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Linear Algebra: Notes for MATH 341

Mitchel T. Keller, Nicholas Long

Section 5.3 Orthonormal Bases

The elementary vectors of \(\mathbb{R}^n\text{,}\) \(\{ \vec{e_1},\vec{e_2}, ... ,\vec{e_n}\}\text{,}\) form a basis for \(\mathbb{R}^n\text{.}\) Even better than that, the basis has only unit vectors and is orthogonal as a set (each pair of vectors is orthogonal to each other). These properties are very fundamental to how you worked with vectors before you started this class and why \(\mathbb{R}^n\) has such nice geometric intuition built in. The fundamental idea of this section is understanding a procedure for how to build a basis that is an orthogonal set and has vectors of “length” one.

Question 5.10.

In this question, you will build an orthonormal basis of \(\mathbb{R}^3\) from the ordered set \(\beta = \{ \colvec{1\\ 2\\ 2}, \colvec{3\\ 0\\ 0},\colvec{-1\\ -1\\ 0} \}\text{.}\) Orthonormal means that the set is orthogonal and contains only unit vectors.

(a)

We will construct the orthonormal basis \(\gamma = \{ \vec{\gamma_1}, \vec{\gamma_2} , \vec{\gamma_3} \}\) by going through the elements in \(\beta\) in order. In other words, we will consider \(\vec{\beta_1} = \colvec{1\\ 2\\ 2}\) first. Find \(\gamma_1\text{,}\) a unit vector in the direction of \(\vec{\beta_1}\text{.}\) This will be our first unit basis vector in \(\gamma\text{.}\)

(b)

We now want to consider \(\vec{\beta_2}= \colvec{3\\ 0\\ 0}\text{.}\) Is \(\vec{\beta_2}\) orthogonal to \(\vec{\gamma_1}\text{?}\)

(c)

We didn’t get lucky, so we will have to take out the part of \(\vec{\beta_2}\) that is not orthogonal to \(\vec{\gamma_1}\text{.}\) In other words, we need to find the projection of \(\vec{\beta_2}\) onto \(\vec{\gamma_1}\text{.}\) Compute \(\proj_{\vec{\gamma_1}} \vec{\beta_2}\text{.}\)

(d)

In order to take out the part of \(\vec{\beta_2}\) that is not orthogonal to \(\vec{\gamma_1}\text{,}\) we should subtract \(\proj_{\vec{\gamma_1}} \vec{\beta_2}\) from \(\vec{\beta_2}\text{.}\) Find \(\vec{\beta_2} - \proj_{\vec{\gamma_1}} \vec{\beta_2}\) and verify that this difference is orthogonal to \(\vec{\gamma_1}\text{.}\)

(e)

Since \(\vec{\beta_2} - \proj_{\vec{\gamma_1}} \vec{\beta_2}\) is orthogonal to \(\vec{\gamma_1}\text{,}\) we define \(\vec{\gamma_2}\) be the unit vector in the direction of \(\vec{\beta_2} - \proj_{\vec{\gamma_1}} \vec{\beta_2}\text{.}\) Write out the set \(\{ \vec{\gamma_1},\vec{\gamma_2} \}\text{.}\)

(f)

All that’s left to do is take \(\vec{\beta_3}\) and make \(\vec{\gamma_3}\text{,}\) a unit vector that is orthogonal to both \(\vec{\gamma_1}\) and \(\vec{\gamma_2}\text{.}\) Find the appropriate projections of \(\vec{\beta_3}\) in order to subtract out the parts of \(\vec{\beta_3}\) that is not orthogonal to \(\vec{\gamma_1}\) and \(\vec{\gamma_2}\text{.}\) Then find the unit vector in the direction of the difference to get \(\vec{\gamma_3}\text{.}\)

(g)

Verify that \(\gamma = \{ \vec{\gamma_1}, \vec{\gamma_2} , \vec{\gamma_3} \}\) is an orthonormal basis for \(\mathbb{R}^3\text{.}\)

Question 5.11.

Go through the same process above to create an orthonormal basis of \(\mathbb{P}_2\) from the basis \(\beta = \{ 1,t,t^2\}\) using the inner product and projection formula given by
\begin{equation*} \langle f,g \rangle = \int_0^1 f(t)g(t) \, dt \end{equation*}
and
\begin{equation*} \proj_{\vec{y}} \vec{x} = \left( \frac{\langle \vec{x},\vec{y}\rangle}{\langle \vec{y},\vec{y}\rangle} \right) \vec{y} \end{equation*}
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