NoteNextra-origin/pages/Math429/Math429_L13.md

Lecture 13

Chapter III Linear maps

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

Matrices 3C

Theorem 3.63

A linear map is invertible if and only if it is injective and surjective.

Example

Consider T:\mathscr{P}(\mathbb{F})\to \mathscr{P}(\mathbb{F}), T(f)=xf

T is injective but not surjective. Since you cannot get constant from multiply x. So it is not invertible.

Theorem 3.65

Let V and W be finite-dimensional with the same dimension, and T\in\mathscr{L}(V,W), then T is invertible, if and only if T is injective if and only if, T is surjective.

Proof:

Suppose T is injective, then null\ T={0}, i.e dim(null\ T)=0, since dim\ V=dim\ null\ T+dim\ range\ T, we have dim\ V=dim\ range\ T but dim\ V\dim\ W, so dim\ W=dim\ range\ T. Thus W=range\ T. This shows that T\ injective \implies T\ surjective.

If T is surjective, then dim\ range\ T=dim\ W but then dim\ V=dim\ null\ T+dim\ W\implies dim\ null\ T=0, so T is injective, T\ surjective\implies T\ injective.

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Theorem 3.68

Suppose V,W finite dimensional dim\ V=dim\ W, then for T\in\mathscr{L}(V,W) and S\in \mathscr{L}(W,V), then ST=I\implies TS=I

Example 3.67

Show that for a polynomial q with degree m, there exists a unique polynomial p of degree m such that ((x^2+5x+7)p)''=q

Solution:

Let T:\mathscr{P}_m(\mathbb{F})\to \mathscr{P}_m(\mathbb{F}) given by T(p)=((x^2+5x+7)p)'' by T is injective since (x^2+5x+7) has degree \geq 2 for p\neq 0, therefore, p is surjective. (by Theorem 3.68)

Isomorphisms

Definition 3.69

An isomorphism of vector spaces is a invertible linear map. Two vector spaces V,W are isomorphic if there exists an isomorphism between them.

Notation: V\cong W means V and W are isomorphic. (Don't use very often, no map is included.)

Example:

\mathscr{P}_m(\mathbb{F}) and \mathbb{F}^{m+1} are isomorphic. T:\mathbb{F}^{m+1}\to \mathscr{P}_m(\mathbb{F}): T((a_0,...,a_m))=a_0+a_1x+...+a_n x^n

Theorem 3.70

Two finite dimensional vector spaces V,W are isomorphic if and only if dim\ V= dim\ W

Ideas of Proof:

\Rightarrow use fundamental theorems of linear map

\Leftarrow Let v_1,...,v_m\in V and w_1,...,w_n\in W be bases. Then define T:V\to W by T(v_k)=w_k for 1\leq k\leq n

Show T is invertible by showing T is injective and surjective.

Theorem 3.71

Let V,W be finite dimensional, let v_1,...,v_n\in V and w_1,...,w_m\in W be bases. Then the map

M(-,(v_1,...,v_n),(w_1,...,w_m)):\mathscr{L}(V,W)\to \mathbb{F}^{m,n}

T\mapsto M(T) or M(-,(v_1,...,v_n),(w_1,...,w_m)) is an isomorphism (M:\mathscr{L}(V,W)\to \mathbb{F}^{m,n})

Sketch of Proof:

Need to show M is surjective and injective.

• Injective: i.e need to show if M(T)=0, then T=0. M(T)=0\implies Tv_k=0, 1\leq k\leq n
• Surjective: i.e let A\in F^{m,n} define T:V\to W given by Tv_k=\sum_{j=1}^m A_{j,k} w_j you cna check that M(T)=A

Corollary 3.72

dim \mathscr{L}(V,W)=(dim\ V)(dim\ W)

Definition 3.73

v\in V, v_1,...,v_n a basis, then $M(v)=\begin{pmatrix} b_1\ ...\ b_n \end{pmatrix}, v=a_1v_1,...,a_nv_n$

Proposition 3.75, 3.76

M(T)_{\cdot,k}=M(Tv_k)

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