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Let $V$ be a vector space on finite dimension, and $L,M\subseteq V$ are subspaces such that $M^0=L^0$. Prove $L=M$.

My try:

By the dimension theorem: $\text{dim}V=\text{dim}M+\text{dim}M^0$, we get $\text{dim}M=\text{dim}L$.

Let $\{m_1,\dots,m_n\}$ be a base of $M$ and $\{l_1,\dots,l_n\}$ be a base of $L$, so $M=Span\{m_1,\dots,m_n\}$ and $L=Span\{l_1,\dots,l_n\}.$

By the annihilator's attribute, we get $M^0=(Span\{m_1,\dots,m_n\})^0.$

$M^0=L^0\ \Rightarrow M^0=(Span\{m_1,\dots,m_n\})^0=(Span\{l_1,\dots,l_n\})^0=L^0 $.

Not quite sure how to continue from here.

Any help appreciated.

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    @Bye_World How can I do that? – 2017-02-26
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    Exactly like I did in my answer. – 2017-02-26

1 Answers 1

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Extend $\{m_1, \dots, m_n\}$ to a basis $\{m_1, \cdots, m_n, v_{n+1}, \dots, v_k\}$ of $V$. Let $\{\varphi_1, \dots, \varphi_k\}$ be its dual basis in $V^*$. It's easy to show that $M^0 = \operatorname{span}(\varphi_{n+1},\dots, \varphi_k)$. But then, because $L^0 = M^0$, $L^0 = \operatorname{span}(\varphi_{n+1},\dots, \varphi_k)$. Let $\{l_1, \dots, l_n\}$ be a basis for $L$. Then $$l_i = a_1m_1 + \cdots + a_kv_k$$ for some scalars $\{a_1, \dots, a_k\}\in \Bbb F$. Now consider $\varphi_{n+1}(l_i)$. It must equal $0$, because $\varphi_{n+1} \in L^0$. Thus $a_{n+1} = 0$. The exact same argument shows that $a_{j}=0$ for all $j = n+1, \dots, k$. Thus $l_i$ is a linear combination of $\{m_1, \dots, m_n\}$. I.e. $l_i\in M$ for any $i$. Thus $L\subseteq M$. But $L$ and $M$ have the same dimension, this implies $L=M$.