Is the set of symmetric matrices in $M_k(n,\mathbb{R})$ a smooth submanifold of $M_k(n,\mathbb{R})$? Let $M_k(n,\mathbb{R})$ be the set of all $n \times n$ real matrices of rank $k$. This forms a smooth manifold (see Proposition 1.14 on page 133 of Optimization and Dynamical Systems by Helmke and Moore).
Is the set $S$ of symmetric matrices in $M_k(n,\mathbb{R})$ a smoothly embedded submanifold of $M_k(n,\mathbb{R})$?
 A: Let $S_k\subset\mathbb R^{n\times n}$ stand for the set of rank $k$ symmetric $n\times n$ real matrices. We will see that $S_k$ is indeed a smooth submanifold of $\mathbb R^{n\times n}$.
Domains of coordinates.
First, for every choice of indices $1\le i_1<\dots<i_k\le n$ denote $S_k(i_1,\dots,i_k)$ the set of all $A\in S_k$ whose columns $i_1,\dots,i_k$ are independent. This set is open, as it is defined by the condition that there is some non-zero $k\times k$ minor involving the chosen columns. Clearly $S_k$ is the union of these open sets, and it is enough to see that each one is a smooth submanifold of $\mathbb R^{n\times n}$. Since a permutation of columns, which is a linear isomorphism of $\mathbb R^{n\times n}$, transforms $S_k(i_1,\dots,i_k)$ onto $\varSigma=S_k(1,\dots,k)$, we are reduced to see that this $\varSigma$ is a smooth submanifold.
Description of matrices in $\varSigma$. Let us characterize when a matrix $A\in\mathbb R^{n\times n}$ belongs to $\varSigma$. We write $A=(U\ |\, V)$, with $U\in\mathbb R^{n\times k}$ and $V\in\mathbb R^{n\times(n-k)}$
(1) The first $k$ columns of $A$ are independent and rk$(A)=k$ if and only if rk$(U)=k$ and there is a matrix $\varLambda\in\mathbb R^{k\times(n-k)}$ such that $V=U\varLambda$.
Set
$$
U=\begin{pmatrix}X\\Y\end{pmatrix}\text{ with } X\in\mathbb R^{k\times k} \text{ and } Y\in\mathbb R^{(n-k)\times k},\text{ so that }
A=\begin{pmatrix}X&X\varLambda\\Y&Y\varLambda\end{pmatrix}.
$$
Then:
(2) That $A$ is symmetric means $X=X^t$ and $Y=(X\varLambda)^t=\varLambda^tX,\ $ because these two conditions imply
$$
(Y\varLambda)^t=\varLambda^tY^t=\varLambda^tX\varLambda=Y\varLambda.
$$
(3) Back to rank$=\!k$ we get
$$
k=\text{rk}(U)=\text{rk}\begin{pmatrix}X\\Y\end{pmatrix}=
\text{rk}\begin{pmatrix}X\\ \varLambda^tX\end{pmatrix}=\text{rk}(X),
$$
hence $X$ is a regular symmetric $k\times k$ matrix: $\det(X)\ne0$.
Parametrizacion of $\varSigma$. After this preparation we consider the linear space $L\subset\mathbb R^{k\times k}$ of all symmetric $k\times k$ matrices, which has dimension $\tfrac{1}{2}k(k+1)$, and the open subset $\varOmega\subset L$ defined by $\det\ne0$. We define the mapping
$$
\varphi:\varOmega\times\mathbb R^{k\times(n-k)}\to\mathbb R^{n\times n}:(X,\varLambda)\mapsto A=\begin{pmatrix}X&X\varLambda\\ \varLambda^tX&\varLambda^tX\varLambda\end{pmatrix}.
$$
This is a diffeomorphism onto $\varSigma$. Indeed, it is clearly smooth, and it is a bijection by (1), (2) and (3). Furthermore its inverse is the smooth mapping
$$
\varphi^{-1}:\begin{pmatrix}X&P\\Y&Q\end{pmatrix}\mapsto (X,X^{-1}Y^t),
$$
(this is well defined because $\varSigma$ is contained in the open subset of $\mathbb R^{n\times n}$ given by $\det(X)\ne0$).
Finally, we can compute dimensions:
$$
\dim(S_k)=\dim(\varSigma)=\dim(\varOmega)+k(n-k)=\tfrac{1}{2}k(k+1)+k(n-k)=\tfrac{1}{2}k(2n-k+1)).
$$
Remark. Although the equations $A=A^t$ define $S_k$ on the manifold $M_k(n,\mathbb R)$, they are difficult to handle to prove $S_k$ is a manifold. The rank restriction makes many of those equations redundant, as shown by the argument above. Indeed, there we see that the symmetry to consider is that of the first $k$ columns and rows. Consequently, from all $\tfrac{1}{2}n(n-1)$ equations in $A=A^t$ only $r=\tfrac{1}{2}k(k-1)+k(n-k)$ are needed. Now those $r$ equations define a linear mapping $L:\mathbb R^{n\times n}\to\mathbb R^r$ and the problem is to check that the restriction $L|:M_k(n.\mathbb R)\to\mathbb R^r$ is a submersion. That assumed, then $S_k=L^{-1}(0)\cap M_k(n,\mathbb R)$ would be a smooth manifold of dimension
$$
\dim(S_k)=\dim(M_k(n,\mathbb R))-r=\big(kn+k(n-k)\big)-\big(\tfrac{1}{2}k(k-1)+k(n-k)\big)
$$
$$
=kn-\tfrac{1}{2}k(k-1)=\tfrac{1}{2}k(2n-k+1))
$$
as we had found before. Now to see $L|$ is indeed a submersion one must check $d_AL|:T_A\to\mathbb R^r$ is onto for every tangent space $T_A$ to $M_k(n,\mathbb R)$ at $A\in S_k$. Since $L$ is linear, $d_AL=L$, hence that restriction is onto iff
$$
r=\dim(T_A)-\dim\big(\ker(L)\cap T_A\big)=\dim\big(\ker(L)+T_A\big)-\dim\big(\ker(L)\big),
$$
or
$$ 
\dim\big(\ker(L)+T_A\big)=r+\dim\big(\ker(L)\big)=n^2,
$$
that is $\ker(L)+T_A=\mathbb R^{n\times n}$. This is what takes this submersion approach: to check the kernel of $L$ and the tangent space generate $\mathbb R^{n\times n}$.
