3 Applied Linear Algebra questions, Linear Transformations

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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.

problem_2.1.png
problem_2.4.png
problem_2.5.png
chegg_ans_for_2.1.png

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Borys_S
School: University of Virginia

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 πΏ: π β π is one-to-one if and only if πΎππ(πΏ) = 0π .
Proof. If and only if means two things:
1) If πΏ is one-to-one then πΎππ(πΏ) = 0π and 2) If πΎππ(πΏ) = 0π then πΏ is one-to-one.
Both will get proof by a contradiction.
1) Let πΏ be one-to-one but πΎππ(πΏ) β  0π , i.e. there is a nonzero vector π£0 β πΎππ(πΏ), i.e.
πΏ(π£0 ) = 0π . But then πΏ(0π ) = πΏ(π₯0 ) while 0π β  π₯0 ,
which is a contradiction with πΏ to be one-to-one, Q.E.D.
2) Let πΎππ(πΏ) = 0π but πΏ is not one-to-one. This means there are π₯1 β  π₯2 : πΏ(π₯1 ) = πΏ(π₯2 ).
But then πΏ(π₯1 β π₯2 ) = πΏ(π₯1 ) β πΏ(π₯2 ) = 0, so π₯1 β π₯2 β πΎππ(πΏ), but π₯1 β π₯2 β  0π .

2.4.
a.) Each one-to one mapping π: π β π has its inverse map π β1 : πΌπ(π) β π such that
π β1 β π = πΌπ. We need to prove that 1) πΌπ(πΏ) = βπ and 2) πΏβ1 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 βπ with dimension π (it is
entire βπ itself).
2) Prove that πΏβ1 (π1 π¦1 + π2 π¦2 ) = π1 πΏβ1 (π¦1 ) + π2 πΏβ1 (π¦2 ).
Indeed, apply πΏ to both sides and get πΏ[πΏβ1 (π1 π¦1 + π2 π¦2 )] = πΏ[π1 πΏβ1 (π¦1 ) + π2 πΏβ1 (π¦2 )],
which is true because both parts are equal to π1 π¦1 + π2 π¦2 .
Because πΏ is one-to-one, this means the arguments of πΏ are the same, Q.E.D.

b.), c.) Denote matrix for πΏ as π΄ and matrix for πΏβ1 as π΅ and prove that
matrix for πΏβ1 πΏ is π΅π΄. This will prove that π΅π΄ = πΌ, so π΄ is invertible
and that the matrix for πΏβ1 is π΄β1 .
In other words, we need to prove that mu...

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Anonymous
Thanks, good work

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