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Math Help - systems of linear equations

  1. #1
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    systems of linear equations

    Let A be some m \times n matrix with entries in \mathbb{R}. Prove the following:
    a) A system of equations with Ax=0 with m<n has a non-trivial solution.

    Firstly does a non-trivial solution (in this case) mean a solution where x \neq 0?
    As for working out the question all I can think of is that if m<n, then there are more equations than there are variables, with the variables being (x_1,x_2,...,x_n). But I can't quite see how this would give a non-zero solution.
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  2. #2
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    You are correct in that a non-trivial solution is a nonzero vector x that satisfies the system. However, if m < n, then there are more variables than equations, not the other way around. See here for a very similar thread. That is, look at my post # 2. How could you adapt that idea for your problem here?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by worc3247 View Post
    Let A be some m \times n matrix with entries in \mathbb{R}. Prove the following:
    a) A system of equations with Ax=0 with m<n has a non-trivial solution.

    Firstly does a non-trivial solution (in this case) mean a solution where x \neq 0?
    As for working out the question all I can think of is that if m<n, then there are more equations than there are variables, with the variables being (x_1,x_2,...,x_n). But I can't quite see how this would give a non-zero solution.
    There is another way to do this assuming you switched m and n. It's a little hight powered so, if it's unfamiliar looking ignore it:

    Spoiler:
    Let our equations be represented by \varphi_1(v)=0,\cdots,\varphi_m(v)=0 where \varphi_k\in\text{Hom}\left(\mathbb{R}^n,\mathbb{R  }\right)

    (the dual space). Note then that since m<n we have that

    \dim \text{Ann}\left(\text{span}\{\varphi_1,\cdots,\var  phi_m\}\right)=n-m>0

    (where \text{Ann}\left(\text{span}\{\varphi_1,\cdots,\var  phi_m\}\right)\subseteq\left(\text{Hom}\left(\math  bb{R}^n,\mathbb{R}\right)\right)^{\ast} is the set of all linear functionals on

    \text{Hom}\left(\mathbb{R}^n,\mathbb{R}\right) which map \text{span }\{\varphi_1,\cdots,\varphi_m\} to \{0\})and so there exists some non-zero

    \Psi\in\text{Ann}\left(\text{span}{\varphi_1,\cdot  s,\varphi_m\}\right). Thus, \Psi(\varphi_1)=\cdots=\Psi(\varphi_m)=0. Recall though that the natural identification

    f:\mathbb{R}^n\to\left(\text{Hom}\left(\mathbb{R}^  n,\mathbb{R}\right)\right)^{\ast}:v\mapsto \Psi_v

    (where \Psi_v is the evaluation functional \Psi_v(\varphi)=\varphi(v)) is an isomorphism and so in particular

    \Psi=\Psi_{v_0} for some v_0\in\mathbb{R}^n. In particular we have that \varphi_1(v_0)=\cdots=\varphi_m(v_0)=0 as

    required.
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