In this question, we consider the Hilbert space of vectors in $\mathbb{R}^4$ with one of its basis $\{v_0, v_1, v_2, v_3\}$. Given the first three basis vectors $$v_0=\left[\begin{array}{c} 1/2\\ 1/2\\ 1/2\\ 1/2 \end{array}\right]v_1=\left[\begin{array}{c} 1/2\\ 1/2\\ -1/2\\ -1/2 \end{array}\right] v_2=\left[\begin{array}{c} 1/2\\ -1/2\\ 1/2\\ -1/2 \end{array}\right]$$

How many possibilit(y/ies) are there for $v_3$, such that $\{v_0, v_1, v_2, v_3\}$ is an orthogonal basis of $\mathbb{R}^4$? Give one example of $v_3$ please.

  • $\begingroup$ Hint $\endgroup$
    – A_A
    Jan 22, 2018 at 9:02
  • $\begingroup$ @A_A, the G-S process is good for creating an orthogonal basis from a non-orthogonal one. It does not provide candidates for missing basis vectors. See my answer for how to find a fourth vector that is orthogonal to any three linearly independent ones, and the other answers that solve it for this particular case. Since the first three are already orthogonal , G-S doesn't do much for you. $\endgroup$ Jan 22, 2018 at 14:24

5 Answers 5


let $v_3 = \left[\begin{array}{vector} a\\ b\\ c\\ d\end{array}\right]$, then $<v_0, v_3>=0, <v_1, v_3>=0, <v_2, v_3>=0, <v_3, v_3>=1$, reduce to


and $a^2+b^2+c^2+d^2 = 1$

so there are many solutions, one example can be

$ \left[\begin{array}{arr} \frac{1}{2}\\ -\frac{1}{2}\\ -\frac{1}{2}\\ \frac{1}{2}\end {array} \right]$


First read the response of Yicheng Ye

\begin{align} a + b + c + d &= 0\tag{1}\\ a + b - c - d &= 0\tag{2}\\ a - b + c - d &= 0\tag{3}\\ a^2 + b^2 + c^2 + d^2 &= 1\tag{4} \end{align}

from (2) $$a+b = c+d$$,

use (1) $$2(c+d) = 0 => c+d=0, a+b=0$$,

from (3) $$a+c = b+d$$ use (1) $$2(b+d) = 0 => b+d = 0, a+c = 0$$

if $b+d = 0$ and $c+d = 0 => b=c => d = -c, a = -c$

use (4) we substitute terms \begin{align} a^2 + b^2 + c^2 + d^2 &= 1 \\ (-c)^2 + c^2 + c^2 + (-c)^2 &= 1 \\ 4c^2 = 1 \\ c = ± 1/2 \end{align}

=> $c=1/2$ or $c=-1/2$ all the other variables are depend on $c$

so we have 2 solutions for the system of equations.

  • $\begingroup$ what can I do? be more specific $\endgroup$
    – canbax
    Jan 20, 2018 at 21:49
  • 1
    $\begingroup$ @canbax, Nice solution! For math mode, search on "Mathjax" for syntax help. Start each equation with two dollar signs, use Latex syntax, end with two dollar signs. $\endgroup$ Jan 21, 2018 at 15:31

A generic approach: the first thing to check is whether your three vectors are linearly independent. They are, so they span a 3D space. Thus, a well-chosen fourth one could complement them into a four-dimensional basis. There would be a infinity of choices: any vector that is not in the 3D space will do the job.

Indeed, the first three vectors are orthogonal, with unit norm as $\sum_1^4 \left(\pm \frac{1}{2}\right)^2 = 1$. Hence, they are pairwise orthonormal. As above, they define a subspace of dimension $3$. Its orthogonal supplement is a 1D vector space, uniquely defined by one non-null vector, which can be scaled by any non-zero scalar.

So if you just want orthogonality, you have an infinity of choices. But in some cases, people uses orthogonal as a proxy for orthonormal. Indeed, your three vectors $v_0$, $v_1$, $v_2$ are of unit norm too.

So in this case, there are only two vectors $v_3$ with unit norm answering your question:

$$ \left[\begin{array}{arr} \frac{1}{2}\\ -\frac{1}{2}\\ -\frac{1}{2}\\ \frac{1}{2}\end {array} \right]$$

and its opposite:

$$ \left[\begin{array}{arr} -\frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ -\frac{1}{2}\end {array} \right]\,.$$

As you can see, $v_0$ exhibits no sign changes, $v_1$ has one, and three for $v_2$. And $v_3$ has two sign changes. You just have rediscovered, up to a factor, the $4$-dimensional Hadamard (or Walsh, or Paley) orthogonal basis:

$$H_4=\left[ \begin{array}{rrrr} 1 & 1 & 1 & 1\\ 1 & -1 & -1 & 1\\ 1 & 1 & -1 & -1\\ 1 & -1 & -1 & 1\\ \end{array}\right]$$

You can find more information about them in:

  • 1
    $\begingroup$ I understand the reasoning, but how to solve the 4 equations ? Also, how did you know that the number of solutions are infinite ? $\endgroup$ Dec 26, 2016 at 18:17
  • $\begingroup$ I have reformulated the answer $\endgroup$ Dec 2, 2020 at 22:28

In 4D, you can use the generalized cross product to calculate a vector that is orthogonal to a set of three other linearly independent vectors. The other vectors don't need to be orthogonal for this method to work. The length of the vector is the volume of the parallelepiped formed by the three vectors.

Calculate the determinate of this matrix where the first row is composed of unit vectors. $$ \left| \begin{array}{cccc} \vec i & \vec j & \vec k & \vec l \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \end{array} \right| $$ In your example, this is probably more difficult to calculate than some of the other answers, but still, it is good to know.



Maybe, some day, when the old teachers have all died and modern (differential) geometry finds its way into higher education this answer will be appreciated:

$$v_3=\ast\left(v_0 \wedge v_1 \wedge v_2 \right)$$

This expression uses the exterior algebra of $\mathbb{R}^4$ with the metric-dependent Hodge-Star $\ast$.

The Exterior algebra of a vector space is an extremely powerful concept, that is unfortunately known only to far too few experts. It finds its generalisation in Clifford/Geometric algebras and is at the center of coordinate-free differential geometry. It's also not hard to learn, or at least much easier than the traditional coordinate-based approaches.

One can dream, at least.


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