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I have the same question as posted here . However I don't understand the proof given.

The question is, for a Vector Space $V$ with a subspace $U$, prove that: $\text{dim} U + \text{dim}U^{0}=\text{dim}V$. Where $U^{0}$ is the annihilator of U.

The proof as given was:

Assuming you are dealing with finite dimensional spaces, you can just use a dual basis argument: Suppose $\{v_1, v_2, \ldots, v_m\}$ is a basis for $U$, which can be extended to a basis $\{v_1, v_2, \ldots, v_m, v_{m+1}, \ldots, v_n\}$ for $V$.

Let $\{\varphi_1, \varphi_2, \ldots, \varphi_n\}$ be a dual basis for $V^{\ast}$, then $$ \{\varphi_{m+1}, \varphi_{m+2}, \ldots, \varphi_n\} \subset U^{\circ} $$ Now check that this set forms a basis for $U^{\circ}$.

I understand that: $\{\varphi_1, \varphi_2, \ldots, \varphi_n\}$ can be assigned as a dual basis for $V^{\ast}$, however I don't understand why that implies that $ \{\varphi_{m+1}, \varphi_{m+2}, \ldots, \varphi_n\} \subset U^{\circ}$. If you could expand on that, I would be very grateful. Thanks.

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  • $\begingroup$ possible duplicate of Proof: dimension of annihilator $\endgroup$ – Stella Biderman Mar 9 '14 at 21:26
  • $\begingroup$ Sorry, I wasn't if my comment would become buried, since the other post is a few months old. $\endgroup$ – user123429 Mar 9 '14 at 21:27
  • $\begingroup$ @user123429 There is no problem is asking this as a question. You linked to the relevant question, but you should make your question self contained, as people shouldn't have to open links to be able to answer. $\endgroup$ – Git Gud Mar 9 '14 at 21:29
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We have

$$\varphi_k(v_j) = \begin{cases} 1 & \text{if } k = j\\ 0 & \text{if } k \ne j\text{} \end{cases}$$

So every element of $\{\varphi_{m+1}, \dots, \varphi_{n}\}$ is zero on each basis element of $U$, hence $\{\varphi_{m+1} , \dots,\varphi_n \} \subseteq U^\circ$.

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