# 3 Applied Linear Algebra questions, Linear Transformations

*label*Mathematics

*timer*Asked: Oct 18th, 2018

*account_balance_wallet*$15

### Question Description

There are three questions I need you help

Just in case you confused. On problem 2.4 it need problem 2.1(which is the first picture I uploaded, its question 21 from my textbook)

I upload the Chegg's answer for problem 2.1, but I need you revises it as detail as you can because I did learn that much in class.

### Unformatted Attachment Preview

## Tutor Answer

The solutions are ready. Please read them and ask if something is unclear. I wrote the proof for 2.1 in some other words, maybe it will be simpler for you.

2.1. A linear operator πΏ: π β π is one-to-one if and only if πΎππ(πΏ) = 0π .

Proof. If and only if means two things:

1) If πΏ is one-to-one then πΎππ(πΏ) = 0π and 2) If πΎππ(πΏ) = 0π then πΏ is one-to-one.

Both will get proof by a contradiction.

1) Let πΏ be one-to-one but πΎππ(πΏ) β 0π , i.e. there is a nonzero vector π£0 β πΎππ(πΏ), i.e.

πΏ(π£0 ) = 0π . But then πΏ(0π ) = πΏ(π₯0 ) while 0π β π₯0 ,

which is a contradiction with πΏ to be one-to-one, Q.E.D.

2) Let πΎππ(πΏ) = 0π but πΏ is not one-to-one. This means there are π₯1 β π₯2 : πΏ(π₯1 ) = πΏ(π₯2 ).

But then πΏ(π₯1 β π₯2 ) = πΏ(π₯1 ) β πΏ(π₯2 ) = 0, so π₯1 β π₯2 β πΎππ(πΏ), but π₯1 β π₯2 β 0π .

This is a contradiction, Q.E.D.

2.4.

a.) Each one-to one mapping π: π β π has its inverse map π β1 : πΌπ(π) β π such that

π β1 β π = πΌπ. We need to prove that 1) πΌπ(πΏ) = βπ and 2) πΏβ1 is linear.

1) is true because the image of a linear one-to-one transformation has the same dimension

as the initial space (i.e. π). And there is only one subspace of βπ with dimension π (it is

entire βπ itself).

2) Prove that πΏβ1 (π1 π¦1 + π2 π¦2 ) = π1 πΏβ1 (π¦1 ) + π2 πΏβ1 (π¦2 ).

Indeed, apply πΏ to both sides and get πΏ[πΏβ1 (π1 π¦1 + π2 π¦2 )] = πΏ[π1 πΏβ1 (π¦1 ) + π2 πΏβ1 (π¦2 )],

which is true because both parts are equal to π1 π¦1 + π2 π¦2 .

Because πΏ is one-to-one, this means the arguments of πΏ are the same, Q.E.D.

b.), c.) Denote matrix for πΏ as π΄ and matrix for πΏβ1 as π΅ and prove that

matrix for πΏβ1 πΏ is π΅π΄. This will prove that π΅π΄ = πΌ, so π΄ is invertible

and that the matrix for πΏβ1 is π΄β1 .

In other words, we need to prove that mu...

*flag*Report DMCA

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors