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Can anyone help me to prove this by using rref:

if A is a square matrix not invertible, then the equation Ax=0 has a non trivial solution.

Thank you.

Amine El Bouzidi
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    If the linear application $f_A$ associated to $A$ is not invertible it is not injective, hence $f_A(u)=f_A(v)$ for some $u\neq v$ and $u-v \in \text{ker} A$. – Jack D'Aurizio Aug 15 '18 at 22:48
  • well, i want to show this without using the linear application associate to A, thank you my friend for your answer. – Amine El Bouzidi Aug 15 '18 at 22:52

2 Answers2

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Jack D'Aurizio's comment is correct.

A matrix is invertible if and only if it is injective. If $A$ is not injective, then $Au = Av$ for two distinct vectors $u$ and $v$. By linearity, $Ax=0$ where $x = u - v \neq 0$.

parsiad
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If the matrix $A$ (of dimension $n$) is not invertible, there is a positive integer $r$ such that the rref of $A$ can be transformed (swapping some columns, which amounts to re-numbering the unknowns) into a block triangular matrix: $$\begin{bmatrix} I_{n-r}&B\\0& 0_{r} \end{bmatrix},$$ where $I_{n-r}$ is the unit matrix of dimension $n-r$, $ 0_{r}$ the null square matrix of dimension $r$ and $B$ is a $n-r\times r$ matrix. So we obtain the subspace $S$ of solutions is defined by the equations: \begin{align} &x_1+b_{11}x_{n-r+1}+\dots+b_{1r}x_{n}=x_2+b_{21}x_{n-r+1}+\dots+b_{2r}x_{n}\\{}={}&\dots=x_{n-r}+b_{n-r\:1}x_{n-r+1}+\dots+b_{n-r\:r}x_{n}=0, \end{align} form which we deduce readily $x_1,\dots, x_{n-r}$ as linear functions $\ell_1,\dots,\ell_{n-r}$ of the $r$ free variables $ x_{n-r+1},\dots ,x_{n}$.

This let us define an isomorphism from $K^r$ ($K$ is the base field) to this subspace: \begin{align} K^r&\longrightarrow S\\ (x_{n-r+1},\dots ,x_{n})&\longmapsto (\ell_1,\dots,\ell_{n-r},x_{n-r+1},\dots ,x_{n}). \end{align}

Bernard
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