What is the canonical basis of a dualspace in $\mathbb{R}^3$?

I have the following:
Consider the basis $$B := \{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \}$$ of the $\mathbb{R}^3$. Express the elements of the dual basis of $B$ in $\mathbb{(R^3)}^{*}$ as a linear combination of the canonical basis $(e^*_1,e^*_2, e^*_3)$ of $(\mathbb{R}^3)^{*}$.

Now, first I have to calculate the elements of the Dual Basis by applying $b_i^*(b_j) = \delta_{ij}$, with $b_i^*:\mathbb{R}^3 \rightarrow \mathbb{R}$ and $\delta_{ij} = 1$ for $i = j$ and $0$ else.

So for $b_1^*$ if have for example:

$$b_1^*(b_1) = \delta_{11} = 1 = 1\cdot\lambda_1 + 1\cdot\lambda_2 + 0\cdot\lambda_3$$

$$b_1^*(b_2) = \delta_{12} = 0 = -1\cdot\lambda_1 + 1\cdot\lambda_2 + 2\cdot\lambda_3$$

$$b_1^*(b_3) = \delta_{13} = 0 = 2\cdot\lambda_1 + 2\cdot\lambda_2 + 1\cdot\lambda_3$$

which translates into

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

and solves with $b_1^* = \begin{pmatrix} \frac{5}{2} \\-\frac{3}{2} \\ -2 \end{pmatrix}$

And if the above is correct: After solving the resulting 3 systems of linear equations I get the desired elements $(b_1^*, b_2^*, b_3^*)$ of the dual basis. The task now states to express those as a linear combination of the canonical basis $(e^*_1,e^*_2, e^*_3)$ of $(\mathbb{R}^3)^{*}$. So:

$$b_j^* = \sum_{i = 1}^3 \lambda_ie_i^* \mspace{2cm} j = 1,2,3$$

But my question now is: What are $(e^*_1,e^*_2, e^*_3)$? Are they similar to $(e_1,e_2, e_3)$ in $\mathbb{R}^3$ or do I have to calculate them separately?

If they are similar - would it really be just:

$$\begin{pmatrix} -4 \\5 \\ -2 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\1 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\0 \\ 1 \end{pmatrix}$$

$$\Rightarrow b_1^* = -4e_1^* + 5e_2^* - 2e_3^*$$

With the help of user7530 I now assume that they are similar:

$$e_1^*(e_1) = 1 = 1 \cdot \lambda_1 + 0 \cdot \lambda_2 + 0 \cdot \lambda_3$$ $$e_1^*(e_2) = 0 = 0 \cdot \lambda_1 + 1 \cdot \lambda_2 + 0 \cdot \lambda_3$$ $$e_1^*(e_3) = 0 = 0 \cdot \lambda_1 + 0 \cdot \lambda_2 + 1 \cdot \lambda_3$$

Which only solution is $\lambda_1 = 1$ and thus $e_1^* = \begin{pmatrix} 1 \\0 \\ 0 \end{pmatrix}$.
Same with $e_2^*,e_3^*$

Thank you very much for your help.

FunkyPeanut

• First, you have made a calculation error somewhere, as $b_1^*(b_2) \neq 0$. Second, you can find $e_i^*$ in exactly the same way: solve for them such that $e_i^*(e_j) = \delta_{ij}$. – user7530 Apr 15 '14 at 16:04
• Thank you! Okay - but if that is so I think I got something wrong. I thought that $b_1^{*}(b_2)$ means that $i = 1$ and $j = 2$. So we get $\delta_{12}$ which is 0 since $i \ne j$. I guess that is wrong then... Oh or do you mean the system od linear equations of you insert the solution I presented= – FunkyPeanut Apr 15 '14 at 16:08
• You are supposed to get zero, but if I eyeball it I see that I instead get something odd. – user7530 Apr 15 '14 at 16:25
• True. I have just corrected it – FunkyPeanut Apr 15 '14 at 16:29

Let the matrices $$P=[b_1\;b_2\;b_3]=\left[\begin{matrix}1&-1&2\\1&1&2\\0&2&1\end{matrix}\right]$$ and $$Q=[b_1^*\;b_2^*\;b_3^*]$$ then since $$b_i^*(b_j)=\delta_{ij}$$ we see that $$P^TQ=I_3$$ hence $$Q=\left(P^T\right)^{-1}=\frac12\left[\begin{matrix}-3&-1&2\\5&1&-2\\-4&0&2\end{matrix}\right]$$
• Wow this looks amazing. But I don't quite see how you achieved that. Why do we see that $P^T Q = I_3$? – FunkyPeanut Apr 15 '14 at 16:32
• In your answer you wrote $$\left(\begin{array}{ccc|c} 1 & 1 & 0 & 1 \\ -1 & 1 & 2 & 0 \\ 2 & 2 & 1 & 0 \end{array}\right)$$ hence $$P^T\left(\begin{array}\\ \lambda_1 \\ \lambda_2 \\ \lambda_3\end{array}\right)=\left(\begin{array}\\ 1 \\ 0 \\ 0\end{array}\right)$$ so do the same thing for the other vectors and you find the equality. – user63181 Apr 15 '14 at 16:45