0
$\begingroup$

I read this statement and I would need some help with what it means:

If A is a matrix and $v$ is a vector such that $Av=0$ then there is a non-zero projection $P$ onto the subspace that annihilates $A$ on the right.

Does this mean that there is a $P$ such that $AP=0$?

  • 1
    Yes: but surely you need to assume that $\mathbf{v}\neq \mathbf{0}$? Also: are you translating or extracting this from some larger context? "onto **the** subspace" doesn't really make sense here: there is no subspace specified to which you can refer by the definite singular article.2011-09-27
  • 2
    You need $P^2=P$ as well.2011-09-27
  • 0
    @Arturo Magidin This is the exact version. You can have a look at it [here](http://mural.uv.es/rusanra/Lie%20Algebras%20in%20Particle%20Physics%202%C2%AA%20ed%20-%20From%20Isospin%20to%20Unified%20Theories%20(Georgi,%201999).pdf) (page 13)2011-09-27
  • 0
    @kuchnahi: You omitted a *lot* of context. You are working in the context of Schur's Lemma, with a particular pair of inequivalent irreducible representations of a group.2011-09-27
  • 0
    Turns out you are not right; "that annihilates $A$ on the right" does not refer to $P$, it refers to "subspace". You want the projection $P$ onto the subspace of all $\mathbf{w}$ such that $A\mathbf{w}=\mathbf{0}$.2011-09-27

1 Answers 1

1

You want to find a projection $P$ (that is, a linear transformation $P$ such that $P^2=P$) which is not zero, and such that $AP=0$. But you need to assume that there is a $\mathbf{v}\neq \mathbf{0}$ such that $A\mathbf{v}=\mathbf{0}$. Otherwise, $A$ is invertible, and then $AP=0$ implies $P=0$.

Correction. In context: we have a group $G$, and two inequivalent irreducible representations $D_1$ and $D_2$ of $G$ such that $D_1(g)A = AD_2(g)$ for all $g\in G$. The desired conclusion is to show that $A=0$.

"The subspace that annihilates $A$ on the right" is the kernel of $A$: the collection of all vectors $\mathbf{v}$ such that $A\mathbf{v}=\mathbf{0}$. You want the projection onto the kernel of $A$; there is an implicit assumption that $\mathbf{v}\neq\mathbf{0}$, as the next paragraph argues that if $A$ does not have any vector that annihilates it on either side, then $A$ must be invertible.