## The Cauchy-Schwarz inequality and the Lagrange identity

The classical Lagrange identity is the following:

$\displaystyle \begin{array}{rl} \displaystyle \left(\sum_{i=1}^n a_ib_i\right)^2+\sum _{1\le i

This can be proven by expanding ${\displaystyle \sum _{1\le i and separating the terms into the cross-terms part and the non cross-terms part.

The Lagrange identity implies the Cauchy-Schwarz inequality in ${\mathbb R^n}$. And when ${n=3}$, this can be rephrased as

$\displaystyle \begin{array}{rl} \displaystyle |{a}|^2|{b}|^2= \left({a}\cdot {b}\right) ^2+|{a}\times {b}|^2\end{array}$

for ${{a}, {b}\in \mathbb R^3}$. In general, the term ${\displaystyle \sum _{1\le i can be identified as the norm squared of the wedge product ${{a}\wedge {b}}$.

In this note, we give the less well-known extension of this identity and the corresponding Cauchy-Schwarz type inequality.

Let ${V}$ be an inner product space and denote its inner product by “${\cdot}$” or ${\langle \cdot,\cdot\rangle}$ (whichever is clearer). The ${n}$-fold tensor product space ${V^{\otimes n}}$ can then be naturally endowed with the inner product (extended linearly)

$\displaystyle \begin{array}{rl} \displaystyle (v_1\otimes \cdots \otimes v_n)\cdot(w_1\otimes \cdots \otimes w_n):=(v_1\cdot w_1)\cdots (v_n\cdot w_n).\end{array}$

For simplicity, let us denote ${v_1\otimes \cdots\otimes v_n}$ by ${v_1\cdots v_n}$.

Now consider the ${n}$-fold wedge product space ${\bigwedge^n (V)}$ of ${V}$. For ${v_1\wedge \cdots \wedge v_n, w_1\wedge \cdots \wedge w_n\in \bigwedge^n (V)}$, the inner product is defined by (extended linearly)

$\displaystyle \begin{array}{rl} \displaystyle (v_1 \wedge \cdots \wedge v_n)\cdot(w_1 \wedge \cdots \wedge w_n):=\det(v_i\cdot w_j). \ \ \ \ \ (1)\end{array}$

For example, ${|e_1\wedge \cdots\wedge e_n|=1}$ for an orthonormal basis in ${\mathbb R^n}$. It turns out that if we identify ${v_1\wedge \cdots \wedge v_n}$ with

$\displaystyle \begin{array}{rl} \displaystyle I(v_1\wedge \cdots\wedge v_n):=\sum_{\sigma\in S_n} \textrm{sgn}(\sigma)v_{\sigma(1)}\cdots v_{\sigma(n)}\in V^{\otimes n},\end{array}$

where ${S_n}$ is the permutation group of ${n}$ objects, then ${ (v_1\wedge \cdots \wedge v_n )\cdot(w_1\wedge \cdots \wedge w_n )=\frac{1}{n!} I(v)\cdot I(w)}$. To see this,

