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Show that the product of two reduced row-echelon matrices is also reduced row-echelon.

That's what I think:

  1. A reduced row-echelon matrix has columns like $e_1 =(1, 0, \cdots , 0)^T$ and $e_2 =(0, 1, 0, \cdots , 0)^T$.
  2. For columns in between $e_n$ and $e_{n+1}$, only the first $n$ entries will be non-zero.

By noticing these two, I can 'imagine' that the product should be reduced row-echelon. But I cannot write down a clear proof for that. Or say, I don't even know how to start my proof. Can someone give me a helping hand?

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    $\begingroup$ Your comment basically is the proof. Given $$AB=C$$, we know that for any column in $C$, non-zero entries are determined by the non-zero column entries in $B$, per matrix multiplication. For any row in $C$, the non-zero entries are determined by the non-zero row entries in $A$, as per matrix multiplication. Since a reduced row-rechelon matrix is upper triangular, the product matrix must be as well, by above. Effectively, the output matrix $C$ will only have non-zero entries in the elements which are non-zero in both $A$ and $B$. $\endgroup$ – FundThmCalculus Jun 2 '15 at 14:10

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