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Proof of the Day 1: Surjective Linear Map

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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This entry was posted on Wednesday, April 14th, 2010 at 10:57 am and is filed under POTD. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

5 Responses to “Proof of the Day 1: Surjective Linear Map”

  1. TomHayden3.com Says:
    April 15th, 2010 at 4:55 pm

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  2. Manolis Says:
    April 19th, 2010 at 4:50 pm

    Excuse me but I will just post it as I would write it in LaTeX.

    There is a small mistake in your proof. You cannot assume, that $T(x)=(5x_1-x_2,7x_3-x_4)$, since it’s not the only mapping with null space $\{(x_1,x_2,x_3,x_4):x_1=5x_2,x_3=7x_4\}$. For example, $T=(10x_1-2x_2,-21x_3+3x_4)$ could also work. Moreoer, you can also try to explain more why your arguements hold.

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