3 Applied Linear Algebra questions, Linear Transformations

timer Asked: Oct 18th, 2018
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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.

3 Applied Linear Algebra questions, Linear Transformations
3 Applied Linear Algebra questions, Linear Transformations
3 Applied Linear Algebra questions, Linear Transformations
3 Applied Linear Algebra questions, Linear Transformations

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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 𝐿: 𝑉 β†’ π‘Š is one-to-one if and only if πΎπ‘’π‘Ÿ(𝐿) = 0𝑉 .
Proof. If and only if means two things:
1) If 𝐿 is one-to-one then πΎπ‘’π‘Ÿ(𝐿) = 0𝑉 and 2) If πΎπ‘’π‘Ÿ(𝐿) = 0𝑉 then 𝐿 is one-to-one.
Both will get proof by a contradiction.
1) Let 𝐿 be one-to-one but πΎπ‘’π‘Ÿ(𝐿) β‰  0𝑉 , i.e. there is a nonzero vector 𝑣0 ∈ πΎπ‘’π‘Ÿ(𝐿), i.e.
𝐿(𝑣0 ) = 0π‘Š . But then 𝐿(0𝑉 ) = 𝐿(π‘₯0 ) while 0𝑉 β‰  π‘₯0 ,
which is a contradiction with 𝐿 to be one-to-one, Q.E.D.
2) Let πΎπ‘’π‘Ÿ(𝐿) = 0𝑉 but 𝐿 is not one-to-one. This means there are π‘₯1 β‰  π‘₯2 : 𝐿(π‘₯1 ) = 𝐿(π‘₯2 ).
But then 𝐿(π‘₯1 βˆ’ π‘₯2 ) = 𝐿(π‘₯1 ) βˆ’ 𝐿(π‘₯2 ) = 0, so π‘₯1 βˆ’ π‘₯2 ∈ πΎπ‘’π‘Ÿ(𝐿), but π‘₯1 βˆ’ π‘₯2 β‰  0𝑉 .
This is a contradiction, Q.E.D.

a.) Each one-to one mapping 𝑓: π‘ˆ β†’ 𝑉 has its inverse map 𝑓 βˆ’1 : πΌπ‘š(𝑓) β†’ π‘ˆ such that
𝑓 βˆ’1 ∘ 𝑓 = 𝐼𝑑. We need to prove that 1) πΌπ‘š(𝐿) = ℝ𝑛 and 2) πΏβˆ’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. 𝑛). And there is only one subspace of ℝ𝑛 with dimension 𝑛 (it is
entire ℝ𝑛 itself).
2) Prove that πΏβˆ’1 (π‘Ž1 𝑦1 + π‘Ž2 𝑦2 ) = π‘Ž1 πΏβˆ’1 (𝑦1 ) + π‘Ž2 πΏβˆ’1 (𝑦2 ).
Indeed, apply 𝐿 to both sides and get 𝐿[πΏβˆ’1 (π‘Ž1 𝑦1 + π‘Ž2 𝑦2 )] = 𝐿[π‘Ž1 πΏβˆ’1 (𝑦1 ) + π‘Ž2 πΏβˆ’1 (𝑦2 )],
which is true because both parts are equal to π‘Ž1 𝑦1 + π‘Ž2 𝑦2 .
Because 𝐿 is one-to-one, this means the arguments of 𝐿 are the same, Q.E.D.

b.), c.) Denote matrix for 𝐿 as 𝐴 and matrix for πΏβˆ’1 as 𝐡 and prove that
matrix for πΏβˆ’1 𝐿 is 𝐡𝐴. This will prove that 𝐡𝐴 = 𝐼, so 𝐴 is invertible
and that the matrix for πΏβˆ’1 is π΄βˆ’1 .
In other words, we need to prove that mu...

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Thanks, good work

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