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Exercise 3.F.24

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Proof. Let $u_{1}, \dots, u_{m}$ be a basis of $U$ . It can be extended to a basis $u_{1}, \dots, u_{m}, v_{1}, \dots, v_{n}$ of V. Let $ψ_{1}, \dots, ψ_{m}, φ_{1}, \dots, φ_{n}$ be the dual basis.

Suppose $φ \in span (φ_{1}, \dots, φ_{n})$ . There are $a_{1}, \dots, a_{n} \in 𝔽$ such that

φ = a_{1} φ_{1} + \dots + a_{n} φ_{n}

Let $u \in U$ . We have

φ (u) = (a_{1} φ_{1} + \dots + a_{n} φ_{n}) (u) = 0

Therefore $φ \in U^{0}$ . Hence $span (φ_{1}, \dots, φ_{n}) \subset U^{0}$ .

Now suppose $φ \in U^{0}$ . Because $φ \in V^{'}$ there are $c_{1}, \dots, c_{m}, a_{1}, \dots, a_{n} \in 𝔽$ such that

φ = c_{1} ψ_{1} + \dots + c_{m} ψ_{m} + a_{1} φ_{1} + \dots + a_{n} φ_{n}

For every $j \in 1, \dots, m$ , we have $ψ_{j} (u_{j}) = c_{j}$ . But $φ \in U^{0}$ , that implies $c_{j} = 0$ and, hence, $φ \in span (φ_{1}, \dots, φ_{m})$ . Thus $U^{0} \subset span (φ_{1}, \dots, φ_{m})$ .

Since $φ_{1}, \dots, φ_{m}$ is linearly independent, $\dim (U^{0}) = m$ . We get

\begin{aligned} \dim V & = m + n \\ = \dim U^{0} + \dim U \end{aligned}

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celiopassos

2017-10-06 00:00

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Alternatively, to prove that $U^{0} \subset span (φ_{1}, . . ., φ_{n})$ , we can use contradiction.
Assume for contradiction, $\exists S \in U^{0}$ yet $S \notin span (φ_{1}, . . ., φ_{n})$
That is,

S = a_{1} ψ_{1} + \dots + a_{m} ψ_{m} + b_{1} φ_{1} + \dots + b_{n} φ_{n}

For $a_{1}, . . ., a_{n}, b_{1}, b_{m} \in F$ , and $\exists$ a nonzero element of the $a$ ’s, say $a_{i}$ . Now consider $S (u_{i})$ which obviously is $1$ by definition of S.

Thus a contradiction is reached and we can conclude that $U^{0} \subset span (φ_{1}, . . ., φ_{n})$ .

Arden_Shi

2025-08-26 20:40

Exercise 3.F.24

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