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Math Help - m equations, rank of m, a solution exists?

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    Member Last_Singularity's Avatar
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    m equations, rank of m, a solution exists?

    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.
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    Quote Originally Posted by Last_Singularity View Post
    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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