Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A square matrix is said to be $weakly \ elementary$ if it is either elementary or it is obtained from an identity matrix by replacing one diagonal entry by zero.
Prove that every square matrix is a product of weakly elementary matrices. You may use that every invertible matrix is a product of elementary matrices.

My Attempt
Now if for example we took the case of $3\times 3$ sqaure matrices, weakly dominated implies that either, the matrix is an elementary matrix or one of these three;
$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right)$, $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$,$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$.

We know and are told in the question that all invertible matrices are products of elementary matrices and so are products of weakly elementary bases by definition.

All that left to prove is that non invertible matrices can be made from products of weakly elementary bases....
Im am not sure on how to proceed after this...
For this case, are all $3\times 3$ matrices products of ;
$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right)$, $\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$,$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)$. ??

Any help would be much appreciated.

share|improve this question

1 Answer 1

Every square matrix $A$ is equivalent to a matrix Dg$[I_r,0]$ with $r$ the rank of $A$, or put differently: $A=P$ Dg$[I_r,0]Q$ with $P$ and $Q$ invertible. So $P$ and $Q$ are products of elementary matrices and Dg$[I_r,0]$ is a product of weakly elementary matrices.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.