If Ax = 0 has one solution, show m>=n

[WY]

Hey

I'm not to sure on how to approach this problem - I think I have to use induction but I don't know where to start!
Question: Let A be an mxn matric. Suppose that Ax=0 has only one solution, namely x=0.
Show that m >= n


Thanks for the help in advance

 

[Yegor]

Dimension theorem for Matrices says if matrix has n columns => rank(A)+nullity(A)=n
In your case nullity(A)=dimKer(A)=0
Is it possible rank(A) to be > min(m,n)?

 

[WY]

Thanks for replying :)
But I'm not to sure what you mean - i don't think i am very familiar with that theorem... could you elaborate a little? hehehe

 

[OlderDan]

Thanks for replying :)
But I'm not to sure what you mean - i don't think i am very familiar with that theorem... could you elaborate a little? hehehe

If you are unfamiliar with those formal properties and want to approach the problem from the basics, you could start by writing simple cases

<br /> \left[ {\begin{array}{*{20}c}<br /> a &amp; b \\<br /> \end{array}} \right]\left[ {\begin{array}{*{20}c}<br /> {x_1 } \\<br /> {x_2 } \\<br /> \end{array}} \right] = 0<br />


<br /> \left[ {\begin{array}{*{20}c}<br /> {\begin{array}{*{20}c}<br /> a &amp; b &amp; c \\<br /> \end{array}} \\<br /> {\begin{array}{*{20}c}<br /> d &amp; e &amp; f \\<br /> \end{array}} \\<br /> \end{array}} \right]\left[ {\begin{array}{*{20}c}<br /> {x_1 } \\<br /> {x_2 } \\<br /> {x_3 } \\<br /> \end{array}} \right] = 0<br />

It's pretty easy to show that these equations can be satisfied with non-zero x vectors. From a linear equations perspective, it is matter of having too many unkowns and not enough equations.