Abstract Nonsense

Crushing one theorem at a time

Linear Transformations


Point of post: In this post I discuss the material discussed in sections 31 and 32 of Halmos.

Motivation

We finally begin studying arguably the most important aspect of linear algebra, linear transformations. These are the structure preserving maps for vector spaces. They correspond naturally to coordinate changes, linear shifts, and contraction/dilations. Moreover, theoretically they are interesting because, as it turns out, many maps which aren’t usually thought of as linear transformations are. For example, the differential operator on the space C^{\infty}[\mathbb{R},\mathbb{R}].

Linear Transformations

If \mathscr{V} and \mathscr{W} are F-space, we call a mapping T:\mathscr{V}\to\mathscr{W}linear transformation/linear homomorphism if

T\left(\alpha x+\beta y\right)=\alpha T\left(x\right)+\beta T\left(y\right)

for all x,y\in \mathscr{V} and \alpha,\beta\in F. If \mathscr{V}=\mathscr{W} we call T a linear transformation on \mathscr{V} in which case T is usually called a linear operator or endomorphism.

As stated earlier examples of linear transformations are abound. We give three examples:

Ex.(1): Let \mathscr{V} be a F-space, then if we consider F as a vector space over itself then every linear functional \varphi\in\text{Hom}\left(\mathscr{V},F\right) is a linear transformation from \mathscr{V} to F.

Ex.(2): If \mathscr{V} is a vector space the the maps \bold{0}:\mathscr{V}\to\mathscr{V}:x\mapsto \bold{0} and \text{id}:\mathscr{V}\to\mathscr{V}:x\mapsto x are both linear transformations.

Remark: In this context it is customary to denote \text{id}:\mathscr{V}\to\mathscr{V} by \bold{1}, the reasons for this will be clear soon enough.

Ex.(3): If C^{\infty}\left[\mathbb{R},\mathbb{R}\right] denotes the space of all smooth functions from \mathbb{R} to \mathbb{R} then the differential operator D:C^{\infty}\left[\mathbb{R},\mathbb{R}\right]\to C^{\infty}\left[\mathbb{R},\mathbb{R}\right]:f\mapsto f' is a linear operator. More specifically, if \mathbb{C}_n[z] is the space of all n-1 degree polynomials with complex coefficients  then

\displaystyle D:\mathbb{C}_n[x]\to\mathbb{C}_n[x]:\sum_{j=0}^{n-1}c_j x^j\mapsto \sum_{j=1}^{n-1}jc_j x^{j-1}

is a linear operator.
Remark: Note that the last map was defined algebraically, in a manner independent of the derivative. Thus, we can make sense of the “differential operator” on arbitrary polynomial rings.

Some aspects of linear transformation become apparent as soon as the definitions are laid out. For example, it’s evident that for any linear transformation T that T(\bold{0})=T\left(0\bold{0}\right)=0T\left(\bold{0}\right)=\bold{0}. Slightly less obvious is the following theorem

Theorem: Let T be a linear transformation from \mathscr{V} to \mathscr{W} which is bijective. Then, T^{-1} is a linear transformation.

Proof: We merely note that if x,y\in\mathscr{W} then x=T(x'),y=T(y') for some x',y'\in\mathscr{V} and thus

\begin{aligned}T^{-1}\left(\alpha x+\beta y\right) &=T^{-1}\left(\alpha T(x')+\beta T(y')\right)\\ &=T^{-1}\left(T\left(\alpha x'+\beta y'\right(\right)\\ &=\alpha x'+\beta y'\\ &=\alpha T^{-1}(x)+\beta T^{-1}(y)\end{aligned}

and since x,y\in \mathscr{W} and \alpha,\beta\in F were arbitrary the conclusion follows. \blacksquare

It’s also fairly clear that we may add linear transformations and multiply them by scalars in the obvious way. Namely if T,T':\mathscr{V}\to\mathscr{W} are linear transformations and \alpha,\beta\in F then we can define \alpha T+\beta T' to be the map

\alpha T+\beta T':\mathscr{V}\to\mathscr{W}:x\mapsto \alpha T(x)+\beta T'(x)\quad\mathbf{(1)}

Moreover, with this definition of addition it’s clear that the zero function \bold{0} acts as a additive inverse and the map -T given by (-T)(x)=-T(x) is an additive inverse. Thus, it’s not a quantum leap to realize that with the operations of addition and scalar multiplication given by \mathbf{(1)} that the set of all linear transformations from \mathscr{V} to \mathscr{W}, denoted \text{Hom}\left(\mathscr{V},\mathscr{W}\right), is a vector space. If we only consider endomorphisms on \mathscr{V} we shorten \text{Hom}\left(\mathscr{V},\mathscr{V}\right) to \text{End}\left(\mathscr{V}\right).

