DLA Concepts

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Section 23.3 Concepts

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In this section.

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Subsection 23.3.1 The vector space $\C^n$

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Algebraically,

$n$ -dimensional complex vectors in

$\C^n$ act exactly as

$n$ -dimensional real vectors in

$\R^n$ — we add them component-wise, and we scalar multiply them by multiplying each component by the scalar. The only difference is that when we add or multiply components, we will be adding or multiplying complex numbers, and we specifically allow complex scalars in the scalar multiplication operation. See Example 23.4.1 for examples of carrying out these operations.

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Furthermore, since we treat the new imaginary number

$\ci$ algebraically as if it were a variable representing a real number, the patterns of algebra that emerge when working with

$n$ -dimensional complex vectors are exactly the same as those we identified from our experience working with

$n$ -dimensional real vectors. In particular, the collection

$\C^n$ will satsify all ten axioms in Definition 16.4.1 (which were modelled on the algebra of

$\R^n$ ), where scalar is allowed to mean complex number instead of just real number.

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How to attach geometric meaning to vectors in

$\C^n\text{,}$ similar to how we conceive of vectors in

$\R^n$ as “displacement arrows”, is another story. How long is the two-dimensional vector

$(\ci, \ci)\text{,}$ and in what direction does it point? Does multiplying a complex vector by

$2$ still make it twice as long? How does a complex number behave as a “scale factor”? For example, what effect does multiplying a complex vector by the complex scalar

$\ci$ have on the “length” of the vector? Should we still think of

$\C^2$ as some sort of plane, when in Section A.3 we have already interpreted

$\C^1 = \C$ geometrically as a plane?

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Some of these geometric questions we will answer in a later chapter, and we will discuss the complex plane as a two-dimensional space in Subsection 23.3.4, but for now we will mostly just treat complex vectors algebraically.

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Subsection 23.3.2 Complex vector spaces

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As the space

$\C^n$ satisfies the ten vector space axioms using complex scalars, we again use that example as a model and define a complex vector space by Definition 16.4.1, except that we interpret scalar to mean complex number instead of just real number.

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To recognize a complex vector space, we again look for patterns of addition and scalar multiplication, but now we must ask whether multiplying by a complex scalar could make sense for the types of objects we are considering.

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For example, we know we can add and scalar multiply matrices. However, if we are only considering matrices of a certain size with entries that are real numbers, then multiplying by complex scalars would not make sense, because that would violate Axiom S 1 — a real matrix multiplied by a complex scalar might yield a complex matrix, if the complex scalar doesn't happen to be purely real. If we instead consider all complex matrices of a certain size, then the fact that a real matrix multiplied by a complex scalar might result in a complex matrix is no longer a problem, because we should be thinking of real matrices as complex matrices that just happen to have entries that are purely real.

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All of the concepts and facts we have learned about real vector spaces apply equally to complex vector spaces, with, as always, the caveat that we are now dealing with complex scalars. In particular, everything we learned about subspaces, span, linear dependence/independence, basis, coordinates, and dimension all still applies to complex vectors spaces. Even our discoveries about column space, row space, and null space still work the same for complex matrices as for real matrices. Perhaps the only subtlety to point out now is that, since we are allowing complex scalars, we must also allow coefficients in linear combinations to be complex numbers, and so coordinate vectors made up of those coefficients in linear combinations of basis vectors will be vectors in

$\C^n$ instead of

$\R^n$ (where

$n$ is the dimension of the complex space being considered).

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Subsection 23.3.3 Instances of complex vector spaces

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Bringing complex numbers into the mix lets us convert some familiar examples into new complex examples.

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The vector space $\C^n$ .

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We have already discussed this space in Subsection 23.3.1, and just as we view

$\R^n$ as the model example of the abstract concept of real vector space, we view

$\C^n$ as the model example of the abstract concept of complex vector space. This space still has the familiar zero vector

$\zerovec = (0,0,\dotsc,0)\text{,}$ and still has the familiar standard basis consisting of the vectors

$\uvec{e}_j$ that have all components equal to zero except for a single

$1$ in the

$\nth[j]$ position. Just like

$\R^n\text{,}$ when convenient we will realize vectors in

$\C^n$ as (complex) column matrices.

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Spaces of complex matrices ( $\matrixring_{m \times n}(\C)$ ).

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Just as with

$\C^n$ versus

$\R^n\text{,}$ there is little new in

$\matrixring_{m \times n}(\C)$ versus

$\matrixring_{m \times n}(\R)\text{,}$ except that of course we allow complex entries in the matrices and allow multiplication by complex scalars. The space

$\matrixring_{m \times n}(\C)$ still has the zero matrix as the zero vector, and still has the familiar standard basis consisting of the matrices

$E_{ij}$ that have all entries equal to zero except for a single

$1$ in the

$(i,j)$ entry.

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Spaces of complex polynomials.

