# Natural basis for set of m x n matrices, visualized

I am told to consider the $m \times n$ matrices $E_{pq}$ described by:

$[E_{pq}]_{ij} = \left\{ \begin{array}{ll} 1 & \textrm{if } i=p\textrm {, }j=q \\ 0 & \textrm{otherwise} \end{array} \right.$

Which are supposed to form the natural basis for the set of $m \times n$ matrices. What I'm struggling with is… visualizing these matrices, I suppose. What does one look like? For some reason I can't work that out from the definition above, and here I thought I knew the notation but it just doesn't make sense in this particular case. I'm not sure what exactly $p$, $q$, $i$ and $j$ represent here, it's like maybe one pair of them ought to have been $m$ and $n$ instead, but which?

All I could get was that they all have a single 1 in the lower-right corner and every other entry is 0, like so

$\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]$

but that doesn't seem right. I'm thinking more like it's the set of all $m \times n$ matrices with a single entry 1 (not just the corner) and all others 0, but I can't get that to match the definition above.

Neither Wikipedia, Wolfram|Alpha or Google turned up anything.

-

Your idea is correct. Note that they are $m\times n$ matrices, but their names are $E_{pq}$. When $p=m$ and $q=n$ you get a 1 in the lower-right corner; this is $E_{mn}$, which you depicted. But $E_{11}$ has the 1 in the upper-left corner, and $E_{23}$ has the 1 in the second row and the third column.

Why does $E_{23}$ have zeroes everywhere, except for the 1 in the second row and third column? The definition you wrote is:

$$[E_{pq}]_{ij} = \left\{ \begin{array}{ll} 1 & \textrm{if } i=p\textrm {, }j=q \\ 0 & \textrm{otherwise} \end{array} \right.$$

Let's take $p=2$ and $q=3$ in this definition:

$$[E_{23}]_{ij} = \left\{ \begin{array}{ll} 1 & \textrm{if } i=2\textrm {, }j=3 \\ 0 & \textrm{otherwise} \end{array} \right.$$

This says that element at row $i$ and column $j$ of $E_{23}$ will be 0, unless $i=2$ and $j=3$, when it will be 1.

-
Ah, thank you, that cleared it up. Sometimes it's the little things you don't get. –  ADCoon Mar 22 '12 at 14:11
The problem I was asked was to prove that they form the natural basis. Now that I can picture them this seems perfectly obvious, of course, and so a proof should be no problem. The little thing I was missing was the idea that p and q should be considered fixed, this didn't occur to me. –  ADCoon Mar 22 '12 at 14:25
@ADCoon: Well, it's a little bit subtle. $p$ and $q$ are considered fixed relative to $i$ and $j$, but variable relative to $m$ and $n$. I was writing up a detailed comment about this, but it wouldn't fit in the box. –  MJD Mar 22 '12 at 15:03