I am wondering how to find the dimension of the sub vector space $A=\{u \in L(E) \ \vert \ u\circ v = v\circ u=0\}$ of $L(E)$, where $L(E)$ is the set of endomorphism of $E$ (i.e linear map from $E$ to $E$) and $v \in L(E)$. Thanks !
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For an endomorphism $v\in L(E)$, we need to find those maps $u\in L(E)$ such that $\operatorname{im} u\subset \ker v$ and $\operatorname{im} v\subset \ker u$. Can you approach this now? – Ian Coley Mar 16 '14 at 22:13
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@Ian Thanks, yes I already noticed that but failed to find a solution... – jerry Mar 16 '14 at 22:17
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Fix $v \in L(E)$, let $u \in A$ be arbitrary. Because $v \circ u = 0$, we know that $\text{Image}(u) \subset \ker(v)$. Because $u \circ v = 0$, we know $\text{Image}(v) \subset \ker(u)$.
Let $\alpha_1,\cdots,\alpha_k$ be a basis for $\ker(v)$.
Let $\beta_1,\cdots,\beta_j$ be a basis for $\text{Image}(v)$. Extend this basis to form the basis $\beta_1,\dots,\beta_n$ of $E$.
We note that $u$ must take the span of $\beta_1,\cdots,\beta_j$ to $0$, and its image must fall within the span of $\alpha_1,\cdots,\alpha_k$. However, we may freely choose where within the span of $\alpha_1,\cdots,\alpha_k$ $u$ takes the span of $\beta_{j+1},\dots,\beta_n$.
That is, $$ A \cong \text{Hom}(\text{span}(\beta_{j+1},\dots,\beta_n),\text{span}(\alpha_1,\cdots,\alpha_k)) $$ It follows that $\dim(A) = (n - \dim(\text{Image}(v))) \times \dim(\ker(v))$. By the rank-nullity theorem, we have $$ \dim(A) = \dim(\ker(v))^2 $$
Ben Grossmann
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I think I know this but how can you say "extend this basis to form the basis... of E" – Tyler Hilton Mar 16 '14 at 23:23
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@TylerHilton any linearly independent set of vectors in a vector space can be extended to a basis. Here's an example. – Ben Grossmann Mar 16 '14 at 23:32
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