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Similar to the real case, we write

$\poly(\C)$ for the space of all complex polynomials, and write

$\poly_n(\C)$ for the space of those polynomials of degree

$n$ or less. To help distinguish between the real and complex contexts, we will continue to use

$x$ as the indeterminate in real polynomials (e.g.

$p(x) = 2 x^2 - x + 1$ ), but will use

$z$ as the indeterminate in complex polynomials (e.g.

$p(z) = \ci z^2 - (5 - \ci) z + (1 + 2 \ci)$ ). Again, there is nothing really new here in

$\poly(\C)$ versus

$\poly(\R)\text{,}$ except for the switch to allowing complex coefficients and to considering the indeterminate

$z$ as representing an indeterminate complex number. These spaces still have the familiar zero polynomial

$\zerovec(z) = 0$ as zero vector, and each space

$\poly_n(\C)$ still has familiar standard basis

$\{1,z,z^2,\dotsc,z^n\}\text{.}$

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Subsection 23.3.4 Every complex space is also a real space

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The field of complex numbers

$\C = \C^1$ is a one-dimensional complex space, since it has a basis consisting of a single vector:

$\C = \Span_{\C} \{1\}.$ Since we are allowing complex scalars here (as the subscript

$\C$ on the word

$\Span$ indicates), this expression of

$\C$ as a span just says that every complex number can be expressed as a complex multiple of

$1\text{.}$ This statement follows from the fact that

$\begin{equation*} z = z \cdot 1 \end{equation*}$

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is always true for every complex number

$z\text{.}$

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But in order to let the new imaginary number

$\ci$ interact algebraically with all of our other numbers, we have defined complex numbers to be algebraic combinations of real numbers and the new number

$\ci\text{.}$ And, since we have the identity

$\ci^2 = -1\text{,}$ the only combinations required in order to describe all possible algebraic combinations of real numbers and

$\ci$ are linear combinations of

$1$ and

$\ci\text{:}$

$\begin{equation*} z = a + b \ci = a \cdot 1 + b \cdot \ci \text{.} \end{equation*}$

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Since such a linear combination always uses real coefficients

$a$ and

$b\text{,}$ we can write

$\begin{equation*} \C = \Span_{\R} \{ 1, \ci \} \text{.} \end{equation*}$

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That is, $\C = \C^1$ is both a one-dimensional complex vector space and a two-dimensional real vector space. This second point of view of

$\C$ as a real vector space lets us make sense of the complex plane as a two-dimensional space, with a horizontal real axis that corresponds to (real) scalar multiples of the basis vector

$1\text{,}$ combined with a vertical complex axis that corresponds to (real) scalar multiples of the basis vector

$\ci\text{.}$

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We can do something similar with every complex vector space. Suppose

$V$ is a finite-dimensional complex vector space with basis

$\{ \uvec{v}_1, \uvec{v}_2, \dotsc, \uvec{v}_n \} \text{,}$ so that

$\begin{align*} V \amp = \Span_{\C} \{ \uvec{v}_1, \uvec{v}_2, \dotsc, \uvec{v}_n \} \text{,} \amp \dim_{\C} V \amp = n \text{.} \end{align*}$

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Then every vector

$\uvec{v}$ in the space is somehow a (complex) linear combination of these basis vectors:

$\begin{equation*} \uvec{v} = z_1 \uvec{v}_1 + z_2 \uvec{v}_2 + \dotsb + z_n \uvec{v}_n \text{.} \end{equation*}$

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But each complex scalar can be broken into a combination of real and imaginary parts, and so we can do the same with this linear combination. Writing

$z_j = a_j + b_j \ci$ for each complex scalar, we have

$\begin{align*} \uvec{v} \amp = (a_1 + b_1 \ci) \uvec{v}_1 + (a_2 + b_2 \ci) \uvec{v}_2 + \dotsb + (a_n + b_n \ci) \uvec{v}_n \\ \amp = a_1 \uvec{v}_1 + b_1 \cdot (\ci \uvec{v}_1) + a_2 \uvec{v}_2 + b_2 \cdot (\ci \uvec{v}_2) + \dotsb + a_n \uvec{v}_n + b_n \cdot (\ci \uvec{v}_n)\text{.} \end{align*}$

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Considering this as a linear combination with real coefficients (the

$a_j$ and

$b_j$ ), we can say that every vector in

$V$ is a real linear combination of the vectors

$\begin{equation*} \uvec{v}_1, \ci \uvec{v}_1, \uvec{v}_2, \ci \uvec{v}_2, \dotsc, \uvec{v}_n, \ci \uvec{v}_n \text{.} \end{equation*}$

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That is,

$\begin{equation*} V = \Span_{\R} \{ \uvec{v}_1, \ci \uvec{v}_1, \uvec{v}_2, \ci \uvec{v}_2, \dotsc, \uvec{v}_n, \ci \uvec{v}_n \} \text{,} \end{equation*}$

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with

$\begin{equation*} \dim_{\R} V = 2n \text{.} \end{equation*}$

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From the above analysis, we can conclude that every finite-dimensional complex vector space can also be considered as a real vector space with doubled dimension.