Dimension of \text{Hom}\left(\mathscr{V},\mathscr{W}\right)

So, the obvious question is, what is the dimension of \text{Hom}\left(\mathscr{V},\mathscr{W}\right)? The answer turns out to be \dim_F\text{Hom}\left(\mathscr{V},\mathscr{W}\right)=\dim_F\left(\mathscr{V}\right)\cdot\dim_F\left(\mathscr{W}\right) although it may not be clear immediately why this is so. But first, we prove a small lemma:

Theorem: Let \mathscr{V},\mathscr{W} be finite dimensional F-spaces. Then, if \{v_1,\cdots,v_m\} is a basis for \mathscr{V} then for any \{w_1,\cdots,w_m\}\subseteq\mathscr{W} there is a unique linear transformation T such that T(v_k)=w_k for each k\in[m].

Proof: Let T:\mathscr{V}\to\mathscr{W} be the map given by

\displaystyle T\left(\sum_{j=1}^{m}\alpha_j v_j\right)=\sum_{j=1}^{m}\alpha_j w_j

Clearly then we have that T is a linear map and T\left(v_k\right)=w_k,\text{ }k=1,\cdots,m. Moreover, if T' is also a linear transformation such that T'(v_k)=w_k we see then that

\displaystyle T'\left(\sum_{j=1}^{m}\alpha_j v_j\right)=\sum_{j=1}^{m}\alpha_j T\left(v_j\right)=\sum_{j=1}^{m}\alpha_j w_j=T\left(\sum_{j=1}^{m}\alpha_j v_j\right)

and since every element of \mathscr{V} may be written in the form \displaystyle \sum_{j=1}^{m}\alpha_j v_j  it follows that T'(v)=T(v) for all v\in\mathscr{V}, from where the conclusion follows. \blacksquare

Theorem: Let \mathscr{V},\mathscr{W} be F-spaces of dimension m and n respectively; then the dimension of \text{Hom}\left(\mathscr{V},\mathscr{W}\right) is mn.

Proof: Let \{x_1,\cdots,x_m\} be a basis for \mathscr{V} and \{y_1,\cdots,y_n\} a basis for \mathscr{W}. Then, for each i\in[m] and j\in[n] the above lemma implies that there exists a unique T_{i,j}\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right) such thatT_{i,j}(v_k)=\delta_{i,k}w_j. We claim that \mathscr{B}\overset{\text{def.}}{=}\left\{T_{i,j}:i\in[m]\text{ and }j\in[n]\right\} forms a basis for \text{Hom}\left(\mathscr{V},\mathscr{W}\right). Firstly, to see that \mathscr{B} we suppose that \alpha_{i,j}\in F,\text{ }i\in[m],j\in[n] are such that

\displaystyle \sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_{i,j}T_{i,j}=\bold{0}

then, for a fixed i_0\in[m] we see that the above implies that

\displaystyle \sum_{j=1}^{m}\alpha_{i_0,j}w_j=\sum_{i=1}^{m}\sum_{j=1}^{n}\alpha_{i,j}T_{i,j}(v_{i_0})=0

and since \{w_1,\cdots,w_n\} is linearly independent it follows that \alpha_{i_0,1},\cdots,\alpha_{i_0,n}=0. But, since i_0 was arbitrary it follows that \alpha_{i,j}=0,\text{ }i\in[m],j\in[n]. Now, to prove that \text{span }\mathscr{B}=\text{Hom}\left(\mathscr{V},\mathscr{W}\right) let T\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right).Then, for each i_0\in[m] we have that

\displaystyle T(v_{i_0})=\sum_{j=1}^{n}\beta_{i_0,j}w_j

we claim that

\displaystyle T=\sum_{i=1}^{m}\sum_{j=1}^{n}\beta_{i,j}T_{i,j}

to do this it suffices to show that the right hand side agrees with the left hand side for v_1,\cdots,v_m. To do this though we merely note that for each i_0\in[m]

\displaystyle \sum_{i=1}^{m}\sum_{j=1}^{n}\beta_{i,j}T_{i,j}(v_{i_0})=\sum_{j=1}^{n}\beta_{i_0,j}T_{i_0,j}(v_{i_0})=\sum_{j=1}^{n}\beta_{i_0,j}w_j=T(v_{i_0})

from where it follows that \mathscr{B} is a basis. Noting that \text{card }\mathscr{B}=mn finishes the argument. \blacksquare

Corollary: If \mathscr{V} is a finite dimensional F-space then \dim_F\text{End}\left(\mathscr{V}\right)=\left(\dim_F\mathscr{V}\right)^2

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

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November 19, 2010 - Posted by | Algebra, Halmos, Linear Algebra, Uncategorized | , , ,

8 Comments »

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  2. […] a basis for where is as it was in previous discussions. It’s evidently linear independent, since it’s a subset of a linearly independent set. […]

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