Vector Space Assessments: Advanced Linear Algebra Examination

Verified

Added on  2023/06/14

|6
|844
|424
Homework Assignment
AI Summary
This document presents solutions to several problems related to vector spaces in advanced linear algebra. It covers topics such as the spectrum of a linear transformation, nilpotent operators, and inner products on real vector spaces. Specifically, it addresses questions concerning the properties of the spectrum under different conditions, the trace of nilpotent operators, and the real nature of inner products even when dealing with complex vectors. The solutions provided offer detailed explanations and references to relevant theorems and concepts from standard linear algebra texts. Desklib provides access to more solved assignments and resources for students.
Document Page
Running head: VECTOR SPACE ASSESSMENTS
1
Vector Space Assessments Paper
Student’s Name
Professor’s Name
Affiliation
Date
tabler-icon-diamond-filled.svg

Secure Best Marks with AI Grader

Need help grading? Try our AI Grader for instant feedback on your assignments.
Document Page
VECTOR SPACE ASSESSMENTS
2
Vector Space Assessments
Question 3.7.24
Part 1
Since the spectrum of α is for values of μi where i assumes all positive values.
=ui
S ( )=ui
Sk ( ) =ui 1 , ,uik
For n dimensional vector space we will assume value i=1,…,∞ . Hence, 1 ii , ,i n
SK ()={μi 1 , , μik } for n dimensions i=1,…,K . Hence, 1 ii , ,iK n (Rotman, 2003, p. 172)
Hence
SK ( )= { μi 1 , , μik }1 ii , , iK n
Part 2
Since the spectrum of α is for values of μi where i1<i2
=ui
( )=ui
k ( ) =ui 1 , , uik
For n dimensional vector space we will assume value i=1,…,∞ . Hence, 1 ii <<i n
Document Page
VECTOR SPACE ASSESSMENTS
3
(Rotman, 2003, p. 172).
K ()={ μi 1 , , μik } for n dimensions which will assume value 1 to N for i=1<…<K. Hence,
1 ii <<iK n
Hence
K ()= {μi 1 , , μik }1 ii <<iK n
Question 3.7.25
If we assume the following dim V =n 1
Then we can suppose that is a nilpotent operator i.e. q=0
We can therefore consider that {0} ( ) ( 2 ) ( q )=V
If x ( q ) then q ( x ) =0 so q+ 1 ( x )= ( q ( x ) )= ( 0 )=0
Therefore
if q 1 then=0
As such
tr (¿¿ q ( ))=tr (q ( 0 ) )=0 ¿ CITATION Jos03 \p 383-384 \l 1033 (Rotman, 2003, pp. 383-384)
Document Page
VECTOR SPACE ASSESSMENTS
4
Question 3.7.26
Since V and U have values that are real numbers then their inner products will only assume
combinations of this real numbers Regardless of how many times the values are repeated. Real
numbers assume values between -∞ and ∞ (Roman, 2005). Hence, we can simply this by writing
Quantities (u,´u) and (v,´v) are always real even if V and U are complex vectors (Vinberg, 2003)
¿ . , .>:V U < . ,.>: ( V U ) (U V )
λ λ ( u ,u' ) = ( λu , λ u' )= ( U ( u ) , U ( u' ) ) =(u ,u' )
λ λ ( u ,u' ) = ( λu , λ u' )= ( V ( v ) ,V ( v' ) )=(v , v' )
They are orthogonal (Axler, 2015)
( V ( v ) , V ( v' ) ) = ( v , v' ) =δvv '
(U ( u ) , U ( u' ) )= (u , u' )=δuu '
With regard to positivity
( v , v' ) 0 for all v , v ' V
( u , u' ) 0 for all u , u ' U
Hence
¿ . , .>:V U R thenu , v ,u' , v' V ×U
tabler-icon-diamond-filled.svg

Secure Best Marks with AI Grader

Need help grading? Try our AI Grader for instant feedback on your assignments.
Document Page
VECTOR SPACE ASSESSMENTS
5
References
Axler, S. (2015). Linear Algebra Done Right. Springer: New York.
Hoffman, K. M., & Kunze, . R. (1971). Linear Algebra (2nd Edition). Pearson: London.
Loehr, N. (2014). Advanced Linear Algebra. Florida: Chapman and Hall Publishers/CRC.
Roman, S. (2008). Advanced Linear Algebra. New York: Springer.
Rotman, J. J. (2003). Advanced Modern Algebra. Upper Saddle River: Prentice Hall.
Vinberg, E. B. (2003). A Course in Algebra . Providence: American Mathematical Society.
Document Page
VECTOR SPACE ASSESSMENTS
6
chevron_up_icon
1 out of 6
circle_padding
hide_on_mobile
zoom_out_icon
logo.png

Your All-in-One AI-Powered Toolkit for Academic Success.

Available 24*7 on WhatsApp / Email

[object Object]