How to decompose a bivector into a sum of _orthogonal_ blades? In Geometric Algebra, any bivector $B\in\Lambda^2\mathbb R^n$ is a sum of blades:
$$B = B_1 + B_2 + \cdots$$
$$= \vec v_1\wedge\vec w_1 + \vec v_2\wedge\vec w_2 + \cdots$$
Each blade's component vectors $\vec v$ and $\vec w$, if they're not already orthogonal to each other, can easily be made so by the Gram-Schmidt process:
$$B_1 = \vec v_1\wedge\vec w_1 = \vec v_1\wedge\left(\vec w_1-\Big(\frac{\vec w_1\cdot\vec v_1}{\vec v_1\cdot\vec v_1}\Big)\vec v_1\right) = \vec v_1\wedge\vec w_1'$$
$$\vec v_1\cdot\vec w_1' = 0$$
(This can even be generalized to pseudo-Euclidean space where $\vec v$ may square to zero: project $\vec v$ away from $\vec w$ instead of vice-versa, or if they both square to zero, take $\vec v'=\frac{\vec v+\vec w}{\sqrt2}$ , $\vec w'=\frac{\vec w-\vec v}{\sqrt2}$. Then $\vec v\wedge\vec w=\vec v'\wedge\vec w'$, and $\vec v'\cdot\vec w'=0$.)
But I don't know how to make each blade orthogonal to the other blades. Orthogonal means that their geometric product is their (grade 4) wedge product; all lower-grade parts are zero.
$$B_1 + B_2 = B_1' + B_2'$$
$$B_1'B_2' = (B_1'\cdot B_2')+(B_1'\times B_2')+(B_1'\wedge B_2') = B_1'\wedge B_2'$$
$$B_1'\cdot B_2' = 0 = B_1'\times B_2'$$
From Wikipedia: In $\Lambda^2\mathbb R^4$,

"every bivector can be written as the sum of two simple bivectors. It is useful to choose two orthogonal bivectors for this, and this is always possible to do."


Here's a simple example, with $n = 4$:
$$B_1 = e_1\wedge e_2 = e_1e_2$$
$$B_2 = (e_1 + e_3)\wedge e_4 = e_1e_4 + e_3e_4$$
$$B = B_1 + B_2 = e_1e_2 + e_1e_4 + e_3e_4$$
$$B_1B_2 = -e_2e_4 + e_1e_2e_3e_4$$
$$B_1\cdot B_2 = 0 \neq B_1\times B_2 = -e_2e_4$$
How can I rewrite $B = B_1' + B_2'$ with $B_1'\cdot B_2' = 0 = B_1'\times B_2'$ ?

EDIT1
After doing some algebra, I arrived at these equations:
$$B_1' = \frac{B+Q}{2}$$
$$B_2' = \frac{B-Q}{2}$$
$$Q^2 = B\cdot B - B\wedge B$$
$$B^2 = Q\cdot Q - Q\wedge Q$$
$$B\times Q = 0$$
$$B\wedge Q = 0$$
We only need to solve for $Q$ in terms of $B$. I was able to take a square root of the third equation (by guessing that $Q = xe_1e_2+ye_3e_4$) but I didn't find the specific root that satisfies the other equations.

EDIT2
After doing some more algebra, I find that, if $Q$ is defined as the reflection of $B$ along some unknown vector $v\neq0$,
$$Q=v^{-1}Bv$$
$$B_1=\frac{v^{-1}vB+v^{-1}Bv}{2}=v^{-1}(v\wedge B)$$
$$B_2=\frac{v^{-1}vB-v^{-1}Bv}{2}=v^{-1}(v\cdot B)$$
then $B_1$ and $B_2$ are blades, and $B_1\cdot B_2=0$ regardless of $v$, and $B_1\times B_2=0$ if and only if $v\wedge\big((v\cdot B)\cdot B\big)=0$. This means that $(v\cdot B)\cdot B$ must be parallel to $v$; in other words, $v$ is an eigenvector of the operator $(B\,\cdot)^2$. It follows that $v\cdot B=w$ is also an eigenvector with the same eigenvalue, and $v\cdot w=0$.
Generalizing, it looks like we want to find an orthogonal set of eigenvectors $v_1,v_2,v_3,\cdots$ of $(B\,\cdot)^2$, so that
$$B_1=v_1^{-1}(v_1\cdot B),\quad B_2=v_2^{-1}(v_2\cdot B),\quad B_3=v_3^{-1}(v_3\cdot B),\quad\cdots$$
Of course, all vectors $v$ must also be orthogonal to all $w=v\cdot B$.

