The Unapologetic Mathematician

Elementary Row and Column Operations

Here’s a topic that might be familiar all the way back to high school mathematics classes. We’re going to use the elementary matrices to manipulate a matrix. Rather than work out abstract formulas, I’m going to take the specific matrix

$\displaystyle\begin{pmatrix}A&B&C&D\\E&F&G&H\\I&J&K&L\end{pmatrix}$

and perform example manipulations on it to draw conclusions. The effects of elementary matrices are so, well, elementary that it will be clear how they generalize.

So how do we use the matrices to manipulate matrices? Well, we’re using the elementary matrices to change the input or output bases of the linear transformation represented by the matrix. So to change the output basis we’ll multiply on the left by an elementary matrix, while to change the input basis we’ll multiply on the right by the inverse of an elementary matrix — which itself is an elementary matrix of the same kind.

$\displaystyle\begin{pmatrix}1&0&0\\{0}&0&1\\{0}&1&0\end{pmatrix}\begin{pmatrix}A&B&C&D\\E&F&G&H\\I&J&K&L\end{pmatrix}=\begin{pmatrix}A&B&C&D\\I&J&K&L\\E&F&G&H\end{pmatrix}$

while on the right we might have

$\displaystyle\begin{pmatrix}A&B&C&D\\E&F&G&H\\I&J&K&L\end{pmatrix}\begin{pmatrix}{0}&0&1&0\\{0}&1&0&0\\1&0&0&0\\{0}&0&0&1\end{pmatrix}=\begin{pmatrix}C&B&A&D\\G&F&E&H\\K&J&I&L\end{pmatrix}$

On the left, the action of a swap is to swap two rows, while on the right the action is to swap two columns of the matrix.

Next come the scalings. On the left

$\displaystyle\begin{pmatrix}c&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}A&B&C&D\\E&F&G&H\\I&J&K&L\end{pmatrix}=\begin{pmatrix}cA&cB&cC&cD\\E&F&G&H\\I&J&K&L\end{pmatrix}$

and on the right

$\displaystyle\begin{pmatrix}A&B&C&D\\E&F&G&H\\I&J&K&L\end{pmatrix}\begin{pmatrix}1&0&0&0\\{0}&c&0&0\\{0}&0&1&0\\{0}&0&0&1\end{pmatrix}=\begin{pmatrix}A&cB&C&D\\E&cF&G&H\\I&cJ&K&L\end{pmatrix}$

On the left, the action of a scaling is to multiply a row by the scaling factor, while on the right the effect is to multiply a column by the scaling factor.

Finally, the shears. On the left

$\displaystyle\begin{pmatrix}1&0&c\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}A&B&C&D\\E&F&G&H\\I&J&K&L\end{pmatrix}=\begin{pmatrix}A+cI&B+cJ&C+cK&D+cL\\E&F&G&H\\I&J&K&L\end{pmatrix}$

and on the right

$\displaystyle\begin{pmatrix}A&B&C&D\\E&F&G&H\\I&J&K&L\end{pmatrix}\begin{pmatrix}1&0&0&0\\{0}&1&0&0\\{0}&0&1&0\\{0}&c&0&1\end{pmatrix}=\begin{pmatrix}A&B+cD&C&D\\E&F+cH&G&H\\I&J+cL&K&L\end{pmatrix}$

On the left, the shear $H_{i,j,c}$ adds $c$ times the $j$th row to the $i$th row, while on the right, the shear adds $c$ times the $i$th column to the $j$th column.

So in general we see that acting on the left manipulates the rows of a matrix, while acting on the right manipulates the columns. We call these the “elementary row operations” and “elementary column operations”, respectively. Any manipulation of the form of a matrix we can effect by these operations can be seen as the result of applying a change of basis matrix on the left (output) or right (input) side. And so any two matrices related by these operations can be seen as representing “the same” transformation in two different bases.

August 27, 2009 - Posted by | Algebra, Linear Algebra

1. Quit making it look so easy! I just graduated with a degree in physics, and such simple operations never seemed more obvious! Thanks for the great explanation; keep up the good work (the simple stuff is the best stuff!).

Comment by AndrewB | August 29, 2009 | Reply

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6. Really cool explanation……thanks a lot. Keep it up.

Comment by dpak | December 6, 2009 | Reply

7. Can we have column operations in analogy to row operations?

Comment by Tahsin | March 18, 2010 | Reply

8. See the last paragraph, Tahsin.

Comment by John Armstrong | March 18, 2010 | Reply