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Determine whether the following statement is true or false:

Given $A$ $∈$ $M_{m×n}$($\mathbb{R}$) and the zero vector $0$ ∈ $\mathbb{R}^m$, the system of linear equations $A$$\mathbf{x}$ = $0$ always have more than one solution for x

I want to say this is false, but I don't know how to start this proof. Help would be much appreciated.

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3 Answers 3

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Hint: what if $A$ has an inverse? What's the simplest example thereof?

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  • $\begingroup$ I hate to sound ignorant, but how would the existence of $A^{-1}$ help me? $\endgroup$
    – darylnak
    Jul 6, 2017 at 9:01
  • $\begingroup$ UPDATE: I found the chapter in my textbook to take advantage of your hint! Thank you. $\endgroup$
    – darylnak
    Jul 6, 2017 at 9:07
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If you want to prove the statement is false, you only need to find one matrix $A$ and one setting of $m$ and $n$ such that the system only has one solution.

So my advice is to first try a simple setting. For example, $m=n=1$. It should be easy to find a counterexample there.

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By the rank nullity theorem $$\text{rank } A + \text{nullity }A = n $$ If $m<n$, a nonzero solution always exists (since $\text{rank }A\leq m$ and $n-m> 0$). For $m=n$, if we have an inverse (i.e. full ranked matrix), then the only solution possible is $A^{-1}0=0$. For $m>n$ again by rank nullity theorem we can argue similarly depending upon the rank of the matrix.

I dont think there is enough info. a straight "true" or "false" answer to this

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