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?
 A: 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,
A: Because of two things:


*

*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.


*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).


A: 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 $I$.

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