$\displaystyle \begin{array}{rl} \displaystyle & \displaystyle \frac{1}{n!} I(v_1\wedge \cdots\wedge v_n)\cdot I(w_1\wedge \cdots\wedge w_n) \\ =& \displaystyle \frac{1}{n!} \left\langle \sum_{\sigma\in S_n}\textrm{sgn}(\sigma)v_{\sigma(1)}\cdots v_{\sigma(n)} , \sum_{\tau\in S_n}\textrm{sgn}(\tau)w_{\tau(1)}\cdots w_{\tau(n)}\right\rangle\\ =& \displaystyle \frac{1}{n!}\sum_{\sigma, \tau\in S_n}\textrm{sgn}(\sigma)\textrm{sgn}(\tau)\left(v_{\sigma(1)}\cdot w_{\tau(1)}\right)\cdots \left(v_{\sigma(n)}\cdot w_{\tau(n)}\right)\\ =& \displaystyle \frac{1}{n!}\sum_{\sigma\in S_n}\sum_{\tau\in S_n}\textrm{sgn}(\sigma)\textrm{sgn}(\tau)\left(v_{ 1 }\cdot w_{\sigma^{-1}\tau(1)}\right)\cdots \left(v_{ n }\cdot w_{\sigma^{-1}\tau(n)}\right)\\ =& \displaystyle \frac{1}{n!}\sum_{\sigma\in S_n}\sum_{\sigma^{-1}\tau\in S_n}\textrm{sgn}(\sigma^{-1}\tau)\left(v_{ 1 }\cdot w_{\sigma^{-1}\tau(1)}\right)\cdots \left(v_{ n }\cdot w_{\sigma^{-1}\tau(n)}\right)\\ =& \displaystyle \frac{1}{n!}\sum_{\sigma\in S_n}\sum_{\mu\in S_n}\textrm{sgn}(\mu)\left(v_{ 1 }\cdot w_{\mu(1)}\right)\cdots \left(v_{ n }\cdot w_{\mu(n)}\right)\\ =& \displaystyle \sum_{\mu\in S_n}\textrm{sgn}(\mu)\left(v_{ 1 }\cdot w_{\mu(1)}\right)\cdots \left(v_{ n }\cdot w_{\mu(n)}\right)\\ =& \displaystyle \det (v_i\cdot w_j). \end{array}$

Therefore (1) really defines an inner product. The extended Lagrange identity is then just the expansion

$\displaystyle \begin{array}{rl} \displaystyle |v_1\wedge \cdots \wedge v_n|^2 =& \displaystyle \det(v_i\cdot v_j) \\ =& \displaystyle \sum_{\sigma\in S_n}\textrm{sgn}(\sigma)\left(v_{ 1 }\cdot v_{\sigma(1)}\right)\cdots \left(v_{ n }\cdot v_{\sigma(n)}\right)\\ =& \displaystyle \sum_{\sigma\in A_n}\left(v_{ 1 }\cdot v_{\sigma(1)}\right)\cdots \left(v_{ n }\cdot v_{\sigma(n)}\right) -\sum_{\sigma\in S_n\setminus A_n}\left(v_{ 1 }\cdot v_{\sigma(1)}\right)\cdots \left(v_{ n }\cdot v_{\sigma(n)}\right). \ \ \ \ \ (2)\end{array}$

Here ${A_n}$ is the alternating subgroup of ${S_n}$, which consists of all the even permutations.

The extended Cauchy-Schwarz inequality is just

$\displaystyle \begin{array}{rl} \displaystyle \det (v_i\cdot v_j)\ge 0, \ \ \ \ \ (3)\end{array}$

which is equivalent to

$\displaystyle \begin{array}{rl} \displaystyle \sum_{\sigma\in S_n\setminus A_n}\left(v_{ 1 }\cdot v_{\sigma(1)}\right)\cdots \left(v_{ n }\cdot v_{\sigma(n)}\right)\le\sum_{\sigma\in A_n}\left(v_{ 1 }\cdot v_{\sigma(1)}\right)\cdots \left(v_{ n }\cdot v_{\sigma(n)}\right).\end{array}$

When ${n=2}$, (3) is just the classical Cauchy-Schwarz inequality

$\displaystyle \begin{array}{rl} \displaystyle |v_1|^2 |v_2|^2 -(v_1\cdot v_2)^2 \ge 0.\end{array}$

For ${n=3}$, (2) gives

$\displaystyle \begin{array}{rl} \displaystyle & \displaystyle |v_1\wedge v_2\wedge v_3|^2\\ =& \displaystyle |v_1|^2|v_2\wedge v_3|^2-(v_1\cdot v_2)\langle v_2\wedge v_3 , v_1\wedge v_3 \rangle+(v_1\cdot v_3) \langle v_2\wedge v_3, v_1\wedge v_2\rangle \\ =& \displaystyle |v_1|^2|v_2|^2|v_3|^2-\left(|v_1|^2(v_2\cdot v_3)^2+|v_2|^2(v_1\cdot v_3)^2+|v_3|^2(v_1\cdot v_2)^2\right) +2(v_1\cdot v_2)(v_2\cdot v_3)(v_1\cdot v_3). \end{array}$

For example, if ${v_1, v_2, v_3\in \mathbb R^3}$, then

$\displaystyle \begin{array}{rl} \displaystyle & \displaystyle \det(v_1, v_2, v_3)^2\\ =& \displaystyle |v_1|^2|v_2|^2|v_3|^2-\left(|v_1|^2(v_2\cdot v_3)^2+|v_2|^2(v_1\cdot v_3)^2+|v_3|^2(v_1\cdot v_2)^2\right) +2(v_1\cdot v_2)(v_2\cdot v_3)(v_1\cdot v_3). \ \ \ \ \ (4)\end{array}$

The Cauchy-Schwarz inequality then becomes

$\displaystyle \begin{array}{rl} \displaystyle |v_1|^2(v_2\cdot v_3)^2+|v_2|^2(v_1\cdot v_3)^2+|v_3|^2(v_1\cdot v_2)^2 \le |v_1|^2|v_2|^2|v_3|^2+2(v_1\cdot v_2)(v_2\cdot v_3)(v_1\cdot v_3). \end{array}$

From (4) we also obtain an interesting inequality

$\displaystyle \begin{array}{rl} \displaystyle \det(v_1, v_2, v_3)^2 \le |v_1|^2|v_2|^2|v_3|^2+2(v_1\cdot v_2)(v_2\cdot v_3)(v_1\cdot v_3), \ \ \ \ \ (5)\end{array}$

with the equality case holds if and only if ${v_i}$ are orthogonal to each other or one of them is zero.

It seems that when ${n=4}$, the algebra becomes quite formidable. Let us write ${v_{ij}:=v_i\cdot v_j}$. Then

$\displaystyle \begin{array}{rl} \displaystyle 0\le& \displaystyle |v_1\wedge \cdots \wedge v_4|^2\\ =& \displaystyle v_{11}|v_2\wedge v_3\wedge v_4|^2 -v_{12}\langle v_2\wedge v_3\wedge v_4, v_1\wedge v_3\wedge v_4\rangle\\ & \displaystyle +v_{13}\langle v_2\wedge v_3\wedge v_4, v_1\wedge v_2\wedge v_4\rangle -v_{14}\langle v_2\wedge v_3\wedge v_4, v_1\wedge v_2\wedge v_3\rangle\\ =& \displaystyle v_{11} v_{22} v_{33} v_{44} +v_{11} v_{23} v_{34} v_{42} +v_{11} v_{24} v_{32} v_{43} +v_{12} v_{21} v_{34} v_{43} +v_{12} v_{23} v_{31} v_{44} +v_{12} v_{24} v_{33} v_{41}\\ & \displaystyle +v_{13} v_{21} v_{32} v_{44} +v_{13} v_{22} v_{34} v_{41} +v_{13} v_{24} v_{31} v_{42} +v_{14} v_{21} v_{33} v_{42} +v_{14} v_{22} v_{31} v_{43} +v_{14} v_{23} v_{32} v_{41}\\ & \displaystyle -v_{11} v_{22} v_{34} v_{43} -v_{11} v_{23} v_{32} v_{44} -v_{11} v_{24} v_{33} v_{42} -v_{12} v_{21} v_{33} v_{44} -v_{12} v_{23} v_{34} v_{41} -v_{12} v_{24} v_{31} v_{43}\\ & \displaystyle -v_{13} v_{21} v_{34} v_{42} -v_{13} v_{22} v_{31} v_{44} -v_{13} v_{24} v_{32} v_{41} -v_{14} v_{21} v_{32} v_{43} -v_{14} v_{22} v_{33} v_{41} -v_{14} v_{23} v_{31} v_{42}. \ \ \ \ \ (6)\end{array}$

