Suppose we have a vector space V with basis $e_1 = (1,1,2)$, $e_2 = (1,0,1)$ and $e_3 = (2,1,0)$ on $\mathbb R^3$ over $\mathbb R$. Its dual basis is found to be $e_1' = (−1/3,2/3,1/3)$, $e_2' = (2/3,−4/3,1/3)$, $e_3' = (1/3,1/3,−1/3)$.

Now, given a vector from $V$, how does this dual basis help in finding the mapping $V\rightarrow F$? That is, how is the operation that takes a vector $v$ from $V$ and turn it into a number? Or is this only a basis for linear functionals that will turn a vector from $V$ to a scalar? How do I construct the linear functionals out of this dual basis?

I'm not sure if this is a right question to ask, but I guess I have not understood some key aspect and I hope this question brings that out.

  • $\begingroup$ Welcome to MSE. It is in your best interest that you use MathJax. $\endgroup$ Commented Jun 24, 2018 at 11:35
  • $\begingroup$ Each basis vector in the dual basis produces a scalar for a given vector from V. That is, e_1^', e_2^', e_3^' when acting on (2,1,0) produce 3 different scalars. What is the relation between these scalars? Does it just mean that each linear functional in the dual basis is equivalent to one another? And any combination of these functionals will still span the given dual space? $\endgroup$ Commented Jun 24, 2018 at 12:30

2 Answers 2


$e^i(e_j)=\delta _i^j$, where $\delta _i^j=\begin{cases} 1, i=j\\ 0, i\not =j\end{cases}$ is the Kronecker delta.

Use this fact and linearity to evaluate...

Example: $(-\frac13,\frac23,\frac13)(1,1,2)=-\frac13\cdot 1+\frac23\cdot 1+\frac13\cdot 2=-\frac13 +\frac23+\frac23=\frac13+\frac23=1$

Similarly, $(-\frac13,\frac23,\frac13)(1,0,1)=0$, and $(-\frac13,\frac23,\frac13)(2,1,0)=0$.

In general, $(-\frac13,\frac23,\frac13)(a,b,c)=-\frac13 a+\frac23b+\frac13 c$.

Any linear functional $\phi$ in $V^*$ will be given by a linear combination of the dual basis: $\phi= a_1e^1+\dots+a_ne^n$.

The dual basis is a basis for the dual space. Thus $V$ and $V^*$ have the same dimension (when $V$ is finite dimensional).

  • $\begingroup$ I am not able understand how Kronecker delta gets used here. Could you kindly elaborate? Also, I have added a few more questions in the comment section under the question itself. Could you kindly take a look? $\endgroup$ Commented Jun 24, 2018 at 12:36
  • $\begingroup$ Notice that your basis vectors satisfy $e^ie_j=\delta_i^j$. In the example I gave, i took $e^1$ of $e_1,e_2$ and $e_3$. Notice I got $1,0$ and $0$... $\endgroup$
    – user403337
    Commented Jun 24, 2018 at 12:41
  • $\begingroup$ This isn't a coincidence: the equations (with the Kronecker delta) hold by definition. $\endgroup$
    – user403337
    Commented Jun 24, 2018 at 14:17
  • $\begingroup$ So, I can make linear combinations of the dual basis to get the linear functionals which then do the actual transform from vector space V into F (the scalar field)? Each such functional could give a different scalar value for the same vector. Is this correct? On a side note; I'm finding this topic of "dual spaces" difficult to wrap my mind around. Is this normal? :-) $\endgroup$ Commented Jun 24, 2018 at 14:38
  • $\begingroup$ Correct. I guess it can vary a bit from person to person. Some are good at this sort of thing... these are just the basics, btw. $\endgroup$
    – user403337
    Commented Jun 24, 2018 at 16:24

The $e'_i$ triples are actually the coordinates of the vectors of the dual basis. If I understand you right, $V=\mathbb R^3$, and so the canonical basis (that is, the dual basis of $\{(1,0,0),(0,1,0),(0,0,1)\}\subset \mathbb R^3$, is $\{\phi_1,\phi_2,\phi_3\}$, where $$\phi_1(x_1,x_2,x_3)=x_1,$$ $$\phi_2(x_1,x_2,x_3)=x_2,$$ $$\phi_3(x_1,x_2,x_3)=x_3.$$ So if your $e'_i$ are coordinates in this canonical base, you have $$e'_1=-\tfrac13\cdot \phi_1+\tfrac23\cdot\phi_2+\tfrac13\cdot\phi_3,$$ and so on for $e'_2$ and $e'3$.

So, for instance, $$e'_1(x_1,x_2,x_3)=-\tfrac13\cdot x_1+\tfrac23\cdot x_2+\tfrac13\cdot x_3,$$ and if you wanted—just to say something—$e'_1(2,-1,0)$, that would be $$e'_1(2,-1,0)=-\tfrac13\cdot 2+\tfrac23\cdot (-1)+\tfrac13\cdot 0=-\tfrac 43.$$


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