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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 $AP=0$?

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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. –  Arturo Magidin Sep 27 '11 at 16:45
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You need $P^2=P$ as well. –  lhf Sep 27 '11 at 16:46
    
@Arturo Magidin This is the exact version. You can have a look at it here (page 13) –  kuch nahi Sep 27 '11 at 16:53
    
@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. –  Arturo Magidin Sep 27 '11 at 17:15
    
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 $A\mathbf{w}=\mathbf{0}$. –  Arturo Magidin Sep 27 '11 at 17:20
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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.

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