Problem of the week - Orthogonal bases

Many concepts of Linear Algebra have emerged from geometric problems in R² and R³, and then generalized to non-visual higher-dimensional spaces. Some of the most widely used geometric concepts are length, distance and perpendicularity, which are well known for R² and R³, and thanks to linear algebra are extended to Rⁿ. These concepts provide powerful geometric tools for solving many applied problems, including the least-squares problems.

These all three notions are defined in terms of the inner product of two vectors, which is also the key concept to deal with orthogonal bases, the subject of the problem of this week. Orthogonal bases, and particularly orthonormal bases, are very useful when dealing with projections onto subspaces, among other problems. Some of the advantages of using this kind of bases lie on the fact that the coordinates of a vector relative to an orthonormal basis can be computed by using an explicit and easy formula, where each coordinate only involves the associate vector of the basis.

Find the values of x, y and z, if any, such that the set A is an orthogonal basis for R³.

A = {(x, 2, 2); (1, y, -1); (-4, -1, z)}

A basis for a vector space is an orthogonal basis if every pair of basis vectors are orthogonal, that is, their inner product is 0. It can be easily proved that an orthogonal set is always linearly independent, so, as the set A has 3 vectors (the dimension of the vector space), it's sufficient to prove that the vectors are orthogonal to prove they are also a basis for R³.

For the set of vectors to be an orthogonal set, every pair of vectors have to be orthogonal:

<(x, 2, 2),(1, y, -1)> = x + 2y - 2 = 0
<(x, 2, 2),(-4, -1, z)> = -4x - 2 + 2y = 0
<(1, y, -1),(-4, -1, z)> = -4 - y - z = 0

So, to find the values of x, y and z, such that the set A is an orthogonal basis, we need to solve the system of linear equations:

  x + 2y = 2
-4x + 2z = 2
 -y -  z = 4

which the augmented matrix is:

┌                 ┐
│  1   2   0 |  2 │
│ -4   0   2 |  2 │
│  0  -1  -1 |  4 │
└                 ┘

Transform the augmented matrix to row echelon form.

r2 <———> r2 + 4•r1
┌                 ┐
│ 1   2   0 |   2 │
│ 0   8   2 |  10 │
│ 0  -1  -1 |   4 │
└                 ┘
r2 <———> r3
┌                 ┐
│ 1   2   0 |   2 │
│ 0  -1  -1 |   4 │
│ 0   8   2 |  10 │
└                 ┘
r3 <———> r3 + 8•r2
┌                 ┐
│ 1   2   0 |   2 │
│ 0  -1  -1 |   4 │
│ 0   0  -6 |  42 │
└                 ┘

Then, translate the row echelon form matrix to the associated system of linear equations.

x+2y = 2
-y-z = 4
-6z = 42

Use back substitution to find the solution. Starting by the last equation, isolate the variable and replace it in the equation above. Do this process until the first equation.

-6z = 42
z = -42/6
z = -7

-y-z = 4
y = -4-z
y = -4-(-7)
y = 3

x+2y = 2
x = 2-2y
x = 2-2•(3)
x = -4

So, the solution is:

x = -4
y = 3
z = -7