$V = \Bbb{R}^3$ and has basis $\mathcal{B} = \{\vec{e_1}-\vec{e_2},\vec{e_1}+\vec{e_2},\vec{e_3}\}$

How do I find the dual basis? This is not homework, but an example that I am struggling to grasp. This is a simple question, so I would really appreciate if you wouldn't skip any details no matter the triviality as there may be fundamental gaps in my understanding.

  • $\begingroup$ First, make sure you fully understand the definition of a dual basis. Then you can write the conditions for a dual basis down, and what you end up with, is a system of linear equations (and the solution gives you the dual basis) $\endgroup$ – supinf May 17 '15 at 10:39

Notice that the definition of a dual basis is that, given $\beta = \{v_1, ..., v_n\}$, its dual is $\beta^* = \{f_1, ..., f_n\}$ such that $f_i(v_j) = \delta_{ij}$.

Given $\beta = \{\vec{e_1}-\vec{e_2},\vec{e_1}+\vec{e_2},\vec{e_3}\}$, we want such a basis. Also, since we know that the $f_i$ are linear functionals, we have that $f_i(x_1, x_2, x_3) = ax_1 + bx_2 + cx_3$. As you probably know, if we define the behaviour of $f_i$ in terms of our basis, we completely determine the function. I'll do the first example, the requirements are:

$$f_1(e_1 - e_2) = a - b = 1$$ $$f_1(e_1 + e_2) = a + b = 0$$ $$f_1(e_3) = c = 0$$

Hence, $a = -b$, which implies that $-2b = 1$, hence $b = \frac{-1}{2}$, and $a = \frac{1}{2}$, while $c = 0$. Thus:

$$f_1(x_1, x_2, x_3) = \frac{1}{2} x_1 - \frac{1}{2} x_2$$

Which, as desired, satisfies all the constraints. Just repeat this process for the other $f_i$s and that will give you the dual basis!

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  • $\begingroup$ Why can't we use the definition kronecker delta to define dual basis? $\endgroup$ – ZHU Nov 18 '17 at 21:18
  • $\begingroup$ @ZHU, the first paragraph does precisely that, define what the $f_i$s should be in terms of the kronecker delta. $\endgroup$ – Misguided Nov 19 '17 at 20:26
  • $\begingroup$ This is really a concise & neat answer! (+1) $\endgroup$ – Invisible Jun 15 at 4:55

Let $P$ be the change of basis matrix from the canonical basis $\mathcal C$ to basis $\mathcal B$. It is the matrix of the identity map from $(V,\mathcal B)$ to $(V,\mathcal C)$, and $P^{-1}$ is the matrix of the identity map from $(V,\mathcal C)$ to $(V,\mathcal B)$.

By duality, $\color{red}{{}^\mathrm t\mkern-1.5muP^{-1}}$ is the matrix of the identity map from $(V^*,\mathcal C^*)$ to $(V^*,\mathcal B^*)$. The column vectors of this matrix are the coordinates of vectors of $\mathcal B^*$ in the canonical basis of the dual space $(e_1^*,e_2^*,e_3^*)$.

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I hope you know the definition of a dual basis (otherwise see http://en.wikipedia.org/wiki/Dual_basis).

If you want to get the dual basis of a basis $\{e_1,e_2,e_3\}$ just take the matrix $A=[e_1,e_2,e_3]$. Since its columns are a basis you can invert it, so that $A^{-1}*A = E$.

Remember how you multiply matrices... $E[i,j] = A^{-1}[i,1]*A[1,j]+A^{-1}[i,2]*A[2,j] + A^{-1}[i,3]*A[3,j]$.

So the rows of your inverted matrix are the dual basis.

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  • $\begingroup$ I understand the definition. But the answer for the example is strange: $\mathcal{B}^* = \{\frac 12(\vec{e^*_1}-\vec{e^*_2}),\frac 12(\vec{e^*_1}+\vec{e^*_2}),\vec{e^*_3}\}$ $\endgroup$ – aidandeno May 17 '15 at 10:48
  • $\begingroup$ This notation is odd... $\endgroup$ – user228113 May 17 '15 at 10:50

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