[proofplan]
We verify the vector subspace criterion directly. First, the zero matrix belongs to $\operatorname{Sym}_n(\mathbb{R})$, so the set is nonempty. Then we check closure under addition and scalar multiplication by computing entries of transposes, which proves that symmetry is preserved by the [vector space](/page/Vector%20Space) operations inherited from $\mathbb{R}^{n \times n}$.
[/proofplan]
[step:Verify that the zero matrix is symmetric]
Let $0_n \in \mathbb{R}^{n \times n}$ denote the zero matrix, so each entry of $0_n$ is $0$. For every $i,j \in \{1,\dots,n\}$, the $(i,j)$-entry of $0_n^\top$ equals the $(j,i)$-entry of $0_n$, hence equals $0$. Therefore $0_n^\top = 0_n$, so $0_n \in \operatorname{Sym}_n(\mathbb{R})$. In particular, $\operatorname{Sym}_n(\mathbb{R})$ is nonempty.
[/step]
[step:Show that sums of symmetric matrices are symmetric]
Let $A,B \in \operatorname{Sym}_n(\mathbb{R})$. By definition of $\operatorname{Sym}_n(\mathbb{R})$, we have $A^\top = A$ and $B^\top = B$.
For every $i,j \in \{1,\dots,n\}$, the definition of transpose and entrywise addition gives
\begin{align*}
((A+B)^\top)_{ij} = (A+B)_{ji}.
\end{align*}
Using entrywise addition in $\mathbb{R}^{n \times n}$, we obtain
\begin{align*}
(A+B)_{ji} = A_{ji}+B_{ji}.
\end{align*}
Since $A^\top=A$ and $B^\top=B$, their entries satisfy $A_{ji}=A_{ij}$ and $B_{ji}=B_{ij}$. Hence
\begin{align*}
((A+B)^\top)_{ij} = A_{ij}+B_{ij}.
\end{align*}
Again by entrywise addition,
\begin{align*}
A_{ij}+B_{ij} = (A+B)_{ij}.
\end{align*}
Thus every entry of $(A+B)^\top$ equals the corresponding entry of $A+B$, so $(A+B)^\top=A+B$. Therefore $A+B \in \operatorname{Sym}_n(\mathbb{R})$.
[guided]
We need to prove that adding two symmetric matrices does not destroy symmetry. Let $A,B \in \operatorname{Sym}_n(\mathbb{R})$. This means exactly that $A^\top=A$ and $B^\top=B$. In entry form, those two equalities say that for every pair of indices $i,j \in \{1,\dots,n\}$,
\begin{align*}
A_{ji}=A_{ij}
\end{align*}
and
\begin{align*}
B_{ji}=B_{ij}.
\end{align*}
Now fix arbitrary indices $i,j \in \{1,\dots,n\}$. To prove that $A+B$ is symmetric, we must show that the $(i,j)$-entry of $(A+B)^\top$ equals the $(i,j)$-entry of $A+B$. By the definition of transpose,
\begin{align*}
((A+B)^\top)_{ij} = (A+B)_{ji}.
\end{align*}
By the definition of matrix addition, the right-hand side is
\begin{align*}
(A+B)_{ji}=A_{ji}+B_{ji}.
\end{align*}
Because $A$ and $B$ are symmetric, we may replace $A_{ji}$ by $A_{ij}$ and $B_{ji}$ by $B_{ij}$. Therefore
\begin{align*}
A_{ji}+B_{ji}=A_{ij}+B_{ij}.
\end{align*}
Finally, by the definition of matrix addition again,
\begin{align*}
A_{ij}+B_{ij}=(A+B)_{ij}.
\end{align*}
Combining these equalities gives
\begin{align*}
((A+B)^\top)_{ij}=(A+B)_{ij}.
\end{align*}
Since the indices $i$ and $j$ were arbitrary, all corresponding entries of $(A+B)^\top$ and $A+B$ agree. Hence $(A+B)^\top=A+B$, so $A+B \in \operatorname{Sym}_n(\mathbb{R})$.
[/guided]
[/step]
[step:Show that scalar multiples of symmetric matrices are symmetric]
Let $c \in \mathbb{R}$ and let $A \in \operatorname{Sym}_n(\mathbb{R})$. By definition, $A^\top=A$.
For every $i,j \in \{1,\dots,n\}$, the definition of transpose and scalar multiplication gives
\begin{align*}
((cA)^\top)_{ij} = (cA)_{ji}.
\end{align*}
Since scalar multiplication in $\mathbb{R}^{n \times n}$ is entrywise,
\begin{align*}
(cA)_{ji}=cA_{ji}.
\end{align*}
Because $A^\top=A$, we have $A_{ji}=A_{ij}$. Hence
\begin{align*}
((cA)^\top)_{ij}=cA_{ij}.
\end{align*}
By entrywise scalar multiplication,
\begin{align*}
cA_{ij}=(cA)_{ij}.
\end{align*}
Thus $(cA)^\top=cA$, so $cA \in \operatorname{Sym}_n(\mathbb{R})$.
[/step]
[step:Apply the vector subspace criterion]
We have shown that $\operatorname{Sym}_n(\mathbb{R})$ contains the zero matrix, is closed under addition, and is closed under scalar multiplication by real scalars. Since all of these operations are the operations inherited from the ambient real vector space $\mathbb{R}^{n \times n}$, the vector subspace criterion implies that $\operatorname{Sym}_n(\mathbb{R})$ is a vector subspace of $\mathbb{R}^{n \times n}$. This proves the theorem.
[/step]
