Lagrange's Identity

Lagrange's identity is very important in linear algebra as is draws a distinct relationship between the cross product of two vectors to the dot product of two vectors. Lagrange's identity is as follows:

Theorem (Lagrange's Identity): If $\vec{u}, \vec{v} \in \mathbb{R}^2$ then $\| \vec{u} \times \vec{v} \|^2 = \| \vec{u} \|^2 \| \vec{v} \|^2 - (\vec{u} \cdot \vec{v})^2$.

• Proof: Let $\vec{u}, \vec{v} \in \mathbb{R}^3$. We will prove this by comparing the righthand side to the lefthand side of this equation. When we expand though on the lefthand side we obtain that:

(1)

\begin{align} \quad \| \vec{u} \times \vec{v} \| = \sqrt{(u_{2}v_{3} - u_{3}v_{2})^2 + (-u_{1}v_{3} + u_{3}v_{1})^2 + (u_{1}v_{2} - u_{2}v_{1})^2} \\ \quad \| \vec{u} \times \vec{v} \|^ 2 = (u_{2}v_{3} - u_{3}v_{2})^2 + (-u_{1}v_{3} + u_{3}v_{1})^2 + (u_{1}v_{2} - u_{2}v_{1})^2 \\ \quad \| \vec{u} \times \vec{v} \|^ 2 = (u_{2}v_{3} - u_{3}v_{2})^2 + (-u_{1}v_{3} + u_{3}v_{1})^2 + (u_{1}v_{2} - u_{2}v_{1})^2 \end{align}

• Now let's compare this with the righthand side of Lagrange's identity:

(2)

\begin{align} \quad \|\vec{u}\|^2 \|\vec{v}\|^2 - (\vec{u} \cdot \vec{v})^2 = \|\vec{u}\|^2 \|\vec{v}\|^2 - (\vec{u} \cdot \vec{v})^2 = (u_{1}^2 + u_{2}^2 + u_{3}^2) (v_{1}^2 + v_{2}^2 + v_{3}^2) - (u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3})^2 \\ \quad \|\vec{u}\|^2 \|\vec{v}\|^2 - (\vec{u} \cdot \vec{v})^2 = u_{1}^2v_{1}^2 + u_{1}^2v_{2}^2 + u_{1}^2v_{3}^2 + u_{2}^2v_{1}^2 + u_{2}^2v_{2}^2 + u_{2}^2v_{3}^2 + u_{3}^2v_{1}^2 + u_{3}^2v_{2}^2 + u_{3}^2v_{3}^2 \end{align}

• This can clearly be simplified further, however, we will leave this to the reader to verify. $\blacksquare$

We will later show the usefulness of Lagrange's identity with its applications.