Reduced row echelon form of a square matrix

[Bipolarity]

I am wondering about the relation betwen RRE forms and identity matrices. Consider the reduced row echelon form of any square matrix. Must this reduced row echelon form of the matrix necessarily be an identity matrix?

I would suppose yes, but can this fact be proven? Could anyone provide an outline of the proof, or provide the link? Thanks much.

BiP

 

[Erland]

I am wondering about the relation betwen RRE forms and identity matrices. Consider the reduced row echelon form of any square matrix. Must this reduced row echelon form of the matrix necessarily be an identity matrix?

Of course not. As a trivial example, take a square zero matrix, i.e. a square matrix such that all its elements are zeros. Or, more generally, any square marix with at least one zero row, or column. In fact, you can easily write down lots of square RRE matrices which are not identity matrices.

In general, a square matrix A is row equivalent to (i.e. its RRE is) the identity matrix of he same size if and only if A is invertible.

 

[Bipolarity]

Of course not. As a trivial example, take a square zero matrix, i.e. a square matrix such that all its elements are zeros. Or, more generally, any square marix with at least one zero row, or column. In fact, you can easily write down lots of square RRE matrices which are not identity matrices.

In general, a square matrix A is row equivalent to (i.e. its RRE is) the identity matrix of he same size if and only if A is invertible.

What if I add the condition that the matrix square has no zero rows? Then is it necessarily the case that its RRE form is equivalent to the identity matrix (of the same size)?

BiP

 

[AlephZero]

What if I add the condition that the matrix square has no zero rows? Then is it necessarily the case that its RRE form is equivalent to the identity matrix (of the same size)?

No, for example
$$\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$$

 

[Bipolarity]

No, for example
$$\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$$

But how is that matrix in RRE form? The leading 1 in the second row is not strictly to the right of the leading 1 of the first row?

BiP

 

[AlephZero]

Of course it's not in RRE form!

You asked if a square matrix with no zero rows always has an identity matrix for its RRE. That matrix has no zero rows. Reduce that matrix to RRE form and see what you get.

If you do that yourself, you might see WHY your idea is wrong (and even discover the right idea), which is more useful than just being told "your idea is wrong".

 

[Vargo]

AlephZero is saying to start with that matrix and then do row operations to put it into RRE form. You will find that you end up with a matrix that is not the identity matrix. Since the given matrix has no zero rows, it is a counter example to your modified question.

 

[Bipolarity]

I see! Thanks! The reduction gave me $$\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}$$

What about if the RRE form of the matrix is a square matrix with no zero rows? In that case is the RRE form become an identity matrix?

BiP

 

[Erland]

What about if the RRE form of the matrix is a square matrix with no zero rows? In that case is the RRE form become an identity matrix?

Yes, that's right. It is easily verified if we carefully examine the definition of RRE and its consequences in the case of a square matrix. What about pivot rows and columns and zero rows in that case?