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First, sorry that the title is a bit messy - here is what I want to ask: Let $V$ be a finite dimensional vector space over the field $K$ , and let $S$ be the set of all linear maps (endomorphisms) of $V$ into itself. Show that $V $ is a simple $S$-space, that is the only $S$-invariant subspaces of $V$ are $V$ itself and the zero subspace. Does anyone have any idea how to do this? Thanks

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  • $\begingroup$ Hint: If $\;B=\{v_1,...,v_n\}\;$ is any basis of $\;V\;$ and $\;\{w_1,...,w_n\}\;$ is any set of vectors in $\;V\;$ , then there's a unique $\;T\in S\;$ s.t. $\;Tv_i=w_i\;$ . How does this solve your problem? $\endgroup$ – DonAntonio Jan 6 '17 at 23:30
  • $\begingroup$ I don't understand why it's important that $K$ be algebraically closed.(or why $V$ is finite -dimensionalfor that matter) $\endgroup$ – Jorge Fernández Hidalgo Jan 6 '17 at 23:30
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    $\begingroup$ @JorgeFernándezHidalgo Indeed, none of those is relevant. Finite dimensionality though is required perhaps to avoid using AC here. $\endgroup$ – DonAntonio Jan 6 '17 at 23:36
  • $\begingroup$ Yes you are right. sorry for not noticing before. I will make the edit now $\endgroup$ – P-S.D Jan 7 '17 at 21:31
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Let $W$ be a non-zero $S$-invariant subspace of $V$ and $dimW=r\leq n$. Consider an ordered basis $B_V=\{v_1,\ldots,v_n\}$ of $V$ and an ordered basis $\{w_1,\ldots,w_r\}$ of $W$. We complete $B_W$ into an ordered basis $B_W=\{w_1,\ldots,w_r,w_{r+1},\ldots,w_n\}$ of $V$, where $w_{r+1},\ldots,w_n\not\in W$. Now, the linear map $A$ of the change of basis $B_W$ to $B_V$ belongs to $S$ and $Aw_i=v_i$. By our hypothesis, $W$ is $S$-invariant, so $v_i$ belongs to $W$ for every $i=1,\ldots,r$.

We can apply this argument to every possible ordering of the elements of $B_V$ and of $B_W$ and this way we have shown that $v_i\in W$ for every $i=1,\ldots,n$. Thus $V=W$.

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If by "$S$-invariant subspace" you mean "subspace which is invariant under every $f\in S$," then the statement doesn't even need the finite dimensional hypothesis.

Suppose $A\subset V$ is a nontrivial subspace. Then let $\alpha\in A$ be nonzero, and let $\beta\in V\setminus A$ (note that since $0\in A$ we have $\beta\not=0$ trivially). Since $A$ is a subspace, the set $\alpha,\beta$ is linearly independent (why?). So - by the Axiom of Choice - we may find a basis $B$ of $V$ with $\alpha,\beta\in B$.

Now consider the function $\pi: B\rightarrow B$ swapping $\alpha$ and $\beta$ and leaving all other elements of $B$ fixed. This extends to a unique $f\in S$, since $B$ is a basis of $V$; and $A$ is clearly not $f$-invariant.


Note that this invokes the axiom of choice to get a basis for $V$. Without the axiom of choice, bases for arbitrary vector spaces need not exist. If $V$ is finite-dimensional, though, the axiom of choice is not needed.

I am not sure whether AC is needed for the general case, but I suspect it is.

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  • $\begingroup$ Thank you for your answer. What do you mean by AC in the last sentence? $\endgroup$ – P-S.D Jan 7 '17 at 22:14
  • $\begingroup$ @P-S.D The axiom of choice - that's its standard abbreviation. $\endgroup$ – Noah Schweber Jan 7 '17 at 22:17
  • $\begingroup$ ah ok, i understand. $\endgroup$ – P-S.D Jan 7 '17 at 22:45
  • $\begingroup$ Could you give me a hint as to why a,b is linearly independent? $\endgroup$ – P-S.D Jan 7 '17 at 22:50
  • $\begingroup$ @P-S.D If $\alpha$ and $\beta$ were not linearly independent, what would that mean about $\beta$ and the subspace generated by $\alpha$? What does that tell you about $\beta$ and $A$? $\endgroup$ – Noah Schweber Jan 7 '17 at 22:54

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