Determining the last column so that the resulting matrix is an orthogonal matrix

Question

Determine the last column so that the resulting matrix is an orthogonal matrix $$\begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} & ? \\ \dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{6}} & ? \\ 0 & \dfrac{2}{\sqrt{6}} & ? \end{bmatrix}$$

Can anyone please provide hints to solve this?

Answer 1

Suppose that

$$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & ?\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & ?\\\ 0 & \frac{2}{\sqrt{6}} & ? \end{bmatrix} $$

Ok if you take $ a_{1} \times a_{2} = a_{3} $ it produces an orthogonal vector.

$$ a_{1} \times a_{2} = \begin{vmatrix} i & j & k \\ a_{11}& a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{vmatrix} $$

$$ a_{1} \times a_{2} = \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} i - \begin{vmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{vmatrix}j + \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}k $$

expanding this we get

$$ a_{1} \times a_{2} = (a_{12}a_{23}-a_{13}a_{22} )i - (a_{11}a_{23}-a_{13}a_{23})j + ( a_{11}a_{22}-a_{12}a_{21})k $$

writing this out now

$$a_{1} \times a_{2} = (\frac{1}{\sqrt{2}}\frac{2}{\sqrt{6}}-0\cdot\frac{-1}{\sqrt{6}})i - (\frac{1}{\sqrt{2}}\frac{2}{\sqrt{6}}-0\cdot\frac{2}{\sqrt{6}})j + (\frac{1}{\sqrt{2}}\frac{-1}{\sqrt{6}}-\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{6}})k $$

Getting rid of stuff

$$a_{1} \times a_{2} = (\frac{1}{\sqrt{2}}\frac{2}{\sqrt{6}})i - (\frac{1}{\sqrt{2}}\frac{2}{\sqrt{6}})j + (\frac{1}{\sqrt{2}}\frac{-1}{\sqrt{6}}-\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{6}})k $$

More cleaning up

$$a_{1} \times a_{2} = (\frac{2}{2\sqrt{3}})i - (\frac{2}{2\sqrt{3}})j + (\frac{-1}{2\sqrt{3}}- \frac{1}{2\sqrt{3}})k $$ More cleaning

$$a_{1} \times a_{2} = (\frac{1}{\sqrt{3}})i - (\frac{1}{\sqrt{3}})j + (\frac{-1}{\sqrt{3}})k $$

Checking if they are orthogonal

we need that $$ a_{1} \cdot (a_{1} \times a_{2}) =0 , a_{2} \cdot (a_{1} \times a_{2}) =0$$

$$a_{1} \cdot (a_{1} \times a_{2}) = \frac{1}{\sqrt{2}}\frac{1}{\sqrt{3}} + (\frac{1}{\sqrt{2}}(\frac{-1}{\sqrt{3}}) + (0)\frac{-1}{\sqrt{3}} $$

Note that they are opposite this is zero.

$$a_{2} \cdot (a_{1} \times a_{2}) = (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{3}}) + (\frac{-1}{\sqrt{6}})(\frac{-1}{\sqrt{3}}) + (\frac{2}{\sqrt{6}}(\frac{-1}{\sqrt{3}}) $$

we have

$$a_{2} \cdot (a_{1} \times a_{2}) = (\frac{2}{\sqrt{6}}(\frac{1}{\sqrt{3}})+ (\frac{2}{\sqrt{6}}(\frac{-1}{\sqrt{3}}) =0 $$

Done, finally

$$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} & \frac{-1}{\sqrt{3}}\\\ 0 & \frac{2}{\sqrt{6}} & \frac{-1}{\sqrt{3}} \end{bmatrix} $$

Answer 2

Hint: remember that the cross product of two vectors of length 3 is orthogonal to both argument vectors and that an orthogonal matrix's columns are themselves an orthogonal set of vectors…

Answer 3

A matrix is orthogonal if all the column vectors are unit vectors and any two columns have dot product zero. Now write the unknown last column as $(x\ y\ z)^T$. You will get two equations when when you insist it be orthogonal to the known first and second columns.

So solve a system of 2 equations an $x,y,z$. There will be infinitely many solutions. Now the condition on length will bring down the solutions to 2 (a vector and its negative).

Answer 4

Using SymPy:

>>> from sympy import *
>>> x, y, z = symbols('x y z')
>>> M = Matrix([[1/sqrt(2),  1/sqrt(6), x],
                [1/sqrt(2), -1/sqrt(6), y],
                [        0,  2/sqrt(6), z]])
>>> simplify(M.T * M)
Matrix([
[                1,                       0,       sqrt(2)*(x + y)/2],
[                0,                       1, sqrt(6)*(x - y + 2*z)/6],
[sqrt(2)*(x + y)/2, sqrt(6)*(x - y + 2*z)/6,      x**2 + y**2 + z**2]])

Thus, we obtain the following system of equations

$$\begin{aligned} x + y &= 0\\ x + z &= 0\\ x^2 + y^2 + z^2 &= 1\end{aligned}$$

which is easy to solve. Alternatively, we can take the $2$ known columns (which are orthonormal), form a $3 \times 2$ matrix, transpose it and then compute its null space:

>>> (Matrix([[1/sqrt(2),  1/sqrt(6)],
             [1/sqrt(2), -1/sqrt(6)],
             [        0,  2/sqrt(6)]])).T.nullspace()
[Matrix([
[-1],
[ 1],
[ 1]])]

All that is left to do is to normalize the vector that spans the null space. Switching its sign is legal.

Answer 5

Let $[ x \ y \ z]$ be the vector you are looking for, then it must satisfy $$x \frac{1}{\sqrt{2}} +y \frac{1}{\sqrt{2}}+z (0)= 0$$ which is $y = - x$ and $$x \frac{1}{\sqrt{6}} -y \frac{1}{\sqrt{6}}+z \frac{2}{\sqrt{6}} = 0$$ which is $z = \frac{1}{2}y - \frac{1}{2}x$. Using $y = -x$ in $z = \frac{1}{2}y - \frac{1}{2}x$, we get $$z = -x$$ So the vector you are looking for has the form $[x, \ -x, \ -x]$ Choose $x = \frac{1}{\sqrt{3}}$ (for example), we get $[\frac{1}{\sqrt{3}}, \ -\frac{1}{\sqrt{3}}, \ -\frac{1}{\sqrt{3}}]$, which is a possible last column.