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_1left[begin{matrix}1\1\0\ end{matrix}right]+

c_2left[begin{matrix}0\2\0\ end{matrix}right]+

c_3left[begin{matrix}1\0\1\ end{matrix}right]

=

3left[begin{matrix}5\0\3\ end{matrix}right]+

5left[begin{matrix}2\3\4\ end{matrix}right]+

7left[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

Given a vector space $V$ over field $K,$ the dual space of $V,$ denoted $V^ast,$ is the set of all linear maps $varphi:Vto K.$ For example, consider $V=mathbb{R}timesmathbb{R}$ with addition $+:Vtimes Vto V$ defined as:

$$begin{bmatrix}a \ bend{bmatrix}+begin{bmatrix}c \ dend{bmatrix}=begin{bmatrix}a+c \ b+dend{bmatrix}$$ and vector multiplication $cdot:Ktimes Vto V$ defined as

$$kcdotbegin{bmatrix}a \ bend{bmatrix}=begin{bmatrix}kcdot a \ kcdot bend{bmatrix}$$

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 $varphi:Vto K$ as $varphi!left(begin{bmatrix}a\bend{bmatrix}right)=a+b.$ One can show that this function adds linearly:

$$varphi!left(begin{bmatrix}a\bend{bmatrix}right)+varphi!left(begin{bmatrix}c\dend{bmatrix}right)=a+b+c+d=a+c+b+d=varphi!left(begin{bmatrix}a+c\b+dend{bmatrix}right)$$

and that it scales linearly:

$$kcdotvarphi!left(begin{bmatrix}a\bend{bmatrix}right)=kcdotleft(a+bright)=kcdot a+kcdot b=varphi!left(begin{bmatrix}kcdot a\kcdot bend{bmatrix}right).$$ Since it’s linear, it’s said to be a co-vector. Thus, one can show that this is equivalent to the “row vector” $begin{bmatrix}1 & 1end{bmatrix}$ using matrix multiplication.$begin{bmatrix}1 & 1end{bmatrix}begin{bmatrix}a\bend{bmatrix}=a+b.$ 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.