In this section we define a couple more operations with vectors, and prove a few theorems. At first blush these definitions and results will not appear central to what follows, but we will make use of them at key points in the remainder of the course (such as Section MINM, Section OD). Because we have chosen to use $\complexes$ as our set of scalars, this subsection is a bit more, uh, … complex than it would be for the real numbers. We will explain as we go along how things get easier for the real numbers ${\mathbb R}\text{.}$ If you have not already, now would be a good time to review some of the basic properties of arithmetic with complex numbers described in Section CNO. With that done, we can extend the basics of complex number arithmetic to our study of vectors in $\complex{m}\text{.}$

# SubsectionCAVComplex Arithmetic and Vectors¶ permalink

We know how the addition and multiplication of complex numbers is employed in defining the operations for vectors in $\complex{m}$ (Definition CVA and Definition CVSM). We can also extend the idea of the conjugate to vectors.

##### DefinitionCCCVComplex Conjugate of a Column Vector

Suppose that $\vect{u}$ is a vector from $\complex{m}\text{.}$ Then the conjugate of the vector, $\conjugate{\vect{u}}\text{,}$ is defined by \begin{align*} \vectorentry{\conjugate{\vect{u}}}{i} &=\conjugate{\vectorentry{\vect{u}}{i}} &&1\leq i\leq m\text{.} \end{align*}

With this definition we can show that the conjugate of a column vector behaves as we would expect with regard to vector addition and scalar multiplication.

##### Proof

These two theorems together tell us how we can “push” complex conjugation through linear combinations.

##### DefinitionIPInner Product

Given the vectors $\vect{u},\,\vect{v}\in\complex{m}$ the inner product of $\vect{u}$ and $\vect{v}$ is the scalar quantity in $\complexes$ \begin{equation*} \innerproduct{\vect{u}}{\vect{v}}= \conjugate{\vectorentry{\vect{u}}{1}}\vectorentry{\vect{v}}{1}+ \conjugate{\vectorentry{\vect{u}}{2}}\vectorentry{\vect{v}}{2}+ \conjugate{\vectorentry{\vect{u}}{3}}\vectorentry{\vect{v}}{3}+ \cdots+ \conjugate{\vectorentry{\vect{u}}{m}}\vectorentry{\vect{v}}{m} = \sum_{i=1}^{m}\conjugate{\vectorentry{\vect{u}}{i}}\vectorentry{\vect{v}}{i}\text{.} \end{equation*}

This operation is a bit different in that we begin with two vectors but produce a scalar. Computing one is straightforward.

In the case where the entries of our vectors are all real numbers (as in the second part of Example CSIP), the computation of the inner product may look familiar and be known to you as a dot product or scalar product. So you can view the inner product as a generalization of the scalar product to vectors from $\complex{m}$ (rather than ${\mathbb R}^m$).

Note that we have chosen to conjugate the entries of the first vector listed in the inner product, while it is almost equally feasible to conjugate entries from the second vector instead. In particular, prior to Version 2.90, we did use the latter definition, and this has now changed to the former, with resulting adjustments propogated up through Section CB (only). However, conjugating the first vector leads to much nicer formulas for certain matrix decompositions and also shortens some proofs.

There are several quick theorems we can now prove, and they will each be useful later.

##### Proof

If treating linear algebra in a more geometric fashion, the length of a vector occurs naturally, and is what you would expect from its name. With complex numbers, we will define a similar function. Recall that if $c$ is a complex number, then $\modulus{c}$ denotes its modulus (Definition MCN).

##### DefinitionNVNorm of a Vector

The norm of the vector $\vect{u}$ is the scalar quantity in $\complexes$ \begin{equation*} \norm{\vect{u}}= \sqrt{ \modulus{\vectorentry{\vect{u}}{1}}^2+ \modulus{\vectorentry{\vect{u}}{2}}^2+ \modulus{\vectorentry{\vect{u}}{3}}^2+ \cdots+ \modulus{\vectorentry{\vect{u}}{m}}^2 } = \sqrt{\sum_{i=1}^{m}\modulus{\vectorentry{\vect{u}}{i}}^2}\text{.} \end{equation*}

Computing a norm is also easy to do.

Notice how the norm of a vector with real number entries is just the length of the vector. Inner products and norms are related by the following theorem.

##### Proof

When our vectors have entries only from the real numbers Theorem IPN says that the dot product of a vector with itself is equal to the length of the vector squared.

##### Proof

Notice that Theorem PIP contains three implications \begin{align*} \vect{u}\in\complex{m}&\Rightarrow\innerproduct{\vect{u}}{\vect{u}}\geq 0 & \vect{u}=\zerovector&\Rightarrow\innerproduct{\vect{u}}{\vect{u}}=0 & \innerproduct{\vect{u}}{\vect{u}}=0&\Rightarrow\vect{u}=\zerovector\text{.} \end{align*}

The results contained in Theorem PIP are summarized by saying “the inner product is positive definite.”

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##### SageCNIPConjugates, Norms and Inner Products
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Orthogonal is a generalization of perpendicular. You may have used mutually perpendicular vectors in a physics class, or you may recall from a calculus class that perpendicular vectors have a zero dot product. We will now extend these ideas into the realm of higher dimensions and complex scalars.

##### DefinitionOVOrthogonal Vectors

A pair of vectors, $\vect{u}$ and $\vect{v}\text{,}$ from $\complex{m}$ are orthogonal if their inner product is zero, that is, $\innerproduct{\vect{u}}{\vect{v}}=0\text{.}$

We extend this definition to whole sets by requiring vectors to be pairwise orthogonal. Despite using the same word, careful thought about what objects you are using will eliminate any source of confusion.

