1
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Thuc Hoang
    Feb 26, 2020 at 8:01
  • $\begingroup$ I see the problem with my counter example. Plugging in $v_j$ will show that $a_j$ must be zero. Thank you a lot! $\endgroup$
    – Thuc Hoang
    Feb 26, 2020 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.