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Thread: systems of linear equations

  1. #1
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    systems of linear equations

    Let A be some $\displaystyle m \times n$ matrix with entries in $\displaystyle \mathbb{R}$. 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 $\displaystyle x \neq 0$?
    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 $\displaystyle (x_1,x_2,...,x_n)$. 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 $\displaystyle m \times n$ matrix with entries in $\displaystyle \mathbb{R}$. 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 $\displaystyle x \neq 0$?
    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 $\displaystyle (x_1,x_2,...,x_n)$. But I can't quite see how this would give a non-zero solution.
    There is another way to do this assuming you switched $\displaystyle m$ and $\displaystyle n$. It's a little hight powered so, if it's unfamiliar looking ignore it:

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

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

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

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

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

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

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

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

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

    required.
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