##### DefinitionOSVOrthogonal Set of Vectors

Suppose that $S=\set{\vectorlist{u}{n}}$ is a set of vectors from $\complex{m}\text{.}$ Then $S$ is an orthogonal set if every pair of different vectors from $S$ is orthogonal, that is $\innerproduct{\vect{u}_i}{\vect{u}_j}=0$ whenever $i\neq j\text{.}$

We now define the prototypical orthogonal set, which we will reference repeatedly.

##### DefinitionSUVStandard Unit Vectors

Let $\vect{e}_j\in\complex{m}\text{,}$ $1\leq j\leq m$ denote the column vectors defined by \begin{equation*} \vectorentry{\vect{e}_j}{i}= \begin{cases} 0&\text{if }i\neq j\\ 1&\text{if }i=j \end{cases}\text{.} \end{equation*} Then the set \begin{align*} \set{\vectorlist{e}{m}}&=\setparts{\vect{e}_j}{1\leq j\leq m} \end{align*} is the set of standard unit vectors in $\complex{m}\text{.}$

Notice that $\vect{e}_j$ is identical to column $j$ of the $m\times m$ identity matrix $I_m$ (Definition IM) and is a pivot column for $I_m\text{,}$ since the identity matrix is in reduced row-echelon form. These observations will often be useful. We will reserve the notation $\vect{e}_i$ for these vectors. It is not hard to see that the set of standard unit vectors is an orthogonal set.

So far, this section has seen lots of definitions, and lots of theorems establishing un-surprising consequences of those definitions. But here is our first theorem that suggests that inner products and orthogonal vectors have some utility. It is also one of our first illustrations of how to arrive at linear independence as the conclusion of a theorem.

##### Proof

The Gram-Schmidt Procedure is really a theorem. It says that if we begin with a linearly independent set of $p$ vectors, $S\text{,}$ then we can do a number of calculations with these vectors and produce an orthogonal set of $p$ vectors, $T\text{,}$ so that $\spn{S}=\spn{T}\text{.}$ Given the large number of computations involved, it is indeed a procedure to do all the necessary computations, and it is best employed on a computer. However, it also has value in proofs where we may on occasion wish to replace a linearly independent set by an orthogonal set.

This is our first occasion to use the technique of mathematical induction for a proof, a technique we will see again several times, especially in Chapter D. So study the simple example described in Proof Technique I first.

##### Proof

One final definition related to orthogonal vectors.

##### DefinitionONSOrthoNormal Set

Suppose $S=\set{\vectorlist{u}{n}}$ is an orthogonal set of vectors such that $\norm{\vect{u}_i}=1$ for all $1\leq i\leq n\text{.}$ Then $S$ is an orthonormal set of vectors.

Once you have an orthogonal set, it is easy to convert it to an orthonormal set — multiply each vector by the reciprocal of its norm, and the resulting vector will have norm 1. This scaling of each vector will not affect the orthogonality properties (apply Theorem IPSM).

We will see orthonormal sets again in Subsection MINM.UM. They are intimately related to unitary matrices (Definition UM) through Theorem CUMOS. Some of the utility of orthonormal sets is captured by Theorem COB in Subsection B.OBC. Orthonormal sets appear once again in Section OD where they are key in orthonormal diagonalization.

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##### 1

Is the set \begin{equation*} \set{\colvector{1\\-1\\2},\,\colvector{5\\3\\-1},\,\colvector{8\\4\\-2}} \end{equation*} an orthogonal set? Why?

##### 2

What is the distinction between an orthogonal set and an orthonormal set?

##### 3

What is nice about the output of the Gram-Schmidt process?

# SubsectionExercises

##### C20

Complete Example AOS by verifying that the four remaining inner products are zero.

##### C21

Verify that the set $T$ created in Example GSTV by the Gram-Schmidt Procedure is an orthogonal set.

##### M60

Suppose that $\set{\vect{u},\,\vect{v},\,\vect{w}}\subseteq\complex{n}$ is an orthonormal set. Prove that $\vect{u}+\vect{v}$ is not orthogonal to $\vect{v}+\vect{w}\text{.}$

##### T10

Prove part 2 of the conclusion of Theorem IPVA.

##### T11

Prove part 2 of the conclusion of Theorem IPSM.

##### T20

Suppose that $\vect{u},\,\vect{v},\,\vect{w}\in\complex{n}\text{,}$ $\alpha,\,\beta\in\complexes$ and $\vect{u}$ is orthogonal to both $\vect{v}$ and $\vect{w}\text{.}$ Prove that $\vect{u}$ is orthogonal to $\alpha\vect{v}+\beta\vect{w}\text{.}$

Solution
##### T30

Suppose that the set $S$ in the hypothesis of Theorem GSP is not just linearly independent, but is also orthogonal. Prove that the set $T$ created by the Gram-Schmidt procedure is equal to $S\text{.}$ (Note that we are getting a stronger conclusion than $\spn{T}=\spn{S}$ — the conclusion is that $T=S\text{.}$) In other words, it is pointless to apply the Gram-Schmidt procedure to a set that is already orthogonal.

##### T31

Suppose that the set $S$ is linearly independent. Apply the Gram-Schmidt procedure (Theorem GSP) twice, creating first the linearly independent set $T_1$ from $S\text{,}$ and then creating $T_2$ from $T_1\text{.}$ As a consequence of Exercise O.T30, prove that $T_1=T_2\text{.}$ In other words, it is pointless to apply the Gram-Schmidt procedure twice.