2017 - Page 11 of 43 - Shawn Zhong

Math 375 – 11/20

$Eigenvalues and Eigenvectors • Definition ○ T:V→V linear, for {█(x∈V@λ∈ℂ)┤, (x≠0) ○ We say x is an eigentvector for T with eigenvalue λ if Tx=λx • Note ○ Tx=λx ○ ⇒Tx−λx=0 ○ ⇒(T−λI)x=0 ○ ⇒x∈N(T−λI) Find all eigenvalues and eigenvectors • T=I ○ Tx=1x, ∀x∈V ○ Eigenvalue = 1 with eigenvectors of all elements in V • T=0 ○ Tx=0x, ∀x∈V ○ Eigenvalue = 0 with eigenvectors of all elements in V • T=[■(c_1&&@&⋱&@&&c_n )], (c_i≠c_j for i=j) ○ [■(c_1&&@&⋱&@&&c_n )] e_i=c_i e_i ○ Eigenvalue = c_i with eigenvector of te_i, (t∈R, t≠0) • T=[■8(1&2@2&1) ○ det⁡(T−λI)=0 § |■8(1−λ&2@2&1−λ)|=0 § (λ−3)(λ+1)=0 ○ λ=3 § [■8(1−3&2@2&1−3)][█(x@y)]=[█(0@0)]⇒x=y § Eigenvector: [█(t@t)](t∈R, t≠0) ○ λ=−1 § [■8(2&2@2&2)][█(x@y)]=[█(0@0)]⇒x+y=0 § Eigenvector: t[█(1@−1)](t∈R, t≠0) • T=[■8(0&−1@1&0) ○ det⁡(T−λI)=0 § |■8(−λ&−1@1&−λ)|=0 § λ^2+1=0 ○ λ=i § [■8(−i&−1@1&−i)][█(x@y)]=[█(0@0)]⇒y=−ix § Eigenvector: t[█(1@−i)](t∈ℂ, t≠0) ○ λ=−i § [■8(i&−1@1&i)][█(x@y)]=[█(0@0)]⇒y=ix § Eigenvector: t[█(1@i)](t∈ℂ, t≠0) Multiplicity of Eigenvalues • T=[■(3&&@&3&@&&4)] ○ Eigenvalues: λ=3 or λ=4 ○ dim⁡〖N(T−λI)={■8(2&λ=3@1&λ=4@0&otherwise)┤〗$

Math 375 – Homework 10

Math 375 – 11/15

$Theorem • Statement ○ If dim⁡V=dim⁡W<∞, then for linear map T:V→W ○ injective ⟺ surjective ⟺ bijective • Proof ○ By Rank-Nullity Theorem § dim⁡W=dim⁡V=dim⁡N(T)+dim⁡Range(T) ○ If T is injective § ⇒dim⁡N(T)=0 § ⇒dim⁡W=dim⁡Range(T) § ⇒ T is surjective § ⇒T is bijective ○ If T is not injective § ⇒dim⁡N(T)>0 § ⇒dim⁡W≠dim⁡Range(T) § ⇒T is not surjective § ⇒T is not bijective Left Inverse and Right Inverse • If both left inverse and right inverse exists • Then they are the same • Suppose ○ f:V→W ○ g,h:W→V ○ gf=id_V (i.e. g is the left inverse of T) ○ fh=id_w (i.e. h is the right inverse of T) • Then ○ g=g(fh=(gf)h=h Injective and Null Space • Proof: T injective⇒N(T)={0} ○ If T is injective ○ then the only one element mapped to 0 is 0 itself. ○ Therefore N(T)={0} • Proof: N(T)={0}⇒T injective ○ If T(x)=T(y), then ○ T(x)−T(y)=T(x−y)=0 ○ So x−y∈N(T) ○ ⇒x=y ○ Therefore T is injective$