Let $m \geq 2$. The subset $X$ of $m \times 2$ matrices with rank $1$ is a (smooth) submanifold of $\mathbb{R}^{m\times 2}$.
Let $A$ be in $X$. I know from a more general statement that the tangent space at $A$ is the set of matrices $B$ such that $Bx$ is in the image of $A$ if $x$ is in the kernel of $A$.
Is there a stronger way to describe the tangent space in this case?
Also, how can I compute the normal space at $A$? (I'm using the Frobenius inner product, if that matters.)
Every $m\times 2$ matrix of rank $1$ has the form $$ \begin{bmatrix} \lambda\textbf{u} & \mu\textbf{u} \end{bmatrix} $$ where $\lambda,\mu\in\mathbb{R}$ are not both zero, and $\textbf{u}\in\mathbb{R}^m$ is a unit (column) vector. Note that this representation is not unique: we can negate $\textbf{u}$, $\lambda$, and $\mu$ without changing the matrix.
The tangent space to the manifold at this matrix is the $(m+1)$-dimensional subspace consisting of all matrices of the form $$ \begin{bmatrix}\alpha\textbf{u} + \lambda\textbf{v} & \beta\textbf{u} + \mu\textbf{v}\end{bmatrix} $$ where $\alpha,\beta\in\mathbb{R}$ and $\textbf{v}$ is any vector perpendicular to $\textbf{u}$. Note that this vector space is indeed $(m+1)$-dimensional. Moreover, each of these vectors really is a tangent vector, since $$ \begin{bmatrix}\alpha\textbf{u} + \lambda\textbf{v} & \beta\textbf{u} + \mu\textbf{v}\end{bmatrix} \;=\; \frac{d}{dt}\biggl(\begin{bmatrix}(\lambda+\alpha t)(\textbf{u}+t\textbf{v}) & (\mu+\beta t)(\textbf{u}+t\textbf{v})\end{bmatrix}\biggr)\Biggr|_{t=0} $$ and the path on the right lies entirely on the manifold.
The normal space consists of all matrices of the form $\begin{bmatrix} \mu\textbf{v} & -\lambda\textbf{v} \end{bmatrix}$, where $\textbf{v}$ is any vector perpendicular to $\textbf{u}$.