Find the vector $v_2$ so that $v=v_1+v_2$, where $v_1$ is in $\operatorname{Span}\{u_1,u_2\}$ and $v_2$ is orthogonal to $\operatorname{Span}\{u_1,u_2\}$. Note that $u_1$ and $u_2$ are orthogonal to each other.

I had tried just orthogonally projecting $v$ onto the span of $u_1$ and $u_2$, but that didn't work.

Given $$\mathbf u_1 = \begin{bmatrix}2\\2\\-1\\2\end{bmatrix}, \quad \mathbf u_2 = \begin{bmatrix}2\\1\\2\\-2\end{bmatrix}, \quad \mathbf v = \begin{bmatrix}3\\-1\\-3\\3\end{bmatrix}, \quad $$ find the vector $\mathbf v_2$ so that $\mathbf v = \mathbf v_1 + \mathbf v_2$, where $\mathbf v_1$ is in $\operatorname{Span}\{\mathbf u_1,\mathbf u_2\}$ and $\mathbf v_2$ is orthogonal to $\operatorname{Span}\{\mathbf u_1,\mathbf u_2\}$. Note that $\mathbf u_1$ and $\mathbf u_2$ are orthogonal to each other.

$\mathbf v_2 = \left\langle 2-\frac 7 {13}\cdot 2,2 - \frac 7 {13}, -1-\frac 7 {13}\cdot 2,2+2\cdot \frac 7 {13}\right\rangle$


Vectors: For $[1,2,3]$ or $\begin{bmatrix}1\\2\\3\end{bmatrix}$, input $\langle 1, 2, 3 \rangle$.

  • $\begingroup$ So you projected $v$ onto the subspace spanned by $u_1, u_2$. that should give you $v_1$. You should then find $v_2 = v - v_1$. Is that what you did? $\endgroup$
    – Muphrid
    Commented Jun 23, 2013 at 1:58
  • $\begingroup$ Thank you so much !! But could you explain the reasoning behind this? Why isn't the orthogonal projection of v onto the subspace spanned by u1 & u2 v2, but instead v1? $\endgroup$
    – ahn
    Commented Jun 23, 2013 at 2:01
  • $\begingroup$ It could be "orthogonal projection" means something else in your course/text, but to me, it says, "I have a vector that does not necessarily lie in this plane; when I project the vector, I am finding the part of it that is in the plane." $\endgroup$
    – Muphrid
    Commented Jun 23, 2013 at 2:02
  • $\begingroup$ Ah I think I misunderstood the meaning of 'orthogonal projection', your explanation makes a lot of sense, thanks again! $\endgroup$
    – ahn
    Commented Jun 23, 2013 at 2:11

2 Answers 2


It sounds like you've already been led basically to the answer. Here I outline an alternative method to directly find $v_2$, using Clifford algebra.

In Clifford algebra, we can directly represent the subspace spanned by $u_1, u_2$ using what's called a geometric product. The geometric product has the following properties:

$$e_1 e_1 = e_2 e_2 = e_3 e_3 = e_4 e_4 = 1, \quad e_1 e_2 = -e_2 e_1, \quad e_1 e_3 = -e_3 e_1, \ldots$$

Along with being associative. The geometric product is a very, very useful product. We can directly represent the subspace using the geometric product.

$$B = \langle u_1 u_2 \rangle_2$$

This angle bracket notation means "keep only terms that have two orthogonal basis vectors in them". That is, throw away any scalar terms. This is exactly what I'll do.

$$\begin{align*}B &=\langle (2 e_1 + 2 e_2 - e_3 + 2 e_4) (2 e_1 + e_2 + 2 e_3 - 2 e_4) \rangle_2 \\ &= -2 e_1 e_2 + 6 e_1 e_3 - 8 e_1 e_4 + 5 e_2 e_3 - 6 e_2 e_4 -2 e_3 e_4\end{align*}$$

Again, this is a direct representation of the planar subspace spanned by the two vectors. Vitally, we need its magnitude. Again, this can be found using a geometric product:

$$B^2 = -4 - 36 - 64 - 25 - 36 - 4 = -169$$

Yes, this is a negative quantity; this is a common property of Euclidean bivectors. They actually naturally square to negative numbers. Clifford algebra naturally provides us with objects that "act like" imaginary units, in the form of Euclidean bivectors.

We can directly find $v_2$ by finding the "rejection" of $v$ in $B$. The rejection is so named to be the "opposite" of the projection. The basic idea is as follows: form a parallelepiped with $v$ and $B$, and then find the part of this volume orthogonal to $B$. The formula is

$$v_2 = \langle v B \rangle_3 B^{-1}$$

where $B^{-1} = B/B^2$. The final result is guaranteed to be a vector. This is, however, usually denoted $(v \wedge B) B^{-1}$, suggesting a relationship with exterior algebra. I will not explore this here.

The geometric product $vB$ has as its grade-3 components

$$\langle vB\rangle_3 = 27 e_1 e_2 e_3 - 32 e_1 e_2 e_4 - 12 e_1 e_3 e_4 - e_2 e_3 e_4$$

All that remains is to compute the final product. I will not do this here (as that would solve the entire problem). But I hope that this shows you some of the power of Clifford algebra, especially in the way it can directly represent subspaces as objects in their own right, as opposed to mere spans of vectors. Clifford algebra makes possible all these operations between general subspaces beyond vectors.


Given $u_1=[2,2,-1,2]^T$, $u_2=[2,1,2,-2]$ and $v=[3,-1,-3,3]$. We need to find $v=v_1+v_2$ where $v_1\in\operatorname{span}\{u_1,u_2\}$ and $v_2$ is orthogonal to this subspace.

Since $v_2$ is orthogonal to the subspace, it must be orthogonal to $v_1$, i.e. $v_1\cdot v_2=0$. This means that $v_1=\operatorname{proj}_U(v)$, where $U=\operatorname{span}\{u_1,u_2\}$.

Now, note that $u_1\cdot u_2=0$, so an orthonormal basis for $U$ is $\{\hat{u}_1,\hat{u}_2\}$, where $\hat{u}_1=u_1/\sqrt{13}$ and $\hat{u}_2=u_2/\sqrt{13}$. Then $v_1=\langle v,\hat{u}_1\rangle\hat{u}_1+\langle v,\hat{u}_2\rangle\hat{u}_2$, where $\langle\cdot,\cdot\rangle$ denotes the usual inner product.

This gives $v_1=\sqrt{13}\hat{u}_1-\dfrac{7}{\sqrt{13}}\hat{u}_2=\left[\dfrac{12}{13},\dfrac{19}{13},\dfrac{-27}{13},\dfrac{40}{13}\right]^T.$ Then, $v_2=v-v_1=\left[\dfrac{27}{13},\dfrac{-32}{13},\dfrac{-12}{13},\dfrac{-1}{13}\right]^T$.


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