# Find $v_3$ such that $\{v_0, v_1, v_2, v_3\}$ is an orthogonal basis of $\mathbb{R}^4$?

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.

• Hint – A_A Jan 22 '18 at 9:02
• @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. – Cedron Dawg Jan 22 '18 at 14:24

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

a+b=0
a+c=0
a-d=0


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 &= 0\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 $c+d = 0 => b=c => d = -c, a = -c$

\begin{align} (a+b)^2 + (c+d)^2 &= 0\\ &= a^2 + b^2 + 2ab + c^2 + d^2 + 2cd \end{align}

use (4) so $0 = 1 + 2(ab + dc) = 1 + 2((-c)c + c(-c))$

=> $1/4 = c^2$

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

so we have 2 solutions for the system of equations.

• what can I do? be more specific – canbax Jan 20 '18 at 21:49
• @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. – Cedron Dawg Jan 21 '18 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:

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

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.