NoteNextra-origin/pages/Math429/Math429_L10.md

Lecture 10

Chapter III Linear maps

Assumption: U,V,W are vector spaces (over \mathbb{F})

Vector Space of Linear Maps 3A

Review

Theorem 3.21 (The Fundamental Theorem of Linear Maps, Rank-nullity Theorem)

Suppose V is finite dimensional, and T\in \mathscr{L}(V,W), then range(T) is finite dimensional (W don't need to be finite dimensional). and

dim(V)=dim(null (T))+dim(range(T))

Proof:

Let u_1,...,u_m be a basis for null(T), then we extend to a basis of V given by u_1,...,u_m,v_1,...,v_m, we have dim(V)=m+n. Claim that Tv_1,...,Tv_n forms a basis for range (T). Need to show

• Linearly independent. (in Homework 3)

• These span range(T).

  Let w\in range(T) the there exists v\in V such that Tv=W, u_1,...,u_m,v_1,...,v_m are basis so \exists a_1,...,a_m,b_1,...,b_n such that v=a_1u_1+...+a_mu_m+b_1v_1+...+b_n v_n. Tv=a_1Tu_1+...+a_mTu_m+b_1Tu_1+...+b_nTv_n.

  Since u_k\in null(T), So Tv_1,...,Tv_n spans range T and so form a basis. Thus range(T) is finite dimensional and dim(range(T))=n. So dim(V)=dim(null (T))+dim(range(T))

Theorem 3.22

Suppose V,W are finite dimensional with dim(V)>dim(W), then there are no injective maps from V to W.

Theorem 3.24

Suppose V,W are finite dimensional with dim(V)<dim(W), then there are no surjective maps from V to W.

ideas of Proof: relies on Theorem 3.21 dim(null(T))>0

Linear Maps and Linear Systems 3EX-1

Suppose we have a homogeneous linear system * with m equation and n variables.

A_{11} x_1+ ... + A_{1n} x_n=0\\
...\\
A_{m1} x_1+ ... + A_{mn} x_n=0

which is equivalent to

A\begin{bmatrix}
    x_1\\...\\x_n
\end{bmatrix}=\vec{0}

also equivalent to

T(v)=0,\textup{ for some }T

T(x_1,...,x_n)=(A_{11} x_1+ ... + A_{1n},...,A_{m1} x_1+ ... + A_{mn} x_n),T\in \mathscr{L}(\mathbb{R}^n,\mathbb{R}^m)

Solution to * is null(T).

Proposition 3.26

A homogeneous linear system with more variables than equations has non-zero solutions.

Proof:

Using T as above, note that since n>m, use Theorem 3.22, implies that T cannot be injective. So, null (T) contains a non-zero vector.

Proposition 3.28

An in-homogenous system with more equations than variables has no solutions for some choices of constants. (A\vec{x}=\vec{b} for some \vec{b} this has no solution)

Matrices 3A

Definition 3.29

For m,n>0 and m\times n matrix A is a rectangular array with elements of the \mathbb{F} given by

A=\begin{pmatrix}
    A_{1,1}& ...&A_{1,n}\\
    ... & & ...\\
    A_{n,1}&...&A_{m,n}\\
\end{pmatrix}

Operations on matrices

Addition:

A+B=\begin{pmatrix}
    A_{1,1}+B_{1,1}& ...&A_{1,n}+B_{1,n}\\
    ... & & ...\\
    A_{n,1}+A_{n,1}&...&A_{m,n}+B_{m,n}\\
\end{pmatrix}

for A+B, A,B need to be the same size

Scalar multiplication:

\lambda A=\begin{pmatrix}
    \lambda A_{1,1}& ...& \lambda A_{1,n}\\
    ... & & ...\\
    \lambda A_{n,1}&...& \lambda A_{m,n}\\
\end{pmatrix}

Definition 3.39

\mathbb{F}^{m,n} is the set of m by n matrices.

Theorem 3.40

\mathbb{F}^{m,n} is a vector space (over \mathbb{F}) with dim(\mathbb{F}^{m,n})=m\times n

Matrix multiplication 3EX-2

Let A be a m\times n matrix and B be an n\times s matrix

(A,B)_{i,j}= \sum^n_{r=1} A_{i,r}\cdot B_{r,j}

Claim:

This formula comes from multiplication of linear maps.

Definition 3.44

Linear maps to matrices, let V, W, Tv_i written in terms of w_i.

M(T)=\begin{pmatrix}
    Tv_1\vert Tv_2\vert ...\vert Tv_n
\end{pmatrix}