What is dual vector and covector? I tried to learn dual vectors online but failed to exactly understand it, I know that it could be understood using change of basis. Below is a example for change of basis, kindly help me with this.
Let V be a space over $\Bbb{R}^3$ and the basis be 
$\begin{Bmatrix}
  \begin{bmatrix}1\\1\\0\\ \end{bmatrix},&
  \begin{bmatrix}0\\2\\0\\ \end{bmatrix},&
  \begin{bmatrix}1\\0\\1\\ \end{bmatrix}
\end{Bmatrix}$ and U be another space over $\Bbb{R}^3$ with basis 
$\begin{Bmatrix}
  \begin{bmatrix}5\\0\\3\\ \end{bmatrix},&
  \begin{bmatrix}2\\3\\4\\ \end{bmatrix},&
  \begin{bmatrix}1\\6\\2\\ \end{bmatrix}
\end{Bmatrix}$, $\begin{bmatrix}3\\5\\7\\ \end{bmatrix}$ is a vector in U and we want this vector's coefficient in V. The approach would be:
$$c_1\left[\begin{matrix}1\\1\\0\\ \end{matrix}\right]+
c_2\left[\begin{matrix}0\\2\\0\\ \end{matrix}\right]+
c_3\left[\begin{matrix}1\\0\\1\\ \end{matrix}\right]
=
3\left[\begin{matrix}5\\0\\3\\ \end{matrix}\right]+
5\left[\begin{matrix}2\\3\\4\\ \end{matrix}\right]+
7\left[\begin{matrix}1\\6\\2\\ \end{matrix}\right]
$$
Where $\mathbf{c}$ are coefficients of $V$'s basis.
$$\left[\begin{matrix}1 & 0 &1 \\ 1&2&0 \\0& 0 & 1\\ \end{matrix}\right]
\left[ \begin{matrix}c_1\\c_2\\c_3\\ \end{matrix}\right]= 
\left[\begin{matrix}5 &2&1\\ 0&3&6 \\3& 4 & 2\\ \end{matrix}\right]
\left[ \begin{matrix}3\\5\\7\\ \end{matrix}\right]
$$
$$
\left[ \begin{matrix}c_1\\c_2\\c_3\\ \end{matrix}\right]= 
\left[\begin{matrix}1 & 0 &1 \\ 1&2&0 \\0& 0 & 1\\ \end{matrix}\right]^{-1}\left[\begin{matrix}5 &2&1\\ 0&3&6 \\3& 4 & 2\\ \end{matrix}\right]
\left[ \begin{matrix}3\\5\\7\\ \end{matrix}\right]= \left[ \begin{matrix}-11\\34\\43\\ \end{matrix}\right]
$$
Now I want to know from  above example:


*

*What is/are dual vector?

*What is/are covector?

*What is dual space?


Let me know if I have understood it completely wrong. Thanks
 A: Given a vector space $V$ over field $K,$ the dual space of $V,$ denoted $V^\ast,$ is the set of all linear maps $\varphi:V\to K.$ For example, consider $V=\mathbb{R}\times\mathbb{R}$ with addition $+:V\times V\to V$ defined as:
$$\begin{bmatrix}a \\ b\end{bmatrix}+\begin{bmatrix}c \\ d\end{bmatrix}=\begin{bmatrix}a+c \\ b+d\end{bmatrix}$$ and vector multiplication $\cdot:K\times V\to V$ defined as
$$k\cdot\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}k\cdot a \\ k\cdot b\end{bmatrix}$$
The $+$ and $\cdot$ inside represent field addition and multiplication, but I'll just use the same symbol because I'm lazy. Define a linear mapping $\varphi:V\to K$ as $\varphi\!\left(\begin{bmatrix}a\\b\end{bmatrix}\right)=a+b.$ One can show that this function adds linearly:
$$\varphi\!\left(\begin{bmatrix}a\\b\end{bmatrix}\right)+\varphi\!\left(\begin{bmatrix}c\\d\end{bmatrix}\right)=a+b+c+d=a+c+b+d=\varphi\!\left(\begin{bmatrix}a+c\\b+d\end{bmatrix}\right)$$
and that it scales linearly:
$$k\cdot\varphi\!\left(\begin{bmatrix}a\\b\end{bmatrix}\right)=k\cdot\left(a+b\right)=k\cdot a+k\cdot b=\varphi\!\left(\begin{bmatrix}k\cdot a\\k\cdot b\end{bmatrix}\right).$$ Since it's linear, it's said to be a co-vector. Thus, one can show that this is equivalent to the "row vector" $\begin{bmatrix}1 & 1\end{bmatrix}$ using matrix multiplication.$\begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=a+b.$ When this collection of maps comes equipped with addition and multiplication that behaves linearly, the set itself also becomes a vector space over $K$ called the vector dual space that contains dual vectors. 
