Are the dual basis functionals not linear maps?

Question

Sheldon Axler's Linear Algebra Done Right defines the dual basis as follows:

Specifically, the dual basis of a basis $v_1, \ldots, v_n$ is the list $\delta_1 \ldots \delta_n \in V'$. However, I cannot see how each $\delta_i$ is a linear functional.

Consider $\delta_i(v_i) + \delta_i(v_j)$. By the given definition of $\delta_i$, we have $\delta_i(v_i) + \delta_i(v_j) = 1 + 0 = 1$.

However, consider $\delta_i(v_i + v_j)$. Because $v_i + v_j \neq v_i$, this evaluates to $0$, which violates the additivity property that all linear maps/functionals must have.

Doesn't this show that $\delta_i$ isn't a linear functional at all? In that case, how could it be an element of $V'$, as claimed in the definition?

Answer 1

$\delta_i$ is equal to $0$ at all the other vectors from the given basis, not on all the vectors in $V$ which are not $v_i$. By definition, this is the unique linear functional $\delta_i:V\to F$ which acts on the vectors from the basis $v_1,...,v_n$ just as you described. (of course, it has to be checked that such a functional indeed exists and unique) So $\delta_i$ is linear simply by definition. A general vector in $V$ has the form $\sum\limits_{j=1}^n \lambda_jv_j$ where $\lambda_j\in F$, and by linearity we get:

$\delta_i(\sum\limits_{j=1}^n\lambda_jv_j)=\sum\limits_{j=1}^n\lambda_j\delta_i(v_j)=\lambda_i$

So $\delta_i$ gives you the coefficient of $v_i$ of the vector on which you apply it. As a special case, $\delta_i(v_i+v_j)=1$.

Answer 2

Either there is a piece of exposition that your book is missing, or you just missed it, but when an author defines a function only on a basis, they may uniquely extend that function to the remaining elements of the vector space by simply declaring it to be linear.

You are not being asked to prove that the $\delta_i$ are linear, if you like, you may try to prove that the following function is well defined on $V$, and that it is the unique linear extension to the function defined on just the $v_i$

Let $x=\sum a_jv_j$, then define $\delta_i(x)= \sum a_j\delta_i(v_j)$.