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

1. May 6, 2016

### 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.

Last edited: May 6, 2016
2. May 6, 2016

### 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.

3. May 7, 2016

### joej24

Thank you, I understand now. When you say
this means that all the $v$ perpendicular to the $k$-dimensional plane satisfy $\overline{A} v = 0$.