Question

Given the basis {(1, - 1, 3), (0, 1, - 1),
(0, 3, - 2)} of R3, determine its dual basis.

Answer 1

Let $e^1=(1,0,0),e^2=(0,1,0),e^3=(0,0,1)$ be the standard basis, and $E_1,E_2,E_3$ be the its dual basis. Then $\delta^i_j=E_i(e^j)$, where $\delta^i_j=\begin{cases} 1,&\text{if $i=j$}\\ 0,&\text{if $i\neq j$} \end{cases}$

Let $f^1=(1,-1,3)=(f^1_1,f^1_2,f^1_3),$

$f^2=(0,1,-1)=(f^2_1,f^2_2,f^2_3), f^3=(0,3,-2)=(f^3_1,f^3_2,f^3_3)$ be the given basis, and $F_1=F_1^1E_1+F_1^2E_2+F_1^3E_3,$

$F_2=F_2^1E_1+F_2^2E_2+F_2^3E_3,$

$F_3=F_3^1E_1+F_3^2E_2+F_3^3E_3$ be a its dual basis.

We have $E_i(f^j)=E_i(f^j_1e^1+f^j_12e^12+f^j_3e^3)=f^j_i$.

Then $\delta_i^j=F_i(f^j)=(F_i^1E_1+F_i^2E_2+F_i^3E_3)(f^j)=$

$=F_i^1f^j_1+F_i^2f^j_2+F_i^3f^j_3$.

That is $I=\begin{pmatrix} F_1^1&F_1^2&F_1^3\\ F_2^1&F_2^2&F_2^3\\ F_3^1&F_3^2&F_3^3 \end{pmatrix}\begin{pmatrix} f_1^1&f_1^2&f_1^3\\ f_2^1&f_2^2&f_2^3\\ f_3^1&f_3^2&f_3^3 \end{pmatrix}$, so $\begin{pmatrix} F_1^1&F_1^2&F_1^3\\ F_2^1&F_2^2&F_2^3\\ F_3^1&F_3^2&F_3^3 \end{pmatrix}=\begin{pmatrix} f_1^1&f_1^2&f_1^3\\ f_2^1&f_2^2&f_2^3\\ f_3^1&f_3^2&f_3^3 \end{pmatrix}^{-1}$

We obtain $\begin{pmatrix} F_1^1&F_1^2&F_1^3\\ F_2^1&F_2^2&F_2^3\\ F_3^1&F_3^2&F_3^3 \end{pmatrix}=\begin{pmatrix} 1&0&0\\ -1&1&3\\ 3&-1&-2 \end{pmatrix}^{-1}$ .

Find $\begin{pmatrix} 1&0&0\\ -1&1&3\\ 3&-1&-2 \end{pmatrix}^{-1}$.

$\left(\begin{array}{lll|lll} 1&0&0&1&0&0\\ -1&1&3&0&1&0\\ 3&-1&-2&0&0&1 \end{array}\right)\to$

$\to\left(\begin{array}{lll|lll} 1&0&0&1&0&0\\ 0&1&3&1&1&0\\ 0&-1&-2&-3&0&1 \end{array}\right)\to$

$\to\left(\begin{array}{lll|lll} 1&0&0&1&0&0\\ 0&1&3&1&1&0\\ 0&0&1&-2&1&1 \end{array}\right)\to$

$\to\left(\begin{array}{ccc|ccc} 1&0&0&1&0&0\\ 0&1&0&7&-2&-3\\ 0&0&1&-2&1&1 \end{array}\right)$

So $\begin{pmatrix} F_1^1&F_1^2&F_1^3\\ F_2^1&F_2^2&F_2^3\\ F_3^1&F_3^2&F_3^3 \end{pmatrix}=\begin{pmatrix} 1&0&0\\ -1&1&3\\ 3&-1&-2 \end{pmatrix}^{-1}=\begin{pmatrix} 1&0&0\\ 7&-2&-3\\ -2&1&1 \end{pmatrix}$

That is $F_1=F_1^1E_1+F_1^2E_2+F_1^3E_3=E_1,$

$F_2=F_2^1E_1+F_2^2E_2+F_2^3E_3=7E_1-2E_2-3E_3,$

$F_3=F_3^1E_1+F_3^2E_2+F_3^3E_3=-2E_1+E_2+E_3$

Answer: $E_1, 7E_1-2E_2-3E_3, -2E_1+E_2+E_3$