Dual basis

Linear maps: Dual vector spaces

Dual basis

The basis constructed in the following theorem is convenient in the description of coordinates.

Let $\alpha = \vec{a}_1, \ldots , \vec{a}_n$ be a basis of $V$ .

There is exactly one basis $a_1^\star, \ldots , a_n^\star$ of $V^\star$ satisfying
$a_i^\star(\vec{a}_j)= \begin{cases} 1& \quad \text{ if } i = j\\ 0& \quad \text{ if } i\neq j \end{cases}$
In particular, $\dim{V}=\dim{V^\star}$ if $V$ is finite-dimensional.
If $\vec{a} \in V$ , then $\alpha(\vec{a}) = \rv{a_1^\star(\vec{a}), \ldots , a_n^\star(\vec{a})}$ is the coordinate vector of $\vec{a}$ .

The basis $a_1^\star, \ldots , a_n^\star$ of $V^\star$ is called the dual basis of $\vec{a}_1, \ldots , \vec{a}_n$ .

1. According to the theorem Linear map determined by basis, there is, for each $i=1, \ldots , n$ , exactly one linear function $a_i^\star : V\rightarrow \mathbb{R}$ with the property specified in the claim.

It remains to prove that the system $a_1^\star, \ldots , a_n^\star$ is a basis of $V^\star$ . First, we show that the system is independent. Suppose that, for certain scalars $\lambda_1 ,\ldots,\lambda_n$ , we have
$\lambda_1 a_1^\star + \cdots + \lambda_n a_n^\star = 0 \qquad (=\text{the null function})$
Apply the left hand side to the vector $\vec{a}_i$ and do the same for the right hand side. On the left we find $\lambda_i$ ; on the right we find $0$ . As this is the case for $i=1,\ldots , n$ , we conclude that $\lambda_1 = \cdots =\lambda_n=0$ . Therefore, the system is independent.

Next we show that every vector $\varphi \in V^\star$ belongs to the span $\linspan{a_1^\star, \ldots , a_n^\star}$ . It is easy to verify that $\varphi(\vec{a}_1) a_1^\star +\varphi(\vec{a}_2) a_2^\star + \cdots + \varphi(\vec{a}_n) a_n^\star$ and $\varphi$ have the same values at $\vec{a}_1, \ldots , \vec{a}_n$ . Again, because of the theorem Linear map determined by basis, this means that $\varphi= \varphi(\vec{a}_1) a_1^\star +\varphi(\vec{a}_2) a_2^\star + \cdots + \varphi(\vec{a}_n) a_n^\star$ .

2. If $\vec{a} = \lambda_1 \vec{a}_1 + \cdots +\lambda_n \vec{a}_n$ , then
$a_i^\star(\vec{a}) = \lambda_1 a_i^\star(\vec{a}_1) + \cdots + \lambda_n a_i^\star(\vec{a}_n) =\lambda_i$
due to the linearity of $a_i^\star$ .

This basis plays a role somewhat similar to that of an orthonormal basis in an inner product space. If we use the dot product in order to identify a vector space $V$ of finite dimension $n$ with $V^\star$ by means of the correspondence $\begin{array}{rcl}V&\leftrightarrow&V^\star\\ \vec{a}&\leftrightarrow&a^\star (\vec{x}) = \dotprod{\vec{a}}{\vec{x}}\end{array}$ then, for an orthonormal basis. the dual basis coincides with the orthonormal basis.

Let $\basis{\vec{e}_1, \vec{e}_2}$ be the standard basis of $\mathbb{R}^2$ .

The dual basis of the basis $\alpha =\basis{ \vec{e}_1+ \vec{e}_2, \vec{e}_1}$ of $\mathbb{R}^2$ is $\alpha^\star =\basis{ e_2^\star, e_1^\star -e_2^\star}$ , where $\basis{e_1^\star, e_2^\star}$ is the dual basis of $\basis{\vec{e}_1, \vec{e}_2}$ .

The calculation of the $\alpha$ -coordinates of the vector $\rv{3,7}=3\, \vec{e}_1 + 7\, \vec{e}_2$ is then simple: the first coordinate is $e_2^\star (3\, \vec{e}_1 + 7\, \vec{e}_2)= 7$ , the second coordinate is $({e}_1^\star -{e}_2^\star)(3\, \vec{e}_1 + 7\, \vec{e}_2)= 3-7=-4$ .

The map $a_i^\star$ assigns to each vector $\vec{x}$ of $V$ the coefficient $x_i$ of $\vec{a}_i$ in the expression $\vec{x} = x_1\vec{a}_1+\cdots + x_n\vec{a}_n$ of $\vec{x}$ as a linear combination of the basis vectors.

By use of matrix techniques, we can easily determine the dual basis of a basis for $\mathbb{R}^n$ :

Dual basis by means of inverse matrix

Let $\vec{a}_1, \ldots , \vec{a}_n$ be a basis for $\mathbb{R}^n$ , and collecting these vectors as columns in the matrix $A$ . Then, the dual basis of $\vec{a}_1, \ldots , \vec{a}_n$ consists of the rows of the inverse matrix of $A$ .

The matrix $A$ is equal to ${}_{\epsilon}I_{\alpha}$ . Each vector of the dual basis $\vec{a}_1^\star, \ldots , \vec{a}_n^\star$ of $\vec{a}_1, \ldots , \vec{a}_n$ can be written as a linear combination of ${e}_1^\star, \ldots , {e}_n^\star$ , say, ${a}_i^\star = b_{i1}\, {e}_1^\star + \cdots + b_{in}\, {e}_n^\star$ . We collect the $b_{ij}$ in the matrix $B$ . It now follows that ${a}_i^\star (\vec{a}_j)= b_{i1}\cdot a_{1j}+ \cdots + b_{in}\cdot a_{nj}$
The right hand side is the entry $i,j$ of the product matrix $B\,A$ . The left hand side is equal to $1$ if $i=j$ and $0$ otherwise. This means $I=B\,A$ . So $B$ is the inverse of $A$ .
The rows of $B$ provide the coefficients of the vectors with respect to the dual basis.

In order to determine the dual basis of $\vec{e}_1+ \vec{e}_2$ , $\vec{e}_1$ , we invert the matrix
$\left(\begin{array}{cc} 1 & 1\\ 1 & 0\end{array}\right)$
and find
$\left(\begin{array}{cc} 0 & 1\\ 1 & -1\end{array}\right)$
Thus, the dual basis is ${e}_2^\star$ (first row), ${e}_1^\star -{e}_2^\star$ (second row), as we have seen before.

Determine the dual basis of the basis $\basis{\rv{1 , -3 }, \rv{-2 , 7 }}$ for $\mathbb{R}^2$ .

Give your answer in the form of a matrix whose rows are the coordinate vectors with respect to the dual basis of the standard basis.

$\matrix{7 & 2 \\ 3 & 1 \\ }$

In order to determine the dual basis of $\basis{\rv{1 , -3 }, \rv{-2 , 7 }}$ , we invert the matrix $A$ whose columns are the basis vectors.
$\begin{array}{rcl} A &=& \matrix{1 & -2 \\ -3 & 7 \\ }\\ A^{-1} &=& \matrix{7 & 2 \\ 3 & 1 \\ }\end{array}$ According to the theorem Dual basis by use of the inverse matrix, the dual basis consists of the rows of this matrix.

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