Let $V$ be a vector space and let $\mathcal{B}=\{v_1,v_2,v_3,v_4\}$ be a basis of this $4$-dimensional space.

We define $F=\{f\in V^*|f(\text{Span}(v_1+v_2))=0\}$.

1) Is it true that $L= \text{Ann(Span(}v_1-v_2+v_3,2v_1+v_2+v_3+v_4)) \subset F?$

2) Prove that for $v \in V$, $v_3+v_4\in \text{Ann}(F\cap \text{Ann}(v)) \iff v\in \text{Span}(v_1+v_2,v_3+v_4) \wedge v \notin \text{Span}(v_1+v_2)$

My solution:

1) I'll say that $L \subset F \iff v_1+v_2 \in \text{Span}(v_1-v_2+v_3,2v_1+v_2+v_3+v_4)$ so the answer is no but I'm not really sure about this one.

2) It is known that: $\text{Ann}(F \cap \text{Ann}(v))=\text{Ann}(F)+\text{Ann}(\text{Ann}(v))=\text{Ann}(\text{Ann}(v_1+v_2))+\text{Ann}(\text{Ann}(v))=\text{Span}(v_1+v_2)+\text{Span}(v).$

From this we can easily say that if $v\in \text{Span}(v_1+v_2)$ clearly we have that $v_3+v_4 \notin \text{Span}(v_1+v_2)$. Otherwise we have the thesis.

Is it a nice solution? Are there wrong passages or errors? Thanks in advance for your help. If you want to propose another solution I'll be glad to read it.


Let $^o$ denote the annihilator. For $(1)$, consider $f \in V^*$ defined on the basis $\mathcal{B}$ as:

$$f(v_1) = f(v_2) = -1\\ f(v_3) = 0\\ f(v_4) = 3$$

$\DeclareMathOperator{\span}{span}$Then $f(v_1-v_2+v_3) = f(2v_1+v_2+v_3+v_4) = 0$ so $f \in \big(\span\{v_1-v_2+v_3, 2v_1+v_2+v_3+v_4\}\big)^o$, but $f(v_1+v_2) = -2\ne 0$ so $f \notin F$.

Thus, $L \nsubseteq F$.

For $(2)$, I wouldn't throw isomorphisms around so lighly. First of all, assume that $V$ is finite-dimensional.

Let $\phi : V \to V^{**}$ be the canonical isomorphism, $\phi(x) = \hat{x}$, where $\hat{x} : V^* \to \mathbb{F}$ is defined as $\hat{x}(f) = f(x), \forall f \in V^*$.

You correctly concluded that $\big(F\cap \{v\}^o\big)^o = F^o + \{v\}^{oo} = \{v_1+v_2\}^{oo} + \{v\}^{oo}$.

$\fbox{$\implies$}$ Assume $v_3 + v_4 \in \{v_1+v_2\}^{oo} + \{v\}^{oo} \subseteq V^{**}$. By this is actually meant:$$\hat{v_3} + \hat{v_4} = \widehat{v_3 + v_4}\in \{v_1+v_2\}^{oo} + \{v\}^{oo}$$

Since for $M \leq V$ holds $M^{oo} = \phi(M)$ (and thus $\span\{x_1, \ldots, x_n\}^{oo} = \span\{\hat{x_1}, \ldots, \hat{x_n}\}$):

\begin{align}\hat{v_3} + \hat{v_4} \in \{v_1+v_2\}^{oo} + \{v\}^{oo} &\implies \hat{v_3} + \hat{v_4} \in \phi(\span\{v_1+v_2\}) + \phi(\span\{v\})\\ &\implies \hat{v_3} + \hat{v_4} \in \span\{\widehat{v_1+v_2}\} + \span\{\hat{v}\}\\ \end{align} This implies $\hat{v_3} + \hat{v_4} = \alpha(\widehat{v_1 + v_2}) + \beta \hat{v}$, for some $\alpha, \beta \in \mathbb{F}$.

By linearity and bijectivity of $\phi$ we conclude: $$v_3 + v_4 = \alpha(v_1 + v_2) + \beta v$$

If $\beta = 0$, it would imply $v_3 + v_4 = \alpha(v_1 + v_2)$, which is a contradiction with the linear independence of $\mathcal{B}$. Thus $\beta \ne 0$ and:

$$v = \frac{1}{\beta}(-\alpha v_1 - \alpha v_2 + v_3 + v_4) = -\underbrace{\frac{\alpha}{\beta}(v_1 + v_2)}_{\in \span\{v_1+ v_2\}} + \underbrace{\frac{1}{\beta}(v_3 + v_4)}_{\in\span{v_3+v_4}}$$

We see that $v \in \span\{v_1+ v_2, v_3 + v_4\}$ and that $v \notin \span\{v_1+ v_2\}$ since the coefficient with $v_3 + v_4$ is $\frac{1}{\beta} \ne 0$ and $\mathcal{B}$ is linearly independent.

$\fbox{$\impliedby$}$ Assume $v \in \span\{v_1+ v_2, v_3 + v_4\} \setminus \span\{v_1+ v_2\}$.

Thus $v = \alpha(v_1+v_2)+\beta(v_3 + v_4)$, for some $\alpha, \beta \in \mathbb{F}$, with $\beta \ne 0$.

$$v_3 + v_4 = \frac{1}{\beta}(v - \alpha(v_1+v_2))$$

By linearity of $\phi$ we get:

$$\widehat{v_3 + v_4} = \frac{1}{\beta}(\hat{v} - \alpha(\widehat{v_1+v_2})) = \underbrace{\frac{1}{\beta}\hat{v}}_{\in \phi(\span\{v\})} - \underbrace{\frac{\alpha}{\beta}\widehat{v_1+v_2}}_{\in \phi(\span\{v_1+v_2\})}$$

Thus, $\widehat{v_3 + v_4} \in \{v_1+v_2\}^{oo} + \{v\}^{oo}$.

  • $\begingroup$ Thanks, nice explanation! You deserved these points! Thanks again! $\endgroup$ – Alberto Andrenucci Sep 12 '17 at 15:46

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