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

• I
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:

andrewkirk
Homework Helper
Gold Member
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.

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##.