• 0 Votes - 0 Average
• 1
• 2
• 3
• 4
• 5
 Lagrange's identity $\displaystyle{ (a \cdot b)^2 =|| a||^2 || b||^2 - \sum_{1\le i 12-23-2010, 12:21 PM (This post was last modified: 05-02-2012 07:49 AM by admin.) Post: #1  elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0 Lagrange's identity$\displaystyle{ (a \cdot b)^2 =|| a||^2 || b||^2 - \sum_{1\le i<j\le n} (a_i b_j -a_j b_i)^2}$By induction we have$\displaystyle{\left(\sum_{k=1}^n x_k\right) = \sum_{k=1}^n x_k^2 + \sum_{1\le i < j \le n} 2x_i x_j}$. This implies that$\begin{align} \left \| \mathbf{a}\right \|^2 \left \| \mathbf{b}\right \|^2 &=\sum_{k=1}^n a_k^2 \sum_{k=1}^n b_k^2 = \sum_{1\le k,j \le n} a_k^2 b_j^2 = \sum_{k=1}^n a_k^2 b_k^2 + \sum_{1\le k<j \le n} (a_k^2 b_j^2 + a_j^2 b_k^2) \\
& = \sum_{k=1}^n (a_k b_k)^2+ \sum_{1\le k < j\le n} \left(2a_k b_k a_j b_j+\left( a_k^2 b_j^2+a_j^2 b_k^2 -2a_k b_k a_j b_j \right )\right )\\
& = \left(\mathbf{a} \cdot \mathbf{b} \right )^2 + \sum_{1 \le k<j \le n} (a_k b_j - a_j b_k)^2\end{align}\$
 « Next Oldest | Next Newest »

Forum Jump: