Idempotent, Not Impotent.

If Jon were idempotent, then if we multiplied Jon with Jon, we'd end up with Jon. Math is fun!

Remember that time I went to Montreal just a few days ago? Well, as I had mentioned, it wasn’t all just fun, food, and sampling the fine fares of a most excellent city. It was also a time to get my nerd on by writing grant applications, and by chatting with Dr. Steph about the subdistribution hazard function in survival analysis. I know, exciting stuff for sure.

But dear readers, those weren’t the only nerdly things to occur on my trip. You see, whilst perusing the streets of Montreal my friend Jon asked if he could request that I post a particular blog entry. Hells ya, I replied. Okay, maybe I simply nodded my head or said sure. The point is I don’t remember what I did, but I do know that I was absolutely game for a blog request.

Jon and Kathy, or in mathematical terms, Awesome times 2. (Note, I may have stolen this picture from Jon's facebook page. Stolen pictures are the best kind of pictures.)

After Jon asked the question, I had assumed that he’d request a blog entry about Montreal, or perhaps one of the other adventures we had been on together. Or maybe he wanted to see a blog about the Big-Chairs from the night before. Whatever it was, I was unprepared for his reply.

What do you want me to write about, Jon? I asked.

I want to know what Idempotent means to a mathematician. 

Of course, hearing this I immediately had a few reactions. First, I was a bit dumbfounded. Idempotent? Jon’s nerd bar was clearly raised by about 1ooo percent. Second, I couldn’t help but wonder How does Jon know about idempotency? I mean, as far as I know this isn’t a common word. Jon’s intrigue factor may have also just increased 1000 fold. Well played Jon. Finally, I thought SQUEEEEEEEEE! Someone wanting me to write about math? That is FREAKING AWESOME.

Anyway, what the hell is idempotency? Simply put, something that is idempotent doesn’t change when applied to itself. Wait. What? It’s probably easiest to understand this when we think of an example.

Consider the number 1. We know that 1\times 1=1. That is, when we multiply 1 by itself, it doesn’t change. In this case, 1 is idempotent. But that’s a little boring, so how about we try this one on for size. Consider the following 2\times 2 matrix

\left[    \begin{array}{cc}    1 & 1\\    0 & 0    \end{array}    \right].

This is apparently an idempotent matrix. How so? Well, first we need a bit of a refresher on matrix multiplication, because although it is straightforward, it’s not quite the same as multiplying 2 numbers together and recording the product. Anyway, assuming we can multiply 2 matrices together (and I’m not going to get into when we can’t right now), then we’d have

\left[    \begin{array}{cc}    a & b\\    c & d    \end{array}    \right]\times    \left[    \begin{array}{cc}    e & f\\    g & h    \end{array}    \right]=    \left[    \begin{array}{cc}    a\times e+b\times g & a\times f + b\times h\\    c\times e + d\times g & c\times f + d\times h    \end{array}    \right]    .

So, for a matrix to have the idempotent property, we would need

\left[    \begin{array}{cc}    a & b\\    c & d    \end{array}    \right]\times    \left[    \begin{array}{cc}    a & b\\    c & d    \end{array}    \right]=    \left[    \begin{array}{cc}    a^{2} + bc & ab + bd\\    ac+ cd & bc+d^{2}    \end{array}    \right]    ,

to hold true. This means that we have a system of 4 equations in 4 unknowns. Specifically, the elements a,\ b,\ c,\ \mathrm{and}\ d must be selected so that all of the following equations are satisfied:

  • a^{2}+bc=a,
  • ab+bd=b,
  • ac+cd=c, and
  • bc+d^{2}=d.
Jon, on the swing, pondering idempotentcy. (Note: this picture was also stolen from facebook, and we know how I feel about stolen pictures).

Anyway, if we apply the normal matrix multiplication to our matrix, we can see that it is idempotent. That is, the matrix times itself is itself:

\left[    \begin{array}{cc}    1 & 1\\    0 & 0    \end{array}    \right]\times    \left[    \begin{array}{cc}    1 & 1\\    0 & 0    \end{array}    \right] =    \left[    \begin{array}{cc}    1 & 1\\    0 & 0    \end{array}    \right].

Cool! Note that idempotentcy isn’t a characteristic of all matrices. Only a special type of matrix can claim to be idempotent.

So there you have it Jon (and dear readers). Idempotent, not to be confused with impotent, is a rather cool little characteristic from a mathematical point of view.

Thanks for the question Jon; you are a good and nerdly man. Furthermore, dear readers, feel free to offer me any other ideas for blog entries (perhaps in the comments section). I shall do my best to address all those that I can.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s