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I am trying to understand the concept of a new vector space where vectors are instead matrices. In this example, I am looking at space M made of all 3 by 3 matrices. The text is as follows:

The dimension of M is 9; we must choose 9 numbers to specify an element of M. The space M is very similar to $\mathbb{R}^9$. A good choice of basis is: $$ \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right], \ldots\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $$

The text goes on to specify: the subspace of symmetric 3 x 3 matrices S has dimension 6. When choosing an element of S we pick three numbers on the diagonal and three in the upper right, which tell us what must appear in the lower left of the matrix. One basis is the collection: $$ \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right] $$

The text goes on to do the same for the subspace of all upper triangular matrices (also of dimension 6). Lastly, it describes the intersection of symmetric and upper triangular to be the subspace of all 3 x 3 diagonal matrices.

Question: The basis and dimension of M, all 3 x 3 matrices, it makes sense that it would require 9 different matrices to span that space. It makes sense each one of the elements in the basis would be a matrix with 1 in a different position of the matrix. However for the symmetric example, I don't see an easy way to come about the dimension or basis. I am not clear how/why it's quickly apparent that the dimension of all symmetric matrices is 6. Is this just by intuition?

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Let $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}$ be a symmetric matrix. Then symmetry forces the relations \begin{align*} b &= d \\ c &= g \\ f &= h \\ \end{align*} So we are free to choose any $a$, $e$, and $i$ we like. But a choice of any of the remaining entries in the matrix specifies the values of two entries in the matrix. So of the six remaining values, we are only free to choose three of them.

The three choices on the diagonal and the three choices that specify pairs of entries gives only six choices.

This it not the first way I came to this answer (many years ago). Take your thumb and cover up the "$d$", the "$g$", and the "$h$" since each of these is forced to be equal to an uncovered entry. There are six uncovered entries -- you only get six free choices, then everything else is forced by symmetry.

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The dimension is defined as the number of elements of a basis. So in order to know that the dimension is what you think it is it is sufficient to prove that the proposed set of matrices is a basis. One useful criterion for that is that any matrix in the pertinent vector space should be expressible as a linear combination of to-be-basis vectors with uniquely determined coefficients. An argument that this is indeed true has been explained in the answer of Eric Towers.

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