What subspace of $3 \times 3$ matrices is spanned by a given set?

Question

For instance, the question asks what subspace of $3\times 3$ matrices is spanned by the set that is the symmetric matrices. Is the question asking me to determine whether the $3\times 3$ symmetric matrices span the nullspace or column space? I'm confused by the wording of the question and what it means by "spans."

Answer 1

The span is considered in the vector space $${\rm Mat}(3, \Bbb R) = \{ \mbox{order $3$ square matrices with real entries} \}.$$Then $${\rm Sym}(3,\Bbb R) = \{\mbox{order $3$ symmetric matrices with real entries} \}$$is a vector subspace of ${\rm Mat}(3,\Bbb R)$, and it makes sense to wonder what can be a basis for it. If for $i,j \in \{1,2,3\}$ we let $E_{ij}$ denote the matrix with $1$ in the $(i,j)$ and $0$ everywhere else, you can convince yourself that $$\mathcal{B} = \{ E_{11}, E_{12}, E_{13}, E_{21}, E_{22}, E_{23}, E_{31}, E_{32}, E_{33} \}$$is a basis for ${\rm Mat}(3,\Bbb R)$. Can you pick only a few elements from $\mathcal{B}$ to form a basis for ${\rm Sym}(3,\Bbb R)$? They will already be linearly independent, so you'd have to check that they actually span ${\rm Sym}(3,\Bbb R)$. How many of the $E_{ij}$ are symmetric?

And since ${\rm Sym}(3,\Bbb R)$ is already a vector subspace of ${\rm Mat}(3,\Bbb R)$, the span of ${\rm Sym}(3,\Bbb R)$ is itself.

Answer 2

It is the $6-$dimensional subspace that contains the matrices of the form: $$ \begin{bmatrix} x&a&b\\ a&y&c\\ b&c&z \end{bmatrix}= $$ $$ =x\begin{bmatrix} 1&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix}+ y\begin{bmatrix} 0&0&0\\ 0&1&0\\ 0&0&0 \end{bmatrix}+ z\begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&1 \end{bmatrix}+ a\begin{bmatrix} 0&1&0\\ 1&0&0\\ 0&0&0 \end{bmatrix}+ $$ $$ +b\begin{bmatrix} 0&0&1\\ 0&0&0\\ 1&0&0 \end{bmatrix}+ c\begin{bmatrix} 0&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix} $$

so it can be spanned by the six matrices in this linear combination that, being linearly independent, are a basis for this subspace.

Answer 3

The space of real matrices $3\text{x} 3$ has dimension $9$ over $\mathbb R$ which correspond to nine entries $a_{ij}$. The subspace generated by the symmetric ones is given by $$\lambda\begin{bmatrix} x&a&b\\ a&y&c\\ b&c&z \end{bmatrix}+\mu\begin{bmatrix} x'&a'&b'\\ a'&y'&c'\\ b'&c'&z' \end{bmatrix}$$ which is obviously still a symmetric matrix $$\begin{bmatrix} X&A&B\\ A&Y&C\\ B&C&Z \end{bmatrix}$$ where $X=\lambda x+\mu x'$, etc.

Consequently the asked dimension is equal to $6$ which corresponds to the number of reals $X,Y,Z,A,B,C$ for each new matrix.