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Proof of the Day 1: Surjective Linear Map

Wednesday, April 14th, 2010

This is my very first proof of the day, so the proof is probably wrong (although, I hope not).

Source: Linear Algebra Done Right by Sheldon Axler – p.59 Question #9

Prerequisite: Some advanced linear algebra

Claim: Prove that if T is a linear map from F^4 to F^2 such that null T = \{(x_1,x_2,x_3,x_4): x_1 = 5x_2, x_3 = 7x_4 \}, then T is surjective.

Idea: In the problem, they give us a very specific null vector of T.  Since we know this, we can derive some information about T. Then, we can construct a vector, put it through the function and show that it is surjective by construction by looking at the dimension of of the input and output.

Proof: We know that null T = \{(x_1,x_2,x_3,x_4): x_1 = 5x_2, x_3 = 7x_4 \} and that T \in L(\mathbb{R}^4,\mathbb{R}^2), and want to show that T is surjective.  By definition, for T to be surjective, its range must be equal to the codomain of the linear map. Therefore, if range T = \mathbb{R}^2, then T is surjective. To show this, let \bar{x} = (x_1,x_2,x_3,x_4) \in \mathbb{R}^4. Then, we know that T(\bar{x}) = (5x_1 - x_2, 7x_3 - x_4).  Since \bar{x} \in \mathbb{R}^4 and the dimension of the resulting mapping is 2, we have shown that T = \mathbb{R}^2 and hence is surjective.

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My Blog - I finally gave in and created a blog where I can post about whatever I like.

My Professional CV - This site has all of the relevant professional links about me; go here if you're interested in my academics.

Fun SI Projects Using Bidding Networks to Search for Exposure in Auctions - Auction 73 Case - This is some work I did in Fall 2008, as a final project for my Networks course at SI. I'm currently trying to see if this is publishable.

Technological Diffusion with Compatibility - This is based off of a model presented at one of Umichigan's STIET lectures this year.