dimensional, because I only need one real number to specify any of its elements. Furthermore, there is a basis in which this vector space is manifestly one dimensional. If I choose as a basis the vector $e=(0,0,1)$ in the standard basis, then every element of the subspace in question can be written like $$v=z\,e, z \in \mathbb{R}$$or $v=(z)$.
Likewise for the case of symmetric $3\times3$ matrices, yes, every matrix can be represented like $$\begin{bmatrix}a&b&c\\b&d&e\\c&e&f \end{bmatrix}$$ which looks like it could be 9 dimensional. After all you specified a number 9 times. But the point is that this representation has redundant information, seeing as if you take as a basis this set of matrices $$\begin{align*} \begin{bmatrix}1&0&0\\0&0&0\\0&0&0 \end{bmatrix}&,\begin{bmatrix}0&1&0\\1&0&0\\0&0&0 \end{bmatrix}\hspace{-8mm}&,\begin{bmatrix}0&0&1\\0&0&0\\0&0&0 \end{bmatrix}&, \begin{bmatrix}0&0&0\\0&1&0\\0&0&0 \end{bmatrix}\hspace{-8mm}&,\begin{bmatrix}0&0&0\\0&0&1\\0&1&0 \end{bmatrix}&,\begin{bmatrix}0&0&0\\0&0&0\\0&0&1 \end{bmatrix}\end{align*}$$ then (if you label them $A_i$ from $1$ through $6$), then every symmetric matrix $S$ can be written like $$A=aA_1+bA_2+cA_3+dA_4+eA_5+fA_6$$ or $A=(a,b,c,d,e,f)$which is of course a six dimensional vector.
The main point is that the notion of dimension isn't one of how many numbers/ vectors you can use to represent/generate a member of a vector space; after all you can always use an arbitrarily high number of vectors to generate the whole space. No, the dimensionality of a space refers to the minimum amount of vectors/numbers you need; any lower and you "lose information" about that space. But you can of course go higher, at the cost of every new vector/ number you had being redundant in the sense that it is a linear combination of the vectors you already had.