Let $V$ be a $n$-dimensional real vector space and $P$ and $P'$ be subspaces with dimension $k$. I need to prove that it is possible to find a subspace $Q$ of dimension $n-k$ such that $Q$ has trivial intersection with both $P$ and $P'$. I tried to manipulate the bases but cannot find a way... Could anyone please help me?
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Shouldn't the dimension of $Q$ be $n-2k$? I mean, $P$ and $P'$ each of them has dimension $k$, right? – la flaca Jan 13 '18 at 18:59
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Ah ok, $P$ and $P'$are not supposed to have trivial intersection I guess – la flaca Jan 13 '18 at 19:02
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No The dimension of $Q$ is n-k – Keith Jan 13 '18 at 19:02
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Could you please help me? – Keith Jan 13 '18 at 19:04
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1I'm thinking on how we could do it – la flaca Jan 13 '18 at 19:06
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The union of the two spaces $P\cup P'$ is not all of $V$, so there is a vector $v_1\notin P\cup P'$. Now proceed inductively on the spaces $P\oplus \text{span}\{v_1\}$ and $P'\oplus \text{span}\{v_1\}$, generating a vector $v_2$ not in the union $P\oplus \text{span}\{v_1\} \cup P'\oplus \text{span}\{v_1\}$, then a vector $v_3$ not in $P\oplus \text{span}\{v_1,v_2\}\cup P'\oplus \text{span}\{v_1,v_2\}$, etc., until you get $n-k$ vectors $v_1,\dots,v_{n-k}$. Each $v_j$ is not in $\text{span}\{v_1,\dots,v_{j-1}\}$, so these vectors are all linearly independent.
The space $Q:=\text{span}\{v_1,\dots,v_{n-k}\}$ is the space you're looking for. It has trivial intersection with $P$, because otherwise $\sum_j c_j v_j \in P$ for some non-zero $\{c_j\}$, so that letting $J$ be the largest number such that $c_J\neq 0$, we would have $v_J \in P\oplus\text{span}\{v_1,\dots,v_{J-1}\}$, a contradiction. Similarly it has trivial intersection with $P'$.
Yly
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