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

1. Jun 8, 2005

### 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

2. Jun 8, 2005

### Yegor

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

3. Jun 8, 2005

### WY

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

4. Jun 8, 2005

### OlderDan

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

$$\left[ {\begin{array}{*{20}c} a & b \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} {x_1 } \\ {x_2 } \\ \end{array}} \right] = 0$$

$$\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} a & b & c \\ \end{array}} \\ {\begin{array}{*{20}c} d & e & f \\ \end{array}} \\ \end{array}} \right]\left[ {\begin{array}{*{20}c} {x_1 } \\ {x_2 } \\ {x_3 } \\ \end{array}} \right] = 0$$

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.