Multiplication by a matrix in GL rotates a plane's basis?

[joej24]

Let A = (a_{ij}) be a k\times n matrix of rank k.
The k row vectors, a_i are linearly independent and span a k-dimensional plane in \mathbb{R}^n.

In "Geometry, Topology, and Physics" (Ex 5.5 about the Grassmann manifold), the author states that for a matrix g\in \textrm{GL}(k,\mathbb{R}),
\overline{A} = gA defines the same plane as A because g simply rotates the basis within the k-plane.

I'm having trouble seeing this.

 

[andrewkirk]

Let $r_j$ denote the $j$th row of $A$, and consider a vector $v$ that is normal to the $k$-dimensional plane. Note that $v$ must be perpendicular to all $r_j$, so $r_j\cdot n=0$.

Then $Av=0$ since the $j$th component of $An$ is $r_j\cdot v$.

So $\bar Av=(gA)v=g(Av)=g\mathbf 0=\mathbf0$. So $v$ is also normal to the plane defined by $\bar A$. Since that holds for all $(n-k)$ basis vectors of the null space of $A$, and the rank of $\bar A$ is the same as that of $A$, the plane (its rowspace) must be the same.

 

[joej24]

Thank you, I understand now. When you say

Since that holds for all $(n-k)$ basis vectors of the null space of $A$

this means that all the $v$ perpendicular to the $k$-dimensional plane satisfy $\overline{A} v = 0$.