# A counter example of a proof for the independence of dual basis

Below is a proof for the independence of dual basis I found in Linear Algebra Done Righ (3rd edition, page 102):

Suppose $$v_1,...,v_n$$ is a basis of $$V$$. Let $$f_1,...f_n \in V'$$, which is the dual space of $$V$$, such that $$f_i(v_j)= 1$$ if $$i = j$$ and $$f_i(v_j)= 0$$ otherwise. To show that $$f_1,...f_n$$ is a linearly independent list of elements of $$V'$$, suppose $$a_1,...,a_n \in F$$ are such that $$a_1f_1+...+a_nf_n = 0 \quad (1)$$

Now $$a_1f_1+...+a_nf_n(v_j) = a_j$$ for $$j = 1,...,n$$. The equation above thus shows that $$a_1 =...=a_n = 0$$. Hence $$f_1,...f_n$$ is linearly independent.

My counter example for the above proof:

Taking $$v = v_1 - v_2$$ as the input vector for the equation $$(1)$$, we have $$a_1 - a_2 = 0(v_1-v_2) = 0$$ ,which means the list $$a_1,...,a_n$$ is not necessarily all zeros for $$(1)$$ to hold. Is this a valid counter example? Are there other (easier) ways to prove the independence of a dual basis defined as above? Thank you very much.

No , it is not a counterexample. You get $$a_1=a_2$$, thats all ! If you add some further arguments , you will get $$a_1=a_2=0.$$
• There is no restrictions on the values of $a_1$ and $a_2$, so I suppose that we can choose any values for $a_1$ and $a_2$. That's the reason why I say the list $a_1,...,a_n$ is not necessarily all zeros for (1) to hold. Could you help me explain further? Thank you so much. Feb 26, 2020 at 8:01
• I see the problem with my counter example. Plugging in $v_j$ will show that $a_j$ must be zero. Thank you a lot! Feb 26, 2020 at 8:15