I'm having troubling finding an example/proof that the following statement is true.

Row operations on a matrix A can change the linear dependence relations among the rows of A.

Any tips on understanding this would be greatly appreciated.

  • 2
    $\begingroup$ It's not true. Row operations are, by design, reversible linear combinations of the rows of $A$. So the span of the rows remains the same after each operation. $\endgroup$ – Matthew Leingang Apr 11 '17 at 19:06
  • $\begingroup$ What do you mean by the "linear dependence relations"? $\endgroup$ – ancient mathematician Apr 11 '17 at 19:10
  • $\begingroup$ @ancientmathematician I think it means that if the rows are linearly dependent, row operations can make them linearly independent, and vice versa. $\endgroup$ – cdignam Apr 11 '17 at 19:15
  • $\begingroup$ See math.stackexchange.com/q/2236376/265466 for additional context for the highlighted statement. $\endgroup$ – amd Apr 16 '17 at 7:05

What is meant by this is that, even though the span of the rows is unchanged by elementary row operations, the various individual linear dependence relations among subsets of the rows can and do change. A fairly simple example is $$\begin{bmatrix}1&1\\1&0\\0&1\end{bmatrix}\to\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}.$$ In the original matrix, the rows are pairwise independent, but in its rref the first/last and second/last pairs are linearly dependent. Another simple but illustrative example is $$\begin{bmatrix}1&0\\1&0\\0&1\end{bmatrix}\to\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}.$$ The first two rows in the original matrix are linearly dependent but are independent in its rref.

More generally, if the first $k$ rows of the rref are non-zero and so linearly independent, you can’t conclude that the first $k$ rows of the original matrix are as well. This contrasts with the columns, for which elementary row operations leave linear dependency relationships unchanged. This is because an elementary row operation amounts to a change of basis for the column space.


Look at $$\begin{pmatrix} 1 & 0\\ 0 & 0\\ \end{pmatrix}. $$

The only "linear dependence relations" I can see are $c.\text{R}_2=0$ for arbitrary $c$.

But if we swap the two rows (achievable by an elementary row operation), then the only "linear dependence relations" are $c.\text{R}_1=0$ for arbitrary $c$.

There's a similar story for all the types elementary row operations, and for arbitrarily complicated matrices. But usually all that matters is that the rank does not change --- as already commented.


This is false. This is why we have row operations. Row operations are there so that if you have a matrix $A$, and you row reduce it to it's $RREF$, then the matrix $H$, has the same solution set as the matrix $A$.

That is, $Ax=0$ has the same solution set as $Hx=0$.

Try showing this for each of the row operations.

We have $3$ row operations. We can either swap, multiply a row by non-zero value, or add a multiple of a row to another row.

Consider $$A=\begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix} $$

$A$ has solution $x_1=0, x_2=0$.

If we swap, we end up with $A=\begin{bmatrix} 0 & 1\\ 1 & 0\\ \end{bmatrix} $, but this still has same solutions, just a different order.

If we multiply by non-zero value, suppose $2$, then $A=\begin{bmatrix} 2 & 0\\ 0 & 1\\ \end{bmatrix} $. Here we have $2x_1=0$, which is still $x_1=0$, so it's the same solution.

If we add a multiple of another row, suppose we do $2R_1+R_2$, then we have $A=\begin{bmatrix} 1 & 0\\ 2 & 1\\ \end{bmatrix} $. We already know $x_1=0$, and so using that fact, we know that $2x_1+x_2=0\to 0+x_2=0\to x_2=0$.

As you can see these row operations do not change the linear dependence, that is, they maintain the same solutions.


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