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Is this proposition true?

Suppose that the columns of a matrix A are linearly independent. Then Ax=0 has only the trivial solution.

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    $\begingroup$ The following holds in general: $Ax = \sum_i x_ia_i$, where the $a_i$ are the columns of $A$ and the $x_i$ are the entries of $x$. $\endgroup$ Commented Apr 2, 2017 at 9:03
  • $\begingroup$ The best way to think about multiplying a matrix by a vector is this: $Ax$ is a linear combination of the columns of $A$. $\endgroup$
    – littleO
    Commented Apr 2, 2017 at 11:40

2 Answers 2

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$$A\mathbf{x}= 0$$ $$\begin{bmatrix}a_{1,1}&a_{1,2} &... &a_{1,m}\\a_{2,1} &&& a_{2,m} \\ : &&&: \\ a_{n,1} & a_{n,2}& ...& a_{n,m}\end{bmatrix} \begin{bmatrix}x_1\\x_2\\:\\x_m\end{bmatrix} = 0_v$$ Let $\mathbf{a}_j$ be the $j$th column of $A$ and $0_v$ the zero vector. Hence, for all $c_j \ne 0$, $\sum_{j=1}^{n}c_j\mathbf{a}_{j} \ne 0_v$ since the columns are independent. Computing the product above yields, $$x_1\begin{bmatrix} a_{1,1}\\:\\a_{,1} \end{bmatrix} + x_2\begin{bmatrix} a_{1,2}\\:\\a_{n,2} \end{bmatrix} ... x_m\begin{bmatrix} a_{1,m}\\:\\a_{n,m} \end{bmatrix} =0_v$$ Because $\sum_{j=1}^{m}c_j\mathbf{a}_{j} \ne 0_v$ for all $c_j \ne 0$ that means the only possible values for $x_i$ is $0$ for all $x_i$. Therefore the only solution is the trivial solution where $\mathbf{x} = 0_v$

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  • $\begingroup$ Isn't "m" to go above both the Σs in the text, rather than "n"? $\endgroup$
    – Kyotomath
    Commented Apr 2, 2017 at 9:54
  • $\begingroup$ It seems not to be a correct proof. The negation of "all $c_j$s are non-zero" is not "all $c_j$ is zero. $\endgroup$
    – Kyotomath
    Commented Apr 2, 2017 at 10:01
  • $\begingroup$ I think that "for all non-zero constants $c_j$s " should be corrected into "not all $c_j$s are zero". $\endgroup$
    – Kyotomath
    Commented Apr 2, 2017 at 10:07
  • $\begingroup$ @Kyotomath I edited it following your concerns. $\endgroup$ Commented Apr 2, 2017 at 11:36
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If the columns of $A$ are linearly independent than $\det(A)\ne 0$ and $A$ is invertible. So, yes, we have $ A^{-1}Ax=A^{-1}0 \iff x=0$

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    $\begingroup$ Doesn't it have to be proved in the case that A is not a square matrix? $\endgroup$
    – Kyotomath
    Commented Apr 2, 2017 at 9:46

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