# Thread: proof involving rank of a matrix

1. ## proof involving rank of a matrix

I have to prove the following statement, or provide a counterexample:
If the coefficient matrix of a system of m linear equations in n unknowns has rank m, then the system has a solution.
My thought is that this statement is true, and I think the proof would use the following theorem:
Let Ax=b be a system of linear equations. Then the system is consistent (the solution set is nonempty) if and only if rank(A) = rank(A|b).

Is it true that if an mxn matrix has rank m as assumed that the rank of the corresponding augmented matrix (A|b) would ALWAYS have the same rank?

2. Originally Posted by dannyboycurtis
I have to prove the following statement, or provide a counterexample:
If the coefficient matrix of a system of m linear equations in n unknowns has rank m, then the system has a solution.
My thought is that this statement is true, and I think the proof would use the following theorem:
Let Ax=b be a system of linear equations. Then the system is consistent (the solution set is nonempty) if and only if rank(A) = rank(A|b).

Is it true that if an mxn matrix has rank m as assumed that the rank of the corresponding augmented matrix (A|b) would ALWAYS have the same rank?

The claim is false and as an easy example take the map $\displaystyle \mathbb{R}\rightarrow\mathbb{R}^2$ defined by $\displaystyle r\rightarrow\left(\begin{array}{c}r\\r\end{array}\ right)$ . Clearly the system:

$\displaystyle \left(\begin{array}{c}1\\1\end{array}\right)(r)=\l eft(\begin{array}{c}b_1\\b_2\end{array}\right)$

has no solution if $\displaystyle b_1\neq b_2$

Tonio