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Apologies for posting picture instead of typing as there is some problem with keyboard.

I am confused with notation $w\in \Lambda^2 V^*$. Does it mean we consider $\omega$ as $:V\times V\rightarrow \mathbb{R}$ in which case it should be an element of $\Lambda^2V$. so, I guess they consider extension of $\omega$ to $V^*$ i.e., map $V^*\times V^*\rightarrow \mathbb{R}$ but then did not specify what extension it is. I defined $\omega(f,g)=\left< u,v\right>$ where $u$ is such that $f(x)=\left$ for all $x\in V$ and $v\in V$ is such that $g(x)=\left$ for all $x\in V$. But proof says

since $$\omega(w,v)=\left=-\left=-\left=-\omega(v,w)$$

This only says $\omega\in \Lambda^2 V$. Defining $\omega\left$ as i have defined, I proved that $\omega\in \Lambda^2 V^*$. I am confused with the notation. It would be grateful if any one who has used this book can clarify this.

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The book is correct.

A skew-symmetric bilinear map $V\times V \to \mathbb{R}$ descends to a linear map $\bigwedge^2V \to \mathbb{R}$, i.e. an element of $\left(\bigwedge^2V\right)^* \cong \bigwedge^2V^*$.

As for your question in the comments (why is $\omega \in \bigwedge^{1,1}V^*$?), recall that ${\bf I} : \bigwedge^*V_{\mathbb{C}} \to \bigwedge^*V_{\mathbb{C}}$ is the linear operator that acts on $\bigwedge^{p,q}V$ by multiplication with $i^{p-q}$, i.e.

$${\bf I} = \displaystyle\sum_{p,q}i^{p-q}\,\Pi^{p,q}$$

where $\Pi^{p,q} : \bigwedge^*V_{\mathbb{C}} \to \bigwedge^{p,q}V$ is the natural projection.

There is a corresponding operator on the dual space $\bigwedge^*V_{\mathbb{C}}$, also denoted ${\bf I}$, given by ${\bf I}(\alpha)(v_1, \dots, v_k) = \alpha({\bf I}(v_1), \dots, {\bf I}(v_k))$. If $\alpha$ is a $(p, q)$-form, then ${\bf I}(\alpha) = i^{p-q}\alpha$ so again we have

$${\bf I} = \displaystyle\sum_{p,q}i^{p-q}\,\Pi^{p,q}$$

where $\Pi^{p,q}$ now denotes the natural projection $\bigwedge^*V_{\mathbb{C}}^* \to \bigwedge^{p,q}V^*$.

For $\alpha \in \bigwedge^2V_{\mathbb{C}}^*$, we have $\alpha = \alpha^{2,0} + \alpha^{1,1} + \alpha^{0,2}$ where the exponent denotes the bidegree. So

$${\bf I}(\alpha) = i^{2-0}\alpha^{2,0} + i^{1-1}\alpha^{1,1} + i^{0-2}\alpha^{0,2} = -\alpha^{2,0} + \alpha^{1,1} - \alpha^{2,0}.$$

If ${\bf I}(\alpha) = \alpha$, then by comparing bidegrees, we see that $\alpha^{2,0} = 0$ and $\alpha^{0,2} = 0$, so $\alpha = \alpha^{1,1} \in \bigwedge^{1,1}V^*$. In particular, $\omega \in \bigwedge^{1,1}V^*$.

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    I am missing something here.. I understand that $\omega$ is skew symmetric bilinear map on $V$ because $\omega(u,v)=-\omega(v,u)$... The definition that i have for $\Lambda^2 V$ is the space of alternating functions $V\times V\rightarrow \mathbb{R}$.. – 2017-01-26
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    I think i understand something.. Considering $\Lambda^2 V$ as quotient of tensor algebra $V\otimes V$ by subspace generated by $\{x\otimes x : x\in V\}$ we have $\Lambda^2 V =\{x\wedge y : \rm{some conditions}\}$.. Given a symmetric bilinear map $\omega: V\times V\rightarrow \mathbb{R}$ we define $\tau: \Lambda^2(V)\rightarrow \mathbb{R}$ by $\tau(v_1\wedge v_2)=\omega(v_1,v_2)$ which is well defined because $\tau(v\wedge v)=\omega(v,v)=0$. So, we have well defined map. Is this what it is? – 2017-01-26
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    @PraphullaKoushik: That's not the definition that's being used here. The $k^{\text{th}}$ exterior power of a vector space is either the subset (or quotient, depending on the construction) of $\bigotimes^kV$ consisting of skew-symmetric tensors. An element of $\bigwedge^2V$ can be written as a sum of elements of the form $v_1\wedge v_2$ where $v_1, v_2 \in V$. See [this](https://en.wikipedia.org/wiki/Exterior_algebra#Exterior_power) for example. – 2017-01-26
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    Yes Yes i realized that after thinking for some time... With a difference of 20+ seconds :) :) And of course the book is correct :D :D – 2017-01-26
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    I see that $\left(\Lambda^2 V\right)^*\cong \Lambda^2 V^*$.. Can you help me how to see that $\omega\in \Lambda^{1,1}V^*$.. – 2017-01-26
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    @PraphullaKoushik: I have edited my answer to include an explanation of why ${\bf I}(\omega) = \omega$ implies $\omega \in \bigwedge^{1,1}V^*$. – 2017-01-26
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    This is clear now. Thank you :) – 2017-01-27
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    Could you please explain why $\omega(\textbf{I}(v),\textbf{I}(w))= \langle I(I(v)) ,I(w) \rangle $? – 2018-02-08
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    because we ony have $\omega(\textbf{I}(v),\textbf{I}(w))=\langle I(\textbf{I}(v)) ,\textbf{I}(v)\rangle$? Thank you in advance. – 2018-02-08
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    @AnKhuongDoan: There is a natural inclusion $V \hookrightarrow V_{\mathbb{C}}$ given by $v \mapsto v\otimes 1$. Now $\textbf{I}(v\otimes 1) = I(v)\otimes 1$. – 2018-02-12