I have a question and i hope that i will find the answer her,

So we all know that the reduced row echelon form for a invertible matrix is the matrix $I_n$, So what about the reduced row echelon form for a non invertible matrix, i want basically to know if there is a specific form for the last one. Thank you in advance.

  • $\begingroup$ Well, any rref that’s not the identity matrix corresponds to a non-invertible matrix. What does the definition of row-reduced echelon form require of its structure? $\endgroup$ – amd Aug 15 '18 at 22:22
  • $\begingroup$ well, i want to show if a sqaure matrix A is not invertible, then the equation Ax=0 has a non trivial solution and i want to do this by the fact that the rref of A is not equal to $I_n$. $\endgroup$ – Amine El Bouzidi Aug 15 '18 at 22:26
  • $\begingroup$ In the general case the rref is a block-diagonal matrix, where the first block is unit matrix $I_r$ with $r\le n$, and the secon diagonal block is the null matrix of dimension $n-r)$, $\:0_{n-r}$. $\endgroup$ – Bernard Aug 15 '18 at 22:34

RREF of any $n\times n$ non-invertible matrix (with dimension of row or column space $m$) has the form as

$$\begin{bmatrix}1 & 0 &0 &\cdots&0&* &\cdots&*&*\\ 0&1&0&\cdots&0&* &\cdots&*&*\\ 0&0&1 &\cdots&0&* &\cdots&*&*\\ \vdots &\vdots &\vdots &\ddots & \vdots &\vdots &\vdots &\vdots &\vdots\\ 0&0&0&\cdots&1 &*&\cdots&*&* \\0 &0&0 &\cdots &0 &0 &\cdots &0&0\\\vdots &\vdots &\vdots &\ddots & \vdots &\vdots &\vdots &\vdots &\vdots \\0 &0&0 &\cdots &0 &0 &\cdots &0&0 \end{bmatrix} $$

where this matrix has $m$ non zero rows and $*$ can be any number.

This form is modified RREF. We can get this form after arranging columns of the original RREF.

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  • $\begingroup$ Thank you Lev Ban for your answer. $\endgroup$ – Amine El Bouzidi Aug 15 '18 at 23:16

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