# Linear Independence and Row Vectors of a Matrix

Is the following statement true?

Given an $n \times m$ matrix A, if the system $Ax=b$ has a unique solution for all b then the column vectors of A must be linearly independent.

I know that in order for $Ax=b$ to have a solution $b \in C(A)$, where C(A) represents the column space of matrix A.

My attempt: Statement is true because in order to have a unique solution for every b A will have a leading/pivot 1 in row-reduced form, thus implying none of the rows or columns were scalar multiples of each other

That's one way to look at it. Another way that's more immediate from the definitions of linear independence and scalar multiplication is to observe the following fact about matrix multiplication: if we have a matrix $$A = \big(v_1 | v_2 | \ldots |v_n\big)$$ i.e. with column vectors $v_1, \ldots, v_n$, and we multiply by a column vector $$x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n\end{pmatrix},$$ then $Ax = x_1 v_1 + x_2 v_2 + \ldots + x_n v_n$. That is, multiplying a matrix $A$ by a column vector $x$ yields a linear combination of the column vectors of $A$, where the scalars are the entries of $x$.
How can we use this? Well, if $Ax = b$ has a unique solution for every $b$, then $Ax = 0$ has a unique solution. Writing $A$ and $x$ as we did above, this means that the equation $$x_1 v_1 + x_2 v_2 + \ldots + x_n v_n = 0$$ has a unique solution: $x_1 = x_2 = \ldots = x_n = 0$. This is the definition of linear independence.