A question in my textbook, asks the following:

Determine, with proof, whether the following subset of $M(n,n)$ is a subspace: The subset of matrices that are in row echelon form.

To clarify (in case this notation is non standard) $M(n,n)$ is the set of all $n\times n$ matrices.

The answers in the rear of the textbook indicate it is NOT a subspace, but provides zero reasoning. Any help or insight would be appreciated. Thank you.

  • $\begingroup$ Other than being a subspace of $M(n,n)$, does it satisfy any of the requirements for a subspace? $\endgroup$ – Robert Israel Nov 4 '16 at 6:59
  • 1
    $\begingroup$ Look up the definition of subspace. Hint: a subspace is a space by itself. $\endgroup$ – SAJW Nov 4 '16 at 7:00
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Nov 4 '16 at 7:00
  • $\begingroup$ Thanks very much for the formatting tips. Also for the other replies, that's great! $\endgroup$ – Alex Devries Nov 4 '16 at 7:51

Consider $$\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} + \begin{bmatrix}-1 & -1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$$ Both matrices on the left are in row echelon form, but their sum is not.

  • $\begingroup$ So, you just showed a counter example that proves that the sum of two matrices that ARE in row echelon form don't necessarily form a matrix that is itself in row echelon form. To clarify my own understanding, is this because it's not closed under addition, per the criteria of a subspace? $\endgroup$ – Alex Devries Nov 4 '16 at 7:58
  • $\begingroup$ @AlexDevries Yes subspaces need to be closed under addition. $\endgroup$ – stupid_question_bot Nov 4 '16 at 15:33

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