Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
$\sum_{i=1}^{n}\sum_{j=1}^{n}(a_ib_j-a_jb_i)^2 = \sum_{i=1}^{n}a_i^2\sum_{j=1}^{n}b_j^2 + \sum_{i=1}^{n}b_i^2 \sum_{j=1}^{n}a_j^2 - 2\sum_{i=1}^{n}a_ib_i \sum_{j=1}^{n}b_ja_j$
Can someone walk me through this. The different summation indices is throwing me off. The Cauchy-Schwarz inequality is generally written as
$\sum_{i=1}^{n} a_ib_i \le (\sum_{i=1}^{n}a_i^2)^{1/2}(\sum_{i=1}^{n}b_i^2)^{1/2}.$
So I'm not 100% sure what the elements $b_j$ and $a_j$ are.
Hint: start off by expanding the square: $$\sum_{i=1}^{n}\sum_{j=1}^{n}(a_ib_j-a_jb_i)^2 =\sum_{i=1}^{n}\sum_{j=1}^{n}(a_i^2b_j^2-2a_ib_ja_jb_i+a_j^2b_i^2)$$ Now split the sum: $$\sum_{i=1}^{n}\sum_{j=1}^{n}(a_ib_j-a_jb_i)^2 =\sum_{i=1}^{n}\sum_{j=1}^{n}a_i^2b_j^2 -\sum_{i=1}^{n}\sum_{j=1}^{n}2a_ib_ja_jb_i +\sum_{i=1}^{n}\sum_{j=1}^{n}a_j^2b_i^2\ .$$ The reason this RHS equals the previous RHS is that you are adding up the same things, only in a different order. Now for the first sum, think about what happens when $j$ varies from $1$ to $n$. As this happens, the $a_i^2$ never changes, so it is a constant factor and can be taken out of the sum: $$\sum_{i=1}^{n}\sum_{j=1}^{n}a_i^2b_j^2 =\sum_{i=1}^{n}\Bigl(\sum_{j=1}^{n}a_i^2b_j^2\Bigr) =\sum_{i=1}^{n}a_i^2\Bigl(\sum_{j=1}^{n}b_j^2\Bigr)\ .$$ And now the sum over $j$ adds up to a fixed quantity which does not depend on $i$, so it can be taken outside the sum over $i$ to give $$\sum_{i=1}^{n}\sum_{j=1}^{n}a_i^2b_j^2 =\Bigl(\sum_{i=1}^{n}a_i^2\Bigr)\Bigl(\sum_{j=1}^{n}b_j^2\Bigr)\ .$$
The last two steps are really just the same as $$\sum_{j=1}^3ab_j=ab_1+ab_2+ab_3=a(b_1+b_2+b_3)=a\sum_{j=1}^3b_j\ .$$
Hopefully you can now treat the rest of the original expression in the same way. Good luck!
