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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

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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.

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