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Vector Representations

You may have noticed that many questions about elements of abstract vector spaces eventually become questions about column vectors or systems of equations. would be an example of this.

Vector Representations

You may have noticed that many questions about elements of abstract vector spaces eventually become questions about column vectors or systems of equations. would be an example of this. We will make this vague idea more precise in this section.

The vector space of $3\times 5$ matrices, $M_{35}$ is isomorphic to what fundamental vector space?
B=\set{ \colvector{1\\2\\-1},\, \colvector{3\\-1\\2},\, \colvector{1\\1\\1} }
surprise,

Vector Representation

We begin by establishing an invertible linear transformation between any vector space $V$ of dimension $n$ and $\complex{n}$. This will allow us to go back and forth between the two vector spaces, no matter how abstract the definition of $V$ might be.

This definition looks more complicated than it really is, though the form above will be useful in proofs. Simply stated, given $\vect{w}\in V$, we write $\vect{w}$ as a linear combination of the basis elements of $B$. It is key to realize that guarantees that we can do this for every $\vect{w}$, and furthermore this expression as a linear combination is unique. The resulting scalars are just the entries of the vector $\vectrep{B}{\vect{w}}$. This discussion should convince you that $\vectrepname{B}$ is well-defined as a function. We can determine a precise output for any input. Now we want to establish that $\vectrepname{B}$ is a function with additional properties it is a linear transformation.

The proof of provides an alternate definition of vector representation relative to a basis $B$ that we could state as a corollary (): $\vectrepname{B}$ is the unique linear transformation that takes $B$ to the standard unit basis.

Vector representations are most interesting for vector spaces that are not $\complex{m}$.

We will have many occasions later to employ the inverse of vector representation, so we will record the fact that vector representation is an invertible linear transformation.

Informally, we will refer to the application of $\vectrepname{B}$ as coordinatizing a vector, while the application of $\ltinverse{\vectrepname{B}}$ will be referred to as un-coordinatizing a vector.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

Characterization of Vector Spaces

Limiting our attention to vector spaces with finite dimension, we now describe every possible vector space. All of them. Really.

is the first of several surprises in this chapter, though it might be a bit demoralizing too. It says that there really are not all that many different (finite dimensional) vector spaces, and none are really any more complicated than $\complex{n}$. Hmmm. The following examples should make this point.

Example: Two isomorphic vector spaces

The vector space of polynomials with degree 8 or less, $P_8$, has dimension 9 (). By , $P_8$ is isomorphic to $\complex{9}$.

Example: Crazy vector space revealed

The crazy vector space, $C$ of , has dimension 2 by . By , $C$ is isomorphic to $\complex{2}$. Hmmmm. Not really so crazy after all?

Example: A subspace characterized

In we determined that a certain subspace $W$ of $P_4$ has dimension $4$. By , $W$ is isomorphic to $\complex{4}$.

Theorem: Isomorphism of Finite Dimensional Vector Spaces

Suppose $U$ and $V$ are both finite-dimensional vector spaces. Then $U$ and $V$ are isomorphic if and only if $\dimension{U}=\dimension{V}$.

Proof

This is just the statement proved in .

This is the advertised converse of . We will assume $U$ and $V$ have equal dimension and discover that they are isomorphic vector spaces. Let $n$ be the common dimension of $U$ and $V$. Then by there are isomorphisms $\ltdefn{T}{U}{\complex{n}}$ and $\ltdefn{S}{V}{\complex{n}}$.

$T$ is therefore an invertible linear transformation by . Similarly, $S$ is an invertible linear transformation, and so $\ltinverse{S}$ is an invertible linear transformation (). The composition of invertible linear transformations is again invertible () so the composition of $\ltinverse{S}$ with $T$ is invertible. Then $\ltdefn{\left(\compose{\ltinverse{S}}{T}\right)}{U}{V}$ is an invertible linear transformation from $U$ to $V$ and says $U$ and $V$ are isomorphic.

