1
$\begingroup$

This is one of the exercise problems I found in Halmos' 'Finite Dimensional Vector Spaces':

Let $V$ be a finite-dimensional vector space and $M$ and $N$ be two of its subspaces. Then, if $$M \cup N = V$$ then either $M=V$ or $N=V$.

I'm trying a proof by contradiction. I assume the conclusion in false. Then neither $M$ nor $N$ can be of the dimension of $V$ because if it were, then it would be equal to $V$ (this I have proven). So both the subspaces will have a lower dimension than $V$. Now all I need to show is that the union of these two subspaces will miss out some element of $V$. I'm stuck at this point and would be very much obliged if someone could give me a hint. Thanks!

$\endgroup$
1
  • 2
    $\begingroup$ Take $x \in M - N $ and $y \in N - M$. What about $x+y$? $\endgroup$
    – sebigu
    Nov 11, 2014 at 10:13

2 Answers 2

Reset to default
2
$\begingroup$

There has to be $y\in M-N$ and $x\in N-M$, $x,y\neq 0$, otherwise $N\subset M$ or $M\subset N$. Now if $x+y\in N$, then $y=(x+y)-x\in N$, a contradiction. Also if $x+y \in M$, then $x = (x+y)-y\in M$, a contradiction again.

$\endgroup$
3
  • $\begingroup$ Why it is that $e_m$ is in $M$? $\endgroup$
    – Learnmore
    Nov 11, 2014 at 10:38
  • $\begingroup$ @learnmore Thank you for pointing that out, now should be fine. $\endgroup$
    – J.A.L
    Nov 11, 2014 at 11:02
  • $\begingroup$ You have proved iff condition for $M\cup N$ to be a subspace.OP did not ask that $\endgroup$
    – Learnmore
    Nov 11, 2014 at 11:06
0
$\begingroup$

$V=M\cup N$ ( given) Now $M\cup N$ is a subspace of $V$ iff either $M\subseteq N$ or $N\subseteq M$(surely you know that).Here $M\cup N(=V)$ is a subspace of $V$ Now if $M\subseteq N$ then $M\cup N=N$ ie.$N=V$ or if $N\subseteq M$ then $M\cup N=M$ i.e $M=V$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .