Math 511: Linear Algebra

5.3 Orthonormal Bases

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5.3.1 Orthogonal Vectors

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During our work with different bases in $\mathbb{R}^n$, surely you have noticed that using the standard basis $\left\{\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n\right\}$ is much easier than using other bases. This comes from two key properties possessed by the standard basis, all of the vectors are orthogonal to one another and each of them is a unit vector. Generally, bases with these two properties are the most desirable for any inner product space $V$.

Definition of Orthogonal Set

Let $\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n$ be nonzero vectors in an inner product space $V$. If $\ \langle \mathbf{v}_i,\mathbf{v}_j \rangle = 0$ when $i\neq j$, then $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\right\}$ is an orthogonal set of vectors.

Example 1

Consider the vectors

$$ \mathbf{v}_1 = \left[ \begin{array}{r}\ 1\ \\ -1\ \\ \ 1\ \end{array} \right] \qquad\qquad \mathbf{v}_2 = \left[ \begin{array}{r}\ 1\ \\ \ 1\ \\ \ 0\ \end{array} \right] \qquad\qquad \mathbf{v}_3 = \left[ \begin{array}{r} -1\ \\ \ 1] \\ \ 2\ \end{array} \right] $$

These form an orthogonal set since each pair of vectors is orthogonal:

$$ \begin{align*} \langle \mathbf{v}_1,\mathbf{v}_2 \rangle = \left[ \begin{array}{r} 1 & -1 & 1 \end{array} \right]\left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] &= (1)(1) + (-1)(1) + (1)(0) = 0 \\ \\ \langle \mathbf{v}_1,\mathbf{v}_3 \rangle = \left[ \begin{array}{r} 1 & -1 & 1 \end{array} \right]\left[ \begin{array}{r} -1 \\ 1 \\ 2 \end{array} \right] &= (1)(-1) + (-1)(1) + (1)(2) = 0 \\ \\ \langle \mathbf{v}_2,\mathbf{v}_3 \rangle = \left[ \begin{array}{r} 1 & 1 & 0 \end{array} \right]\left[ \begin{array}{r} -1 \\ 1 \\ 2 \end{array} \right] &= (1)(-1) + (1)(1) + (0)(2) = 0 \end{align*} $$

Orthogonal sets are also linearly independent, which makes them usable as a basis for a vector space.

Theorem 5.3.1

Orthogonal Sets are Linearly Independent

If $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\right\}$ is an orthogonal set of nonzero vectors in an inner product space $V$, then $\left\{ \mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n \right\}$ is a linearly independent set.

Proof:

Suppose $\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n$ is a collection of mutually orthogonal vectors. If a linear combination of these vectors is $\mathbf{0}$

$$ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_n\mathbf{v}_n = \mathbf{0} $$

we take the inner product of this linear combination with some $\mathbf{v}_j$ for a $j\in\left\{1,2,\ldots,n\right\}$, then

$$ \begin{align*} c_1\langle\mathbf{v}_j,\mathbf{v}_1 \rangle + c_2\langle\mathbf{v}_j,\mathbf{v}_2 \rangle + \ldots + c_n\langle \mathbf{v}_j,\mathbf{v}_n \rangle &= 0 \\ \\ c_j\left\|\mathbf{v}_j\right\|^2 &= 0 \end{align*} $$

Since this is true for any choice of $j$, all scalars $c_1,c_2,\ldots,c_n$ must be $0$. $\tombstone$

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5.3.2 Orthonormal Sets

Definition of Orthonormal Set

A set of orthogonal unit vectors is an orthonormal set.

The set $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ is orthonormal if and only if

$$ \langle \mathbf{u}_i,\mathbf{u}_j \rangle = \delta_{ij} $$

where

$$ \delta_{ij} = \left\{\begin{matrix} 1, & \quad i= j \\ 0, & \quad i\neq j \end{matrix}\right. $$

is the Kronecker delta.

In addition to being orthogonal, we would like our bases to be composed of unit vectors whenever possible. Any orthogonal set $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\right\}$ may be transformed into an orthonormal set dividing each $\mathbf{v}_j$ by its norm

$$ \mathbf{u}_j = \frac{\mathbf{v}_j}{\left\|\mathbf{v}_j\right\|} $$

The new set $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ is an orthonormal set.

Example 2

In Example 1, we showed the set was orthogonal, but can plainly see that the vectors

$$ \mathbf{v}_1 = \begin{bmatrix}\ \ 1 \\ -1 \\ \ \ 1 \end{bmatrix} \qquad\qquad \mathbf{v}_2 = \begin{bmatrix}\ \ 1 \\ \ \ 1 \\ \ \ 0 \end{bmatrix} \qquad\qquad \mathbf{v}_3 = \begin{bmatrix} -1 \\ \ \ 1 \\ \ \ 2 \end{bmatrix} $$

are not unit vectors. Dividing each vector by its length yields

$$ \mathbf{u}_1 = \begin{bmatrix}\ \ \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \ \ \frac{1}{\sqrt{3}} \end{bmatrix} \qquad\qquad \mathbf{u}_2 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} \qquad\qquad \mathbf{u}_3 = \begin{bmatrix} -\frac{1}{\sqrt{6}} \\ \ \ \frac{1}{\sqrt{6}} \\ \ \ \frac{2}{\sqrt{6}} \end{bmatrix} $$

an orthonormal set.

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5.3.3 Orthonormal Bases

Definition

In an inner product space $V$, if $\mathfrak{B} = \left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ is an orthonormal set, then we say that $\mathfrak{B}$ is an orthonormal basis for subspace $S = \text{span}\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n \right\}$.

The primary benefits of using orthonormal bases are

• finding the coordinates of vector $\mathbf{v}$ with respect to an orthonormal basis is simpler
• projections of $\mathbf{v}$ onto subsets spanned by an orthonormal set are immediate
• inner product and norm may be computed directly from the coordinates of a vector

As such, we tend to do linear algebra using orthonormal bases. The following example features a particularly useful orthonormal basis, one whose application we will discuss later in the section.

Example 3

$\left\{\,\ihat,\,\jhat,\,\khat\,\right\}$ is an orthonormal basis for Euclidean inner product space $\mathbb{R}^3$.

Example 4

The canonical basis $\left\{\,\mathbf{e}_1,\,\mathbf{e}_2,\,\dots,\,\mathbf{e}_n\,\right\}$ is an orthonormal basis for Euclidean inner product space $\mathbb{R}^n$.

Example 5

Consider $C[-\pi,\pi]$ with the inner product

$$ \langle f,g \rangle = \frac{1}{\pi}\int_{-\pi}^\pi f(x)g(x)\, dx $$
and the set $\left\{1,\cos x,\sin x,\cos 2x\,\sin 2x,\ldots,\cos nx,\sin nx\right\}$. This set is orthogonal since for any positive integers $j$ and $k$

$$ \begin{align*} \langle 1,\cos jx \rangle &= \frac{1}{\pi}\int_{-\pi}^\pi \cos jx\, dx = 0 \\ \\ \langle 1,\sin kx \rangle &= \frac{1}{\pi}\int_{-\pi}^\pi \sin kx\, dx = 0 \\ \\ \langle \cos jx,\cos kx \rangle &= \frac{1}{\pi}\int_{-\pi}^\pi \cos jx \cos kx\, dx = 0 \qquad j\neq k \\ \\ \langle \cos jx,\sin kx \rangle &= \frac{1}{\pi}\int_{-\pi}^\pi \cos jx \sin kx\, dx = 0 \\ \\ \langle \sin jx,\sin kx \rangle &= \frac{1}{\pi}\int_{-\pi}^\pi \sin jx \sin kx\, dx = 0 \qquad j\neq k \end{align*} $$
The $\cos jx$ and $\sin kx$ terms for positive integers $j$ and $k$ are already units vectors because

$$ \begin{align*} \langle \cos jx,\cos jx \rangle &= \frac{1}{\pi}\int_{-\pi}^\pi \cos^2 jx \, dx = 1 \\ \\ \langle \sin kx,\sin kx \rangle &= \frac{1}{\pi}\int_{-\pi}^\pi \sin^2 kx \, dx = 1 \end{align*} $$
but $1$ is not, as

$$ \| 1 \|^2 = \langle 1,1 \rangle = \frac{1}{\pi}\int_{-\pi}^\pi 1 \, dx = 2 $$
We divide $1$ by its length to form an orthonormal set

$$ \left\{\frac{1}{\sqrt{2}},\cos x,\sin x,\cos 2x\,\sin 2x,\ldots,\cos nx,\sin nx\right\} $$
in $C[-\pi,\pi]$.

(These integrals may be verified by using the product-to-sum trigonometric identities.)

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5.3.4 Orthonormal Bases and Coordinates

Always choose an orthonormal basis for your coordinate system. This will make computations far easier.

Theorem 5.3.2

The Inner Product of a Vector onto an Orthonormal Basis Vector is its Coordinate

Let $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ be an orthonormal basis for an inner product space $V$. If

$$ \mathbf{v} = \sum_{i=1}^n c_i\mathbf{u}_i = c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + \ldots + c_n\mathbf{u}_n $$

then

$$ c_i = \langle \mathbf{v},\mathbf{u}_i \rangle $$

Proof:

$$ \langle \mathbf{v},\mathbf{u}_i \rangle = \left\langle \sum_{i=1}^n c_i\mathbf{u}_i, \mathbf{u}_i \right\rangle = \sum_{j=1}^n c_j\left\langle \mathbf{u}_j,\mathbf{u}_i \right\rangle = \sum_{j=1}^n c_j\delta_{ij} = c_i $$
$\tombstone$

This means that the coordinates of a vector in the coordinate system defined by an orthonormal basis is exactly the inner product of the vector with each basis vector.

Example 6

Consider the basis $F = \left\{\,\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},\ \begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix},\ \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\,\right\}$ in $\mathbb{R}^3$ and vector $\mathbf{x}\in\mathbb{R}^3$ given in standard coordinates by $\mathbf{x} = \begin{bmatrix} 0 \\ 4 \\ 7 \end{bmatrix}$. What are the coordinates of vector $\mathbf{x}$ in the $F$ coordinate system?

$$ \mathbf{x} = 2\,\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + 1\,\begin{bmatrix} -2 \\ \ \ 1 \\ \ \ 3 \end{bmatrix} + 1\,\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}_F $$

What are the inner products of vector $\mathbf{x}$ with the basis vectors?

$$ \begin{align*} \left\langle\mathbf{x},\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\right\rangle &= 0(1) + 4(1) + 7(1) = 11 \\ \\ \left\langle\mathbf{x},\begin{bmatrix} -2 \\ \ \ 1 \\ \ \ 3 \end{bmatrix}\right\rangle &= 0(-2) + 4(1) + 7(3) = 25 \\ \\ \left\langle\mathbf{x},\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\right\rangle &= 0(0) + 4(1) + 7(2) = 18 \\ \end{align*} $$

Note:

I compute the inner product using standard coordinates. This is very important.

What are the scalar components of vector $\mathbf{x}$ with the basis vectors?

$$ \begin{align*} \text{Comp}_{\mathbf{v}_1}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_1\rangle}{\|\mathbf{v}_1\|} = \dfrac{11}{\sqrt{65}} \\ \\ \text{Comp}_{\mathbf{v}_2}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_2\rangle}{\|\mathbf{v}_2\|} = \dfrac{25}{\sqrt{65}} \\ \\ \text{Comp}_{\mathbf{v}_3}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_3\rangle}{\|\mathbf{v}_3\|} = \dfrac{18}{\sqrt{65}} \\ \end{align*} $$

What are the vector projections of vector $\mathbf{x}$ onto each basis vector?

$$ \begin{align*} \text{Proj}_{\mathbf{v_1}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_1\rangle}{\langle\mathbf{v},\mathbf{v}_1\rangle}\mathbf{v}_1 = \dfrac{11}{65}\mathbf{v}_1 \\ \\ \text{Proj}_{\mathbf{v_2}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_2\rangle}{\langle\mathbf{v},\mathbf{v}_2\rangle}\mathbf{v}_2 = \dfrac{25}{65}\mathbf{v}_1 \\ \\ \text{Proj}_{\mathbf{v_3}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_\rangle}{\langle\mathbf{v},\mathbf{v}_3\rangle}\mathbf{v}_3 = \dfrac{18}{65}\mathbf{v}_1 \\ \end{align*} $$

Example 7

Consider again the basis given in Example 1 and the vector $\mathbf{x}\in\mathbb{R}^3$ given in standard coordinates by $\mathbf{x} = \begin{bmatrix} 0 \\ 4 \\ 7 \end{bmatrix}$. What are the coordinates of vector $\mathbf{x}$ in the $V = \left\{\mathbf{v}_1,\,\mathbf{v}_2,\,\mathbf{v}_3\right\}$ coordinate system?

$$ \mathbf{x} = 1\begin{bmatrix}\ \ 1 \\ -1 \\ \ \ 1 \end{bmatrix} + 2\begin{bmatrix}\ \ 1 \\ \ \ 1 \\ \ \ 0 \end{bmatrix} + 3\begin{bmatrix} -1 \\ \ \ 1 \\ \ \ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}_V $$

What are the inner products of vector $\mathbf{x}$ with the basis vectors?

$$ \begin{align*} \left\langle\mathbf{x},\mathbf{v}_1\right\rangle &= 0(1) + 4(-1) + 7(1) = 3 \\ \\ \left\langle\mathbf{x},\mathbf{v}_2\right\rangle &= 0(1) + 4(1) + 7(0) = 4 \\ \\ \left\langle\mathbf{x},\mathbf{v}_3\right\rangle &= 0(-1) + 4(1) + 7(2) = 18 \\ \end{align*} $$

What are the scalar components of vector $\mathbf{x}$ with the basis vectors?

$$ \begin{align*} \text{Comp}_{\mathbf{v}_1}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_1\rangle}{\|\mathbf{v}_1\|} = \dfrac{3}{\sqrt{3}} = \sqrt{3} \\ \\ \text{Comp}_{\mathbf{v}_2}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_2\rangle}{\|\mathbf{v}_2\|} = \dfrac{4}{\sqrt{2}} = \sqrt{2} \\ \\ \text{Comp}_{\mathbf{v}_3}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_3\rangle}{\|\mathbf{v}_3\|} = \dfrac{18}{\sqrt{6}} = 3\sqrt{6} \\ \end{align*} $$

What are the vector projections of vector $\mathbf{x}$ onto each basis vector?

$$ \begin{align*} \text{Proj}_{\mathbf{v_1}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_1\rangle}{\langle\mathbf{v},\mathbf{v}_1\rangle}\mathbf{v}_1 = \dfrac{3}{3}\mathbf{v}_1 = 1\,\mathbf{v}_1 \\ \\ \text{Proj}_{\mathbf{v_2}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_2\rangle}{\langle\mathbf{v},\mathbf{v}_2\rangle}\mathbf{v}_2 = \dfrac{4}{2}\mathbf{v}_1 = 2\,\mathbf{v}_2 \\ \\ \text{Proj}_{\mathbf{v_3}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_\rangle}{\langle\mathbf{v},\mathbf{v}_3\rangle}\mathbf{v}_3 = \dfrac{18}{6}\mathbf{v}_1 = 3\,\mathbf{v}_3 \\ \end{align*} $$

Notice that with an orthogonal basis, the vector projections produce the linear combination of basis vectors that yields the coordinates of a vector in the coordinate system. This happens because orthogonal vectors

• have no common projection onto each other. (do not overlap each other)

Example 8

Finally consider the basis given in Example 2 and the vector $\mathbf{x}\in\mathbb{R}^3$ given in standard coordinates by $\mathbf{x} = \begin{bmatrix} 0 \\ 4 \\ 7 \end{bmatrix}$. What are the coordinates of vector $\mathbf{x}$ in the $U = \left\{\mathbf{u}_1,\,\mathbf{u}_2,\,\mathbf{u}_3\right\}$ coordinate system?

$$ \mathbf{x} = \sqrt{3}\,\begin{bmatrix}\ \ \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \ \ \frac{1}{\sqrt{3}} \end{bmatrix} + 2\sqrt{2}\,\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} + 3\sqrt{6}\,\begin{bmatrix} -\frac{1}{\sqrt{6}} \\ \ \ \frac{1}{\sqrt{6}} \\ \ \ \frac{2}{\sqrt{6}} \end{bmatrix} = \begin{bmatrix} \sqrt{3} \\ 2\sqrt{2} \\ 3\sqrt{6} \end{bmatrix}_U $$

What are the inner products of vector $\mathbf{x}$ with the basis vectors?

$$ \begin{align*} \left\langle\mathbf{x},\mathbf{u}_1\right\rangle &= 0\left(\frac{1}{\sqrt{3}}\right) + 4\left(-\frac{1}{\sqrt{3}}\right) + 7\left(\frac{1}{\sqrt{3}}\right) = \sqrt{3} \\ \\ \left\langle\mathbf{x},\mathbf{u}_2\right\rangle &= 0\left(\frac{1}{\sqrt{2}}\right) + 4\left(\frac{1}{\sqrt{2}}\right) + 7\left(0\right) = 2\sqrt{2} \\ \\ \left\langle\mathbf{x},\mathbf{u}_3\right\rangle &= 0\left(-\frac{1}{\sqrt{6}}\right) + 4\left(\frac{1}{\sqrt{6}}\right) + 7\left(\frac{2}{\sqrt{6}}\right) = 3\sqrt{6} \\ \end{align*} $$

What are the scalar components of vector $\mathbf{x}$ with the basis vectors?

$$ \begin{align*} \text{Comp}_{\mathbf{u}_1}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{u}_1\rangle}{\|\mathbf{u}_1\|} = \dfrac{\sqrt{3}}{1} = \sqrt{3} \\ \\ \text{Comp}_{\mathbf{u}_2}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{u}_2\rangle}{\|\mathbf{u}_2\|} = \dfrac{2\sqrt{2}}{1} = 2\sqrt{2} \\ \\ \text{Comp}_{\mathbf{u}_3}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{u}_3\rangle}{\|\mathbf{u}_3\|} = \dfrac{3\sqrt{6}}{1} = 3\sqrt{6} \\ \end{align*} $$

What are the vector projections of vector $\mathbf{x}$ onto each basis vector?

$$ \begin{align*} \text{Proj}_{\mathbf{u_1}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{u}_1\rangle}{\langle\mathbf{u},\mathbf{u}_1\rangle}\mathbf{u}_1 = \dfrac{\sqrt{3}}{1}\mathbf{u}_1 = \sqrt{3}\,\mathbf{u}_1 \\ \\ \text{Proj}_{\mathbf{u_2}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{v}_2\rangle}{\langle\mathbf{u},\mathbf{v}_2\rangle}\mathbf{u}_2 = \dfrac{2\sqrt{2}}{1}\mathbf{u}_1 = 2\sqrt{2}\,\mathbf{u}_2 \\ \\ \text{Proj}_{\mathbf{u_3}}\mathbf{x} &= \dfrac{\langle\mathbf{x},\mathbf{u}_\rangle}{\langle\mathbf{u},\mathbf{u}_3\rangle}\mathbf{u}_3 = \dfrac{3\sqrt{6}}{1}\mathbf{u}_1 = 3\sqrt{6}\,\mathbf{u}_3 \\ \end{align*} $$

Notice that with an orthonormal basis and a vector $\mathbf{x}$ in the inner product space and each basis vector $\mathbf{u}_j$, the following are equal

• The inner product of a vector $\mathbf{x}$ and $\mathbf{u}_j$
• The scalar component a vector $\mathbf{x}$ and $\mathbf{u}_j$, and
• The coordinate of a vector $\mathbf{x}$ with respect $\mathbf{u}_j$.

That is, for a vector $\mathbf{x}$ in our inner product space and for each basis vector $\mathbf{u}_j$

$$ \langle\mathbf{x},\,\mathbf{u}_j\rangle = \text{Comp}_{\mathbf{u}_j}\mathbf{x} = \text{the coordinate of vector }\mathbf{x}\text{ for basis vector }\mathbf{u}_j $$

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5.3.5 Invariance of the Inner Product

We have already seen that if we choose a different orientation (basis or coordinate system), then the coordinates (list of numbers) that describes each vector are also different.

However the value of the inner product of two vectors in an inner product space does not change. When one defines an inner product on a vector space, the value of the inner product does NOT depend on the choice of orientation. Consider Euclidean inner product space $\mathbb{R}^n$ endowed with the Euclidean dot product. The definition of dot product relies on the familiar geometry of projections. The value of the inner product of two vectors $\mathbf{u},\ \mathbf{v}\in\mathbb{R}^n$ can be computed

$$ \mathbf{u}\cdot\mathbf{v} = \left\|\mathbf{u}\right\|\,\left\|\mathbf{v}\right\|\,\cos(\theta) $$

This formula makes no mention of the coordinates of the vectors $\mathbf{u}$ and $\mathbf{v}$.

I mentioned earlier in this section, that many of the examples of inner products were defined using the canonical basis, or standard coordinates. In Example 6 one does not use the $F$-coordinates to compute the inner product of vector $\mathbf{x}$ and one of the basis vectors. One uses the canonical basis $\left\{\,\ihat,\,\jhat,\,\khat\,\right\}$.

Can one compute the correct value of the inner product of two vectors using their non-standard representation?

Usually one must convert the coordinates of the vector to standard coordinates first. To compute the inner product $\langle\cdot,\cdot\rangle_F$ in $F$-coordinates

$$ \left\langle\left[\mathbf{x}\right]_F,\,\left[\mathbf{y}\right]_F\right\rangle_F = \left\langle F\left[\mathbf{x}\right]_F,\,F\left[\mathbf{y}\right]_F\right\rangle = \langle\mathbf{x},\,\mathbf{y}\rangle $$

However if one chooses an orthonormal basis, computation is much simpler.

Corollary 5.3.3

The Inner Product of Coordinate Vectors with respect to an Orthonormal Basis is the Sum of the Component-wise Product

Let $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ be an orthonormal basis for a finite dimensional inner product space $V$. If

$$ \mathbf{x} = \sum_{i=1}^n x_i\mathbf{u}_i \qquad\qquad \mathbf{y} = \sum_{i=1}^n y_i\mathbf{u}_i $$

then

$$ \langle \mathbf{x},\mathbf{y} \rangle = \sum_{i=1}^n x_i y_i $$

Proof:

From Theorem 5.5.2,


$$ \begin{align*} \langle \mathbf{x},\mathbf{u}_i \rangle = x_i \qquad i=1,\dots,n \\ \langle \mathbf{y},\mathbf{u}_i \rangle = y_i \qquad i=1,\dots,n \\ \end{align*} $$
so using the properties of inner product


$$ \langle \mathbf{x},\mathbf{y} \rangle = \left\langle \sum_{i=1}^n x_i\mathbf{u}_i, \mathbf{y} \right\rangle = \sum_{i=1}^n x_i \left\langle\mathbf{u}_i, \mathbf{y} \right\rangle = \sum_{i=1}^n x_i \left\langle\mathbf{y}, \mathbf{u}_i \right\rangle = \sum_{i=1}^n x_i y_i $$
$\tombstone$

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5.3.6 Parseval's Identity

Corollary 5.3.4

Parseval's Identity

If $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ is an orthonormal basis for an inner product space $V$ and

$$ \mathbf{x} = \sum_{i=1}^n x_i\mathbf{u}_i $$

then

$$ \|\mathbf{x}\|^2 = \langle \mathbf{x},\mathbf{x} \rangle = \sum_{i=1}^n x_i^2 $$

Proof:

For $\mathbf{x} = \sum_{i=1}^n x_i\mathbf{u}_i$, we have from Corollary 5.5.3

$$ \|\mathbf{x}\|^2 = \langle \mathbf{x},\mathbf{x} \rangle = \sum_{i=1}^n x_i^2 $$ $\tombstone$

Exercise 1

Consider again orthonormal basis $\left\{\frac{1}{\sqrt{2}},\cos x,\sin x,\cos 2x\,\sin 2x,\ldots,\cos nx,\sin nx\right\}$ in $C[-\pi,\pi]$ and the finite dimensional subspace

$$ V = \text{Span}\left\{\frac{1}{\sqrt{2}},\cos x,\sin x,\cos 2x\,\sin 2x,\ldots,\cos nx,\sin nx\right\} $$

Compute the following:

1. $ \|\sin^2 x\|^2 $
2. $ \|2\sin 3x + \sin^2 x\|^2$
3. $ \|2 + \cos^4 x\|^2 $

Check Your Work

$$ \begin{align*} \|\sin^2 x\|^2 &= \frac{3}{4} \\ \\ \|2\sin 3x + \sin^2 x\|^2 &= \frac{19}{4} \\ \\ \|2 + \cos^4 x\|^2 &= \frac{739}{64} \end{align*} $$

Follow Along

We could integrate these directly, but that would not take advantage of our new powerful tool Parseval's identity. Instead, we shall employ trigonometric identities to write each of these functions in terms of our orthonormal basis and then apply Parseval's identity to find the (square of the) norm.

$$ \begin{align*} \|\sin^2 x\|^2 &= \left\|\frac{1}{2}\left(1-\cos 2x\right)\right\|^2 \\ \\ &= \left\|\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} - \frac{1}{2}\cos 2x\right\|^2 \\ \\ &= \left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{3}{4} \\ \\ \|2\sin 3x + \sin^2 x\|^2 &= \left\|\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} - \frac{1}{2}\cos 2x + 2\sin 3x\right\|^2 \\ \\ &= \left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{2}\right)^2 + 2^2 = \frac{19}{4} \\ \\ \|2 + \cos^4 x\|^2 &= \left\| 2 + \left[\frac{1}{2}\left(1+\cos 2x\right)\right]^2\right\|^2 \\ \\ &= \left\| 2 + \frac{1}{4}\left(1+2\cos 2x + \cos^2 2x \right)\right\|^2 \\ \\ &= \left\| \frac{9}{4} + \frac{1}{2}\cos 2x + \frac{1}{8}\left(1+\cos 4x \right)\right\|^2 \\ \\ &= \left\| \frac{19}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right\|^2 \\ \\ &= \left\| \frac{19\sqrt{2}}{8}\cdot\frac{1}{\sqrt{2}} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right\|^2 \\ \\ &= \left(\frac{19\sqrt{2}}{8}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{8}\right)^2 = \frac{739}{64} \end{align*} $$

Example 9

Compute the following integrals using Parseval's identity.

$$ \begin{align*} \displaystyle\int_{-\pi}^{\pi}\sin^4(x)\,dx &= \dfrac{\pi}{\pi}\displaystyle\int_{-\pi}^{\pi}\sin^2(x)\sin^2(x)\,dx \\ \\ &= \pi\,\left\|\sin^2(x)\right\| = \dfrac{3\pi}{4} \end{align*} $$

$$ \begin{align*} \displaystyle\int_{-\pi}^{\pi}\left(2\sin 3x + \sin^2 x\right)^2\, dx &= \dfrac{\pi}{\pi}\displaystyle\int_{-\pi}^{\pi}\left(2\sin 3x + \sin^2 x\right)^2\, dx \\ \\ &= \pi\,\left\| 2\sin 3x + \sin^2 x \right\|^2 = \frac{19\pi}{4} \end{align*} $$

$$ \begin{align*} \displaystyle\int_{-\pi}^{\pi}\left(2 + \cos^4 x\right)^2\, dx &= \dfrac{\pi}{\pi}\displaystyle\int_{-\pi}^{\pi}\left(2 + \cos^4 x\right)^2\, dx \\ \\ &= \pi\,\left\| 2 + \cos^4 x \right\|^2 = \frac{739\pi}{64} \end{align*} $$

Exercise 2

Compute the following integral using subspace $V$

$$ \displaystyle\int_0^{\pi} \cos^3(t)\,dt $$

Check Your Work

$$ \displaystyle\int_0^{\pi} \cos^3(t)\,dt = 0 $$

Follow Along

The function $\cos^3(t)$ is an *even* function, that is $\cos^3(-t) = \cos^3(t)$, so its graph is symmetric with respect to the $y$-axis. Thus the area on the left and right of the $y$-axis are mirror images. Thus $$ \begin{align*} \displaystyle\int_0^{\pi} \cos^3(t)\,dt &= \dfrac{1}{2}\displaystyle\int_{-\pi}^{\pi} \cos^3(t)\,dt \\ \\ &= \dfrac{1}{2}\displaystyle\int_{-\pi}^{\pi} \cos(t)\cos^2(t)\,dt \\ \\ &= \dfrac{1}{2}\displaystyle\int_{-\pi}^{\pi} \cos(t)\dfrac{1 + \cos(2t)}{2}\,dt \\ \\ &= \dfrac{1}{4}\displaystyle\int_{-\pi}^{\pi} \left(\cos(t) + \cos(t)\cos(2t)\right)\,dt \\ \\ &= \dfrac{1}{4}\displaystyle\int_{-\pi}^{\pi} \cos(t)\,dt + \dfrac{1}{4}\displaystyle\int_{-\pi}^{\pi} \cos(t)\cos(2t)\,dt \\ \\ &= \dfrac{\pi}{4}\,\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} 1\cdot\cos(t)\,dt + \dfrac{\pi}{4}\,\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} \cos(t)\cos(2t)\,dt \\ \\ &= \dfrac{\pi}{4}\,\left\langle 1,\,\cos(t)\,\right\rangle + \dfrac{\pi}{4}\,\left\langle\, \cos(t),\,\cos(2t)\,\right\rangle = 0 \end{align*} $$

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5.3.7 Orthogonal Matrices

If the columns of a square matrix $A\in\mathbb{R}^{n\times n}$ forms an orthonormal set, then columns must be an orthonormal basis for $\mathbb{R}^n$. Why?

Check Your Answer

Because the columns of $A$ are a set of $n$ linearly independent vectors in $\mathbb{R}^n$ so they must be a basis. As these vectors are orthonormal, they form an orthonormal basis for $\mathbb{R}^n$, the domain and the codomain of the linear transformation represented by $A$.

Definition of Orthogonal Matrix

A matrix $Q\in\mathbb{R}^{n\times n}$ is an orthogonal matrix if and only if its columns are an orthonormal set in $\mathbb{R}^n$.

Clearly, this means that such a matrix would be the transition matrix from the column basis, a basis for the column space $C(A)$ of matrix $A$

$$ \left\{\mathbf{q}_i,\mathbf{q}_2,\ldots,\mathbf{q}_n\right\} $$
to the standard (canonical) basis

$$ \left\{\mathbf{e}_i,\mathbf{e}_2,\ldots,\mathbf{e}_n\right\}. $$
The transition matrix $Q^{-1}$ from $\left\{\mathbf{e}_i,\mathbf{e}_2,\ldots,\mathbf{e}_n\right\}$ to $\left\{\mathbf{q}_i,\mathbf{q}_2,\ldots,\mathbf{q}_n\right\}$ has a special form for orthogonal matrices.

Theorem 5.3.5

The Transpose of an Orthogonal Matrix is its Inverse

Matrix $Q\in\mathbb{R}^{n\times n}$ is an orthogonal matrix if and only if all of the following are also true.

(i) the columns of $Q$ are an orthonormal basis for $\mathbb{R}^n$

(ii) $ Q^T Q = I $

(iii) $ Q^T = Q^{-1} $

(iv) $ \left\langle Q\mathbf{x},Q\mathbf{y} \right\rangle = \langle \mathbf{x},\mathbf{y}\rangle$

(v) $\left\| Q\mathbf{x} \right\| = \left\| \mathbf{x} \right\|$

(vi) $\det(Q) = \pm 1$

Proof:

(i) The matrix $Q\in\mathbb{R}^n$ is an orthogonal matrix if and only if the columns of matrix $Q$ form an orthonormal set of $n$ vectors. Therefore $Q$ is an orthogonal matrix if and only if the columns of $Q$ form an orthonormal basis for $\mathbb{R}^n$.

(ii) $\implies$ If $Q$ is an orthogonal square matrix and $C=Q^TQ$, then by definition each element of matrix $C$, is the product of the $i^\text{th}$ row of $Q^T$ and $j^\text{th}$ column of $Q$

$$ c_{ij} = \mathbf{q}_i^T\mathbf{q}_j = \delta_{ij} $$

Hence

$$ C = \begin{bmatrix} c_{ij} \end{bmatrix} = \begin{bmatrix} \delta_{ij} \end{bmatrix} = I. $$

That is $Q^T Q = I$. Additionally, inverse matrices are unique, so $Q^T = Q^{-1}$.

$\impliedby$ If $Q^TQ = I$, then

$$ \begin{bmatrix} \delta_{ij} \end{bmatrix} = I = Q^TQ = \begin{bmatrix} \mathbf{q}_i^T\mathbf{q}_j \end{bmatrix} $$

This says that

$$ \langle\mathbf{q}_i,\mathbf{q}_j\rangle = \mathbf{q}_i^T\mathbf{q}_j = \left\{\begin{array}{lcr} 1 &\ & \text{if $i=j$} \\ 0 & \ & \text{if $i\neq j$} \end{array}\right. $$

Thus the column vectors $\left\{\mathbf{q}_1,\dots,\mathbf{q}_n\right\}$ form an orthonormal set. That is matrix $Q$ is an orthogonal matrix.

(iii) Matrix $Q\in\mathbb{R}^{n\times n}$ is a square orthogonal matrix if and only if $QQ^T=I$. Thus $Q$ is an orthogonal matrix if and only if $Q$ is nonsingular and $Q^{-1}=Q^T$.

(iv) $\implies$ If $Q\in\mathbb{R}^{n\times n}$ is an orthogonal matrix, and $\mathbf{x},\mathbf{y}\in\mathbb{R}^n$, then

$$ \left\langle Q\mathbf{x},Q\mathbf{y} \right\rangle = \left(Q\mathbf{x}\right)^T Q\mathbf{y} = \mathbf{x}^T Q^T Q\mathbf{y} = \mathbf{x}^T I \mathbf{y} = \mathbf{x}^T \mathbf{y} = \langle \mathbf{x},\mathbf{y} \rangle $$

$\impliedby$ If $Q\in\mathbb{R}^n$ and for every vectors $\mathbf{x}, \mathbf{y}\in\mathbb{R}^n$ one has that $\left\langle Q\mathbf{x},Q\mathbf{y} \right\rangle = \langle \mathbf{x},\mathbf{y} \rangle$, then consider the canonical basis for $\mathbb{R}^n$, $\left\{\mathbf{e}_1, \dots, \mathbf{e}_n \right\}$. For every $1 \le i,j \le n$ one obtains for the columns of matrix $Q$,

$$ \left\langle\mathbf{q}_i,\mathbf{q}_j\right\rangle = \left\langle Q\mathbf{e}_i, Q\mathbf{e}_j \right\rangle = \left\langle\mathbf{e}_i,\mathbf{e}_j\right\rangle = \delta_{ij} $$

Thus matrix $Q$ is an orthogonal matrix.

(v) If $Q\in\mathbb{R}^{n\times n}$ is an orthogonal matrix, then by property (iv)

$$ \left\| Q\mathbf{x} \right\|^2 = \left\langle Q\mathbf{x},Q\mathbf{x} \right\rangle = \left\langle \mathbf{x},\mathbf{x} \right\rangle = \left\| \mathbf{x} \right\|^2 $$

(vi) If $Q\in\mathbb{R}^{n\times n}$ is an orthogonal matrix, then

$$ 1 = \det\left(I_n\right) = \det\left(Q^TQ\right) = \det(Q^T)\det(Q) = \det(Q)^2 $$ $\tombstone$

We say that orthogonal matrices preserves the inner product since

$$ \left\langle Q\mathbf{x},Q\mathbf{y} \right\rangle = \langle \mathbf{x},\mathbf{y}\rangle $$

We say that orthogonal matrices preserve the norm since the linear transformation represented by matrix $Q$ maps vectors in $\mathbb{R}^n$ to vectors in $\mathbb{R}^n$ with the same norm.

$$ \left\| Q\mathbf{x} \right\| = \left\| \mathbf{x} \right\| $$

Since the angle $\theta$ between two vectors $\mathbf{x},\mathbf{y}\in\mathbb{R}^n$ is the same as the angle between $Q\mathbf{x}$ and $Q\mathbf{y}$

$$ \cos(\theta) = \dfrac{\left\langle \mathbf{x},\mathbf{y} \right\rangle}{\left\|\mathbf{x}\right\|\,\left\|\mathbf{y}\right\|} = \dfrac{\left\langle Q\mathbf{x},Q\mathbf{y} \right\rangle}{\left\|Q\mathbf{x}\right\|\,\left\|Q\mathbf{y}\right\|}, $$

we say that orthogonal matrices preserve angles.

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5.3.8 Reflections

Several linear operators on $\mathbb{R}^n$ with clear geometric interpretations are expressed as orthogonal matrices. These include

• reflections
• rotations
• permutations

Definition

A reflection is a linear operator that fixes an $n-1$ dimensional (hyper)plane through the origin and "flips" or "mirrors" all vectors on either side of that (hyper)plane. In $\mathbb{R}^2$ the hyperplane of reflection is a line and in $\mathbb{R}^3$ it is a two-dimensional plane. A plane in $\mathbb{R}^3$ may be defined by its normal vector, a vector that is orthogonal to each vector in the plane. Using this normal vector, we can construct the matrix representing the linear operator for a reflection across this plane.

Example 10

Definition

The matrix representing the linear operator $R\,:\,\mathbb{R}^n\rightarrow\mathbb{R}^n$ defined by

$$ R := I - 2\mathbf{n}\mathbf{n}^T $$

for a unit vector $\mathbf{n}\in\mathbb{R}^n$ is called a Householder transformation or Householder Reflection.

Suppose we have a unit normal vector to a plane.

$$ \mathbf{n} = \left[\begin{array}{r} -\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array}\right] $$

Then we can define the hyperplane $H$ as the graph of the equation

$$ \mathbf{n}\cdot\mathbf{x} = 0 $$

Now $\mathbf{n}^T\mathbf{n} = \langle\mathbf{n},\mathbf{n}\rangle = \left\|\mathbf{n}\right\|^2 = 1$ because we chose vector $\mathbf{n}$ to be a unit vector. Thus $\left\{ \mathbf{n} \right\}$ is an orthonormal set, and $P = \mathbf{n}\mathbf{n}^T$ is the projection of $\mathbb{R}^{n\times n}$ onto the subspace $S = \text{Span}\left\{\mathbf{n}\right\}$. Furthermore $I_n - P = I_n - \mathbf{n}\mathbf{n}^T$ is the projection of $\mathbb{R}^n$ onto the orthogonal complement of $S$. $H = S^{\perp}$, the hyperplane, and $S$ is the normal line to the hyperplane $H$.

We can also define a Householder reflection $R\,:\,\mathbb{R}^n\rightarrow\mathbb{R}^n$ by

$$ R = I - 2\mathbf{n}\mathbf{n}^T = \left[\begin{array}{rrr} \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{array} \right] $$

The image below shows the effect of this reflection on a polygon $p$ defined by the points $(1,1.5,0.5),(0.5,1.5,1.5),(-1,2,0.5)$ plotted in $\mathbb{R}^3$, and reflected about the plane $H$, with the reflection denoted as $Rp$.

Figure 1

The shapes do not appear to be the same size due to the perspective of the image, but represent the same shape reflected about the pictured plane. One can view this reflection in a GeoGebra Applet, Reflection Across a Plane.

We know that a Householder transformation is an orthogonal transformation because

$$ \begin{align*} RR^T &= \left(I - 2\mathbf{n}\mathbf{n}^T\right)\left(I - 2\mathbf{n}\mathbf{n}^T\right)^T \\ \\ &= \left(I - 2\mathbf{n}\mathbf{n}^T\right)\left(I^T - \left(2\mathbf{n}\mathbf{n}^T\right)^T\right) \\ \\ &= \left(I - 2\mathbf{n}\mathbf{n}^T\right)\left(I - 2\mathbf{n}\mathbf{n}^T\right) \\ \\ &= I - 2\mathbf{n}\mathbf{n}^T - 2\mathbf{n}\mathbf{n}^T + 4\left(\mathbf{n}\mathbf{n}^T\right)\left(\mathbf{n}\mathbf{n}^T\right) \\ \\ &= I - 4\mathbf{n}\mathbf{n}^T + 4\mathbf{n}\left(\mathbf{n}^T\mathbf{n}\right)\mathbf{n}^T \\ \\ &= I - 4\mathbf{n}\mathbf{n}^T + 4\mathbf{n}(1)\mathbf{n}^T = I \\ \end{align*} $$

As the $RR^T=I$, the matrix $R$ is invertible and $R^{-1}=R^T$, so matrix $R$ is an orthogonal matrix. $\tombstone$

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5.3.9 Rotations

Definition

A rotation is a linear operator that fixes a line in space and transforms all other vectors invariantly about that line by rotating all vectors around the line. This means that although vectors are moved from one place in the coordinate system to another,

• a rotation preserves the norm
• a rotation preserves angles

This tells us that a rotation is an orthogonal linear transformation and its matrix representation is an orthogonal matrix.

Example 11

Create the standard matrix representation of a rotation in $\mathbb{R}^2$ counterclockwise $\frac{\pi}{3}$ radian.

Figure 2

In order to create the standard matrix representation $R_{\pi/3}$ one needs only know where $\ihat$ and $\jhat$ are mapped by the rotation. On the unit circle

$$ \ihat\longrightarrow\begin{bmatrix} \cos\left(\frac{\pi}{3}\right) \\ \sin\left(\frac{\pi}{3}\right) \end{bmatrix}\qquad\qquad\jhat\longrightarrow\begin{bmatrix} -\sin\left(\frac{\pi}{3}\right) \\ \cos\left(\frac{\pi}{3}\right)\end{bmatrix} $$

Hence the standard matrix representation of the rotation is given by

$$ R_{\pi/3} = \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} $$

Example 12

Create the standard matrix representation of linear transformation that rotates all vectors in $\mathbb{R}^3$ counterclockwise $\frac{\pi}{4}$ radians about the $x$-axis.

Figure 3

To determine the standard matrix representation of this rotation $R_{\pi/4}$ we need the images of our canonical vectors

$$ \ihat \longrightarrow \ihat\qquad \jhat \longrightarrow \begin{bmatrix} 0 \\ \cos\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right) \end{bmatrix}\qquad \ihat \longrightarrow \begin{bmatrix} 0 \\ -\sin\left(\frac{\pi}{4}\right) \\ \cos\left(\frac{\pi}{4}\right) \end{bmatrix} $$

Thus the standard matrix representation of our rotation about the $x$-axis is given by

$$ R_{\pi/3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} $$

Example 13

If we form a matrix $U$ using $\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3$ from Example 2, $U$ represents a rotation in $\mathbb{R}^3$. In fact, it may be written as a product of two rotations: one about the $y$-axis and one about the $z$-axis. Since $U$ is an orthogonal matrix, this decomposition may be determined using Euler angles. Employing the technique from the linked site, the rotation angles about the $x_2$- and $x_3$-axes are determined to be approximately $\theta=-0.6155$ and $\psi=-0.7854$ radians respectively. Since $U$ is the transition matrix from $\left\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\right\}$ to the standard basis the decomposition has the form

$$ \begin{align*} U &= \left[\begin{array}{rrr} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0\, \ & \frac{2}{\sqrt{6}} \end{array}\right] = \\ \\ &= \left(R_y R_z\right)^{-1} \\ \\ &= \left(\left[ \begin{array}{ccc} \cos\psi & \sin\psi & 0 \\ -\sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{ccc} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \end{array} \right]\right)^{-1} \\ \\ &= R_z^T R_y^T \end{align*} $$

since $R_y$ and $R_z$ are themselves both orthogonal matrices.

Both the original and rotated axes are shown here:

Figure 4

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5.3.10 Permutations

Definition

A permutation is an arrangement of items, an ordering. Permutation matrices are the identity matrix with a different ordering of columns. These matrices represent a reordering of the coordinates of a basis. Let $k_1,k_2,\ldots,k_n$ be an ordering of $1,2,\ldots,n$ and a permutation matrix $P$ be defined by

$$ P = \left[\ \mathbf{e}_{k_1}\ \mathbf{e}_{k_2} \ \ldots\ \mathbf{e}_{k_n}\ \right] $$

Given a matrix $A\in\mathbb{R}^{m\times n}$, right multiplication by a permutation matrix yields a reordering of the columns of $A$:

$$ AP = \left[\ \mathbf{a}_{k_1}\ \mathbf{a}_{k_2} \ \ldots\ \mathbf{a}_{k_n}\ \right] $$

Example 14

Given a matrix

$$ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8\end{bmatrix} $$

and a permutation matrix

$$ P = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} $$

the product
$$ AP = \begin{bmatrix} 2 & 1 & 4 & 3 \\ 6 & 5 & 8 & 7 \end{bmatrix} $$

reorders the columns of $A$. Likewise, the matrix

$$ Q = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$

will reorder the rows of $A$ if we left multiply by it

$$ QA = \begin{bmatrix} 5 & 6 & 7 & 8 \\ 1 & 2 & 3 & 4 \end{bmatrix} $$

In general, a permutation matrix may be formed by reordering either the rows or columns of the identity matrix. To use a permutation matrix to reorder the rows of an arbitrary matrix $A$, left multiplication will reorder the rows and right multiplication will reorder the columns.

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