Example: Multiple isomorphic vector spaces

$\complex{10}$, $P_{9}$, $M_{25}$ and $M_{52}$ are all vector spaces and each has dimension 10. By each is isomorphic to any other.

The subspace of $M_{44}$ that contains all the symmetric matrices () has dimension $10$, so this subspace is also isomorphic to each of the four vector spaces above.

Coordinatization Principle

With $\vectrepname{B}$ available as an invertible linear transformation, we can translate between vectors in a vector space $U$ of dimension $m$ and $\complex{m}$. Furthermore, as a linear transformation, $\vectrepname{B}$ respects the addition and scalar multiplication in $U$, while $\vectrepinvname{B}$ respects the addition and scalar multiplication in $\complex{m}$. Since our definitions of linear independence, spans, bases and dimension are all built up from linear combinations, we will finally be able to translate fundamental properties between abstract vector spaces ($U$) and concrete vector spaces ($\complex{m}$).

Theorem: Coordinatization and Linear Independence

Suppose that $U$ is a vector space with a basis $B$ of size $n$. Then S=\set{\vectorlist{u}{k}} is a linearly independent subset of $U$ if and only if R=\set{\vectrep{B}{\vect{u}_1},\,\vectrep{B}{\vect{u}_2},\,\vectrep{B}{\vect{u}_3},\,\ldots,\,\vectrep{B}{\vect{u}_k}} is a linearly independent subset of $\complex{n}$.

Proof

The linear transformation $\vectrepname{B}$ is an isomorphism between $U$ and $\complex{n}$ (). As an invertible linear transformation, $\vectrepname{B}$ is an injective linear transformation (), and $\ltinverse{\vectrepname{B}}$ is also an injective linear transformation (, ).

Since $\vectrepname{B}$ is an injective linear transformation and $S$ is linearly independent, says that $R$ is linearly independent.

If we apply $\ltinverse{\vectrepname{B}}$ to each element of $R$, we will create the set $S$. Since we are assuming $R$ is linearly independent and $\ltinverse{\vectrepname{B}}$ is injective, says that $S$ is linearly independent.

Here is a fairly simple example that illustrates a very, very important idea.

Example: Coordinatizing in $P_2$

In we needed to know that D=\set{ -2-x+3x^2,\, 1-2x^2,\, 5+4x+x^2 } is a basis for $P_2$. With and this task is much easier.

First, choose a known basis for $P_2$, a basis that forms vector representations easily. We will choose B=\set{1,\,x,\,x^2}

Now, form the subset of $\complex{3}$ that is the result of applying $\vectrepname{B}$ to each element of $D$, F&=\set{\vectrep{B}{-2-x+3x^2},\,\vectrep{B}{1-2x^2},\,\vectrep{B}{5+4x+x^2}}\\ &= \set{ \colvector{-2\\-1\\3},\, \colvector{1\\0\\-2},\, \colvector{5\\4\\1} } and ask if $F$ is a linearly independent spanning set for $\complex{3}$. This is easily seen to be the case by forming a matrix $A$ whose columns are the vectors of $F$, row-reducing $A$ to the identity matrix $I_3$, and then using the nonsingularity of $A$ to assert that $F$ is a basis for $\complex{3}$ (). Now, since $F$ is a basis for $\complex{3}$, and tell us that $D$ is also a basis for $P_2$.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

Symbols used here

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

matrix
A rectangular array of numbers; a linear map.

\det A,\ |A|

determinant
Scaling factor of area/volume under A; zero means singular.

A^{-1},\ A^{T}

inverse, transpose
The matrix that undoes A; A with rows and columns swapped.

\lambda

lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

Questions people ask

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn — it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

Özün sına

Parts of this page are adapted from Beezer, A First Course in Linear Algebra (GFDL 1.2). Condensed and re-explained here; errors are ours.

Daha çox Linear Algebra

Determinants Matrix inverse Eigenvalues and eigenvectors Matrix multiplication Row reduction Vector spaces, span and linear independence Orthogonality, projections and least squares Diagonalisation and matrix powers Linear transformations and change of basis