NoteNextra-origin/content/Math429/Math429_L18.md

Lecture 18

Chapter III Linear maps

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

Duality 3F

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Review

Theorem 3.128, 3.130

Let V,W be a finite dimensional vector space, T\in \mathscr{L}(V,W)

a) null\ T'=(range\ T)^0, dim (null\ T')=dim\ null\ T+dim\ W-dim\ V
b) range\ T'=(null\ T)^0, dim (range\ T')=dim (range\ T) c) dim(range\ T')= dim(range\ T)

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New materials

Theorem 3.129, 3.131

Let V,W be a finite dimensional vector space, T\in \mathscr{L}(V,W)

a) T is injective \iff T' is surjective
b) T is surjective \iff T' is injective

Proof:

T is injective \iff null\ T=\{0\}\iff range\ T'=V'\iff T' surjective

T is surjective \iff range\ T=W\iff null\ T'=0\iff T' injective

Theorem 3.132

Let V,W be a finite dimensional vector space, T\in \mathscr{L}(V,W)

Then M(T')=(M(T))^\top. Where the basis for M(T)' are the dual basis to the ones for M(T)

Theorem 3.133

col\ rank\ A=row\ rank\ A

Proof: col\ rank\ A=col\ rank\ (M(T))=dim\ range\ T=dim\ range\ T'=dim\ range\ T'=col\ rank\ (M(T'))=col\ rank\ (M(T)^\top)=row\ rank\ (M(T))

Chapter V Eigenvalue and Eigenvectors

Invariant Subspaces 5A

Goal: Study maps in \mathscr{L}(V) (linear operations)

Question: Given T\in \mathscr{L}(V) when can I restrict to U\subseteq V such that T\vert_U\in \mathscr{L}(U)

Definition 5.2

Suppose T\in \mathscr{L}(V) and U\subseteq V a subspace is said to be invariant under T if Tu\in U,\forall u\in U

Example:

For any T\in \mathscr{L}(V), the following are invariance subspaces.

1. \{0\}
2. V
3. null\ T, v\in null\ T\implies Tv=0\in null\ T
4. range\ T, v\in range\ T\subseteq V \implies Tv\in range\ T

Definition 5.5

Suppose T\in\mathscr{L}(V), then for \lambda \in \mathbb{F} is an eigenvalue of T if \exists v\in V such that v\neq 0 and Tv=\lambda v.

Definition 5.8

Suppose T\in\mathscr{L}(V) and \lambda \in \mathbb{F} is an eigenvalue of T. The v\in V is an eigenvector of T corresponding to \lambda if v\neq 0 and Tv=\lambda v

Note: if \lambda is an eigenvalue of T and v an eigenvector corresponding to \lambda, then U=Span(V) is an invariant subspace. and T\vert_U is multiplication by \lambda

Proposition 5.7

V is finite dimensional T\in \mathscr{L},\lambda\in \mathbb{F} then the following are equivalent: (TFAE)

a) \lambda is an eigenvalue
b) T-\lambda I is not injective c) T-\lambda I is not surjective d) T-\lambda I is not invertible

Proof:

(a)\iff (b) \lambda is an eigenvalue \iff \exists v\in V such that Tv=\lambda v\iff \exists v\in V, v\neq 0, (T-\lambda I)v=0

Example:

T(x,y)=(-y,x) what are the eigenvalues of T.

If \mathbb{F}=\mathbb{R} rotation by 90\degree, so no eigenvalues.

what if \mathbb{F}=\mathbb{C}? we can solve the system T(x,y)=\lambda (x,y),(-y,x)=\lambda (x,y)

-y=\lambda x \\
x=\lambda y

So

-1=\lambda ^2,\lambda =\plusmn i

when \lambda =-i, v=(1,i), \lambda=i, v=(1,-i)