What is the tangent space to the quadric $x_1^2+x_2^2+\ldots+x_{n-1}^2=x_n^2$ at the point $p=(1,0,\ldots,0,1)$?
The definition of a tangent space that I know is based on the fact that we have a manifold $X$, and then finding a map $f$ from a neighborhood $V$ of $p$ to $\mathbb{R}^k$ such that $f$ is a submersion at all points of $V$.
But here, this quadric is not a manifold. How can I compute its tangent space?
The tangent space to an algebraic variety $V$ defined by $f_1(X)=\dots=f_p(X)=0$ at a smooth point is the kernel of the Jacobian matrix of $(f_1,\ldots,f_p)$. A smooth point is a point of $V$ where the rank of the Jacobian matrix is equal to the codimension of $V$.
Here, the Jacobian matrix is $J(x_1,\ldots, x_n)=\begin{bmatrix} 2 x_1&\dots&2 x_{n-1}&-2 x_n\end{bmatrix}$.
At $p=(1,0,\ldots,0,1)$, we get $J(p)=\begin{bmatrix} 2 & 0 &\dots&0&-2\end{bmatrix}$.
The rank of $J(p)$ is equal to $1$, which is also also the codimension of your hypersurface so $p$ is indeed a smooth point of the quadric. Then the tangent space to your quadric at $p$ can be computed as the kernel of $J(p)$.
