NoteNextra-origin/content/Math429/Math429_L14.md

Lecture 14

Chapter III Linear maps

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

Matrices 3C

Review

Proposition 3.76

M(Tv)=M(T)M(v)

Theorem 3.78

Let V,W be finite dimensional vector space, and T\in \mathscr{L}(V,W) then dim\ range\ T=column\ rank (M(T))=rank(M(T))

Proof:

range=Span\{Tv_1,...,Tv_n\} compare to Span\{M(T)_{\cdot,1},...,M(T)_{\cdot, n}\}=Span\{M(T)M(v_1),...,M(T)M(v_n)\}=Span\{M(Tv_1),...,M(Tv_n)\}

Since M is a isomorphism, then the theorem makes sense.

Change of Basis

Definition 3.79, 3.80

The identity matrix

I=\begin{pmatrix}
    1.& 0\\
    0& '1\\
\end{pmatrix}

The inverse matrix of an invertible matrix A denoted A^{-1} is the matrix such that

AA^{-1}=I=A^{-1}A

Question: Let u_1,...,u_n and v_1,...,v_n be two bases for V. What is M(I,(u_1,...,u_n),(v_1,...,v_n)),I\in \mathscr{L}(V)

Proposition 3.82

Let u_1,...,u_n and v_1,...,v_n be bases of V, then M(I,(u_1,...,u_n),(v_1,...,v_n)),I\in \mathscr{L}(V) and M(I,(v_1,...,v_n),(u_1,...,u_n)),I\in \mathscr{L}(V) are inverse to each other.

Proof:

M(I,(u_1,...,u_n),(v-1,...,v_n)),I\in \mathscr{L}(V) M(I,(v-1,...,v_n),(u_1,...,u_n))=M(I,(u_1,...,u_n),(u_1,...,u_n))

Theorem 3.84 Change of Basis

Let u_1,...,u_n and v_1,...,v_n be two bases for V and T\in \mathscr{L}(v), A=M(T,(u_1,...,u_n)), B=M(T,(v_1,...,v_n)), C=M(I,(u_1,...,u_n),(v_1,...,v_n)), then A=C^{-1}BC

Theorem 3.86

Let T\in \mathscr{L}(v) be an invertible linear map, then M(T^{-1})=M(T)^{-1}

Products and Quotients of Vector Spaces 3E

Goals: To construct vectors spaces from other vector spaces.

Definition 3.87

Suppose V_1,...,V_m vectors spaces over some field \mathbb{F}, then the product is given by

V_1\times ...\times V_n=\{(v_1,v_2,...,v_n)\vert v_1\in V_1, v_2\in V_2,...,v_n\in V_n\}

with addition given by

(v_1,...,v_n)+(u_1,...,u_n)=(v_1+u_1,...,v_n+u_n)

and scalar multiplication

\lambda (v_1,...,v_n)=(\lambda v_1,...,\lambda v_n),\lambda \in \mathbb{F}

Theorem 3.89

If v_1,...,v_n are vectors paces over \mathbb{F} then V_1\times ...\times V_n is a vector space over \mathbb{F}

Example:

V=\mathscr{P}_2(\mathbb{R})\times \mathbb{R}^2=\{(p,v)\vert p\in \mathscr{P}_2(\mathbb{R}), v\in \mathbb{R}^2\}=\{(a_0+a_1x+a_2x,(b,c))\vert a_0,a_1,a_2,b,c\in \mathbb{R}\}

A basis for V would be (1,(0,0)),(x,(0,0)),(x^2,(0,0)),(0,(1,0)),(0,(0,1))

Theorem 3.92

dim(V_1\times ...\times V_n)=dim(V_1)+...+dim(V_n)

Sketch of proof:

take a basis for each V_k, make them vectors in the product then combine the entire list of vector to be basis.

Example:

\mathbb{R}^2\times \mathbb{R}^3=\{((a,b),(c,d,e))\vert a,b,c,d,e\in \R\}

\mathbb{R}^2\times \mathbb{R}^3\cong \mathbb{R}^5,((a,b),(c,d,e))\mapsto(a,b,c,d,e)

Theorem 3.93

Let V_1,...,V_m\subseteq V, define \Gamma: V_1\times...\times V_m\to V_1+...+V_m. \Gamma(v_1,...,v_n)=v_1+...+v_n then \Gamma is always surjective. And it is injective if and only if V_1+...+V_m is a direct sum.

Sketch of the proof:

injective \iff null\ T\{ (0,...,0) \} \iff the only way to write 0=v_1,...,v_m is v_1=...=v_n=0 \iff then V_1+...+V_m is a direct sum

Theorem 3.94

V_1+...+V_m is a direct sum if and only if dim(V_1+...+V_m)=dim(V_1)+...+dim(V_m)

Proof:

Use \Gamma above is an isomorphism \iff V_1+...+V_m is a direct sum

Use \Gamma above is an isomorphism \implies dim(V_1+...+V_m)=dim(V_1)+...+dim(V_m)