m equations, rank of m, a solution exists?

**Last_Singularity** · October 15th 2009, 06:00 PM

Prove or provide counterexample: if the coefficient matrix of a system of $m$ linear equations in $n$ unknowns has rank $m$ , then the system has a solution.

I have an idea for the proof but I am not sure how to formalize it. Some advice, please?

Here's the proof: let $A$ be the coefficient matrix in the system $Ax=b$ . If $A$ has rank of $m$ , then $n \geq m$ . Say that's not true - then $n < m$ , implying that the maximum rank of $A$ which equivalent to the max number of linearly independent columns, has a maximum of $n$ , which is less than $m$ by assumption, meaning that the rank will never equal $m$ . So we must have $n \geq m$ . Then if the rank of $A$ is $m$ , we can reduce the matrix [A|b] using row operations to a form [A'|b'] where all $x_i,i=1,...,m$ have coefficients of $1$ . Then a solution becomes $x_i=b'_i$ for $i=1,...,m$ and for $i=m+1,...,n$ , we have $x_i = 0$ .

**NonCommAlg** · October 15th 2009, 06:46 PM

Originally Posted by Last_Singularity

Prove or provide counterexample: if the coefficient matrix of a system of $m$ linear equations in $n$ unknowns has rank $m$ , then the system has a solution.

I have an idea for the proof but I am not sure how to formalize it. Some advice, please?

Here's the proof: let $A$ be the coefficient matrix in the system $Ax=b$ . If $A$ has rank of $m$ , then $n \geq m$ . Say that's not true - then $n < m$ , implying that the maximum rank of $A$ which equivalent to the max number of linearly independent columns, has a maximum of $n$ , which is less than $m$ by assumption, meaning that the rank will never equal $m$ . So we must have $n \geq m$ . Then if the rank of $A$ is $m$ , we can reduce the matrix [A|b] using row operations to a form [A'|b'] where all $x_i,i=1,...,m$ have coefficients of $1$ . Then a solution becomes $x_i=b'_i$ for $i=1,...,m$ and for $i=m+1,...,n$ , we have $x_i = 0$ .

i guess that's fine. i'd solve the problem like this: suppose the entries of $A$ come from a field $F.$ let $v_1, \cdots , v_n \in F^m$ be the columns of $A.$ without loss of generality we may assume that $v_1, \cdots , v_m$

are linearly independent. now, since $\dim_F F^m = m,$ the vectors $v_1, \cdots , v_m$ span $F^m.$ so if $b \in F^n,$ then $b=c_1v_1 + \cdots + c_m v_m,$ for some $c_j \in F.$ let $x_j = c_j$ for $1 \leq j \leq m$ and $x_j = 0$ otherwise.

clearly the $n \times 1$ vector $x$ with the entries $x_j$ satisfies $Ax=b.$

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