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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.)

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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 $\textbf{u}$.

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  • $\begingroup$ The more general statement I've looked at says that the dimension of the submanifold should be $m+1$. So I think the dimension of the tangent space should also be $m + 1$. $\endgroup$ – SorTheene Jun 13 '15 at 20:21
  • $\begingroup$ Here's that statement: Let $E$ and $F$ be two finite-dimensional vector spaces. Let $r$ be an integer with $0 \leq r \leq \min(\dim(E),\dim(F))$. In the vector space $L(E;F)$ [of linear maps from $E$ to $F$], the set $L_r$ of those $v$ with $\mathrm{rank}(v) = r$ is a submanifold with codimension equal to $(\dim(E) - r)(\dim(F) - r)$. $\endgroup$ – SorTheene Jun 13 '15 at 20:26
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    $\begingroup$ Also: "The tangent space at the point $v$ to the submanifold $L_r$ of $L(E;F)$ consists of the linear maps $w:E \to F$ such that $w(\mathrm{Ker}(v)) \subset \mathrm{Im}(v)$." $\endgroup$ – SorTheene Jun 13 '15 at 21:01
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    $\begingroup$ @SorTheene My apologies. My answer was completely wrong -- I don't know what I was thinking. I've fixed it now. $\endgroup$ – Jim Belk Jun 13 '15 at 21:33
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    $\begingroup$ @SorTheene Only $m-1$ of the dimensions come from $\textbf{v}$, since $\textbf{v}$ must be perpendicular to $\textbf{u}$. The remaining two dimensions are the choices of $\alpha$ and $\beta$. $\endgroup$ – Jim Belk Jun 17 '15 at 23:49

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