The Cauchy-Schwarz inequality in this case is rather complicated:

$\displaystyle \begin{array}{rl} \displaystyle & \displaystyle v_{11} v_{22} v_{34} v_{43} +v_{11} v_{23} v_{32} v_{44} +v_{11} v_{24} v_{33} v_{42} +v_{12} v_{21} v_{33} v_{44} +v_{12} v_{23} v_{34} v_{41} +v_{12} v_{24} v_{31} v_{43}\\ & \displaystyle +v_{13} v_{21} v_{34} v_{42} +v_{13} v_{22} v_{31} v_{44} +v_{13} v_{24} v_{32} v_{41} +v_{14} v_{21} v_{32} v_{43} +v_{14} v_{22} v_{33} v_{41} +v_{14} v_{23} v_{31} v_{42}\\ \le& \displaystyle v_{11} v_{22} v_{33} v_{44} +v_{11} v_{23} v_{34} v_{42} +v_{11} v_{24} v_{32} v_{43} +v_{12} v_{21} v_{34} v_{43} +v_{12} v_{23} v_{31} v_{44} +v_{12} v_{24} v_{33} v_{41}\\ & \displaystyle +v_{13} v_{21} v_{32} v_{44} +v_{13} v_{22} v_{34} v_{41} +v_{13} v_{24} v_{31} v_{42} +v_{14} v_{21} v_{33} v_{42} +v_{14} v_{22} v_{31} v_{43} +v_{14} v_{23} v_{32} v_{41}. \end{array}$

Observe that some of the “odd permutation terms” (those with a negative sign in (6), e.g. ${v_{11}v_{22}v_{34}v_{43}}$) are non-negative, so we can drop these terms and rearrange it to have an inequality similar to (5):

$\displaystyle \begin{array}{rl} \displaystyle & \displaystyle |v_1\wedge \cdots \wedge v_4|^2\\ \le& \displaystyle v_{11} v_{22} v_{33} v_{44}\\ & \displaystyle +[v_{11} (v_{23} v_{34} v_{42} +v_{24} v_{32} v_{43}) +v_{22} ( v_{14} v_{43}v_{31} +v_{13} v_{34} v_{41}) +v_{33} (v_{12} v_{24} v_{41} +v_{14} v_{21} v_{42}) +v_{44}(v_{12} v_{23} v_{31} +v_{13} v_{21} v_{32} )]\\ & \displaystyle +\left(v_{13} v_{31}v_{24} v_{42} +v_{12} v_{21} v_{34} v_{43} +v_{14} v_{41}v_{23} v_{32} \right)\\ & \displaystyle -\left(v_{12} v_{23} v_{34} v_{41} +v_{12} v_{24} v_{43}v_{31} +v_{13} v_{34} v_{42}v_{21} +v_{13} v_{32}v_{24} v_{41} +v_{14} v_{43}v_{32} v_{21} +v_{14} v_{42}v_{23} v_{31}\right) \\ =& \displaystyle v_{11} v_{22} v_{33} v_{44}\\ & \displaystyle +2\left(v_{11} v_{23} v_{34} v_{42} +v_{22} v_{14} v_{43}v_{31} +v_{33} v_{12} v_{24} v_{41} +v_{44} v_{12} v_{23} v_{31}\right)\\ & \displaystyle +\left(v_{13}^2v_{24}^2 +v_{12} ^2v_{34}^2 +v_{14} ^2v_{23}^2\right)\\ & \displaystyle -2\left(v_{12} v_{23} v_{34} v_{41} +v_{12} v_{24} v_{43}v_{31} +v_{13} v_{32}v_{24} v_{41} \right). \end{array}$

The equality holds if and only if these ${v_i}$‘s are orthogonal to each other or one of them is zero.