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

  • $\begingroup$ 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) $\endgroup$ – Yorch Jan 6 '17 at 23:30
  • 1
    $\begingroup$ @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

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 $V=W$.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.