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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.

Thuc Hoang
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No , it is not a counterexample. You get $a_1=a_2$, thats all ! If you add some further arguments , you will get $a_1=a_2=0.$

Fred
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  • 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. – Thuc Hoang Feb 26 '20 at 08:01
  • I see the problem with my counter example. Plugging in $v_j$ will show that $a_j$ must be zero. Thank you a lot! – Thuc Hoang Feb 26 '20 at 08:15