 A: By using the exponential function to rewrite $B^2$ and take its square root (link, section 2.1.1), I found a formula for bivectors in 4D:
$$B_1 = \left(\frac{-B\cdot B+\sqrt{(B\cdot B)^2-(B\wedge B)^2}+B\wedge B}{2\sqrt{(B\cdot B)^2-(B\wedge B)^2}}\right)B$$
$$B_2 = \left(\frac{B\cdot B+\sqrt{(B\cdot B)^2-(B\wedge B)^2}-B\wedge B}{2\sqrt{(B\cdot B)^2-(B\wedge B)^2}}\right)B$$
Obviously, this is undefined if $|B\cdot B|=\lVert B\wedge B\rVert$. That corresponds to an isoclinic rotation, where the two planes of rotation are not unique. In this case, any vector $v\neq0$ is an eigenvector of $(B\,\cdot\,)^2$, so we can just take $B_1=e_1(e_1\cdot B)$.

Applying this to the example problem,
$$B = e_1e_2+e_1e_4+e_3e_4$$
$$B^2 = -3 + 2e_1e_2e_3e_4;\quad B\cdot B = -3,\quad B\wedge B = 2e_1e_2e_3e_4$$
$$B_1 = \frac{(1+\sqrt5)(e_1e_2+e_3e_4)+(3+\sqrt5)e_1e_4-2e_2e_3}{2\sqrt5}$$
$$B_2 = \frac{(-1+\sqrt5)(e_1e_2+e_3e_4)+(-3+\sqrt5)e_1e_4+2e_2e_3}{2\sqrt5}$$
$$B_1\wedge B_1 = 0 = B_2\wedge B_2$$
(The vanishing wedge product means that they are actually blades, though we don't know what vectors they're made of.)
$$B_1\cdot B_2 = 0 = B_1\times B_2$$
$$B_1 + B_2 = B$$

To prove that such a decomposition always exists in Euclidean spaces, note that $(B\,\cdot\,)^2$ is symmetric/self-adjoint:
$$\big((v\cdot B)\cdot B\big)\cdot w=(v\cdot B)\cdot(B\cdot w)=-(v\cdot B)\cdot(w\cdot B)$$
so the spectral theorem applies.

...In fact the result is false for pseudo-Euclidean spaces in general. Take an orthogonal basis $\{\sigma_1,\sigma_2,\tau_1,\tau_2\}$ with $\sigma_1\!^2=\sigma_2\!^2={^+}1,\;\tau_1\!^2=\tau_2\!^2={^-}1$, and consider the bivector
$$J=\sigma_1\frac{\sigma_2+\tau_2}{2}+\tau_1\frac{\sigma_2-\tau_2}{2}\quad=\frac{\sigma_1+\tau_1}{2}\sigma_2+\frac{\sigma_1-\tau_1}{2}\tau_2$$
(The $1/2$ is only to normalize $J^4=1$.)
$$(\sigma_1\cdot J)\cdot J=\frac{-\tau_1}{2},\quad(\tau_1\cdot J)\cdot J=\frac{\sigma_1}{2}$$
$$(\sigma_2\cdot J)\cdot J=\frac{-\tau_2}{2},\quad(\tau_2\cdot J)\cdot J=\frac{\sigma_2}{2}$$
It's easy to see that $(J\,\cdot\,)^2$ has no eigenvectors, so $J$ is not orthogonally decomposable.
(If it were orthogonally decomposable as $J=v_1\wedge w_1+v_2\wedge w_2$, then $v_1$ would be an eigenvector, with eigenvalue $(v_1\wedge w_1)^2$.)
I suspect that the result is still true for Lorentzian spaces (where only one basis vector has a different signature from the others).
