# Why is reduced echelon form of a matrix always the identity matrix

If $A$ is $3\times3$ matrix with linearly independent columns, then why is the reduced echelon form the identity matrix?

• My professor posted this question for our exam review. Feb 12, 2015 at 19:58
• Duplicate of math.stackexchange.com/questions/977819/… Feb 12, 2015 at 20:03
• While this is technically a duplicate, I don't think it should necessarily be marked as such. It is quite possible that the OP doesn't realise that having independent columns means that necessarily the rows are independent - others may not also. (Of course, it is the case, but this does actually take some thinking.) Alternatively, could be marked as such and a note put on the other question... Feb 12, 2015 at 20:17

If the columns of the matrix are linearly independent this means that they span the vector space. We can think of the matrix as a linear transformation T: since the dimImT = dimV, we have that the transformation is onto and injective, thus invertible.

Now, the identity matrix is also a representation of a linear transformation and we can think about it as a vector space with dimension equal to that of T. Thus we have an isomorphism between the identity matrix and the original matrix.

Lastly, define a series of row operations which act on either matrix to achieve the second. These operations are linear and thus do not change the image of the transformations.

Hope this helps,

Because of two things:

1. Row operations preserve column dependencies and independencies in matrices.

• E.g. if $A=[\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3]$ has the column dependency $\mathbf{a}_1+2\mathbf{a}_2=\mathbf{a}_3$, and $B=[\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3]$ is row-equivalent to $A$, then $\mathbf{b}_1+2\mathbf{b}_2=\mathbf{b}_3$ also.

To prove this, we just check it holds each of the row operations.

2. The identity matrix is the only matrix in reduced row echelon form with linearly independent columns.

• In any other reduced row echelon form matrix, any non-zero column without a leading entry can be written as a linear combination of other columns (a zero column is linearly dependent in itself).

## Method 1

Since all the columns are independent, the determinant is non-zero. So we see that there must be at least one non-zero entry. Swap rows and columns so that this is in the top left, then divide the entire row (or column) by this value. This gives a 1 in the top left. We can then 'clear' the top row and left column by standard row/column operations (if entry (1,2) is 3, then subtract 3 times the first column from the second column, etc).

We know that swapping rows/columns only changes the determinant by a sign, and dividing through by a (finite!) number changes the absolute value, but not whether it is zero or not. So the determinant of the 2x2 matrix in the bottom right must be non-zero (as the new determinant is 1 times this). So we can do the same for this, observing that any row/column operations have no effect on the first row/column, since you're adding/subtracting (multiples of) 0.

So do this and we get the identity. :)

Similarly, for any $n$x$n$ matrix with independent rows (non-zero determinant), we have that the reduced echelon form is the identity. We argue this by induction: do it once as explained above, then since the determinant remains non-zero, apply the inductive argument.

## Method 2

We know that the top left of a reduced echelon form is always the identity matrix, then always zeros. That is, we have the form $$\left( \begin{matrix} I & 0 \\ 0 & 0 \end{matrix} \right).$$ We know that the size of the matrix $I$ (that is, 'dimension': number of columns/rows) is the rank of the matrix (and the number of remaining rows/columns is the dimension of the kernel - this is guaranteed by the rank-nullity theorem). But, since we have independent columns, we know that the kernel is trivial, ie just the element $0$, and the image/range is full, ie given any $y$ there exists $u$ such that $Au = y$, where $A$ is the original matrix ($A$ is invertible, so $x = A^{-1}y$). Thus $I$ has 3 rows/columns (maximal) and the final $0$ has dimension 0 (just the point $0$). Thus the reduced echelon form is just $I$.

If anything isn't clear, just give me a shout and I'll elaborate!

If this has helped, then please remember to upvote and/or accept!