2 X 2 Idempotent matrix

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This post is a part of my elective course Mathematical Foundations for Data Science during my M. Tech. in Software Systems with specialization in Data Analytics.

I had to provide an example of an idempotent matrix. That’s the kind of matrix that yields itself when multiplied to itself. Much like 0 and 1 in scalar multiplication (1 x 1 = 1).

It is not so easy to predict the result of a matrix multiplication, especially for large matrices. So, instead of settling with the naïve method of guessing with trial and error, I explored the properties of a square matrix of the order 2.

In this page I state the question and begin to attempt it. I realised that for a matrix to be idempotent, it would have to retain its dimensions (order), and hence be a square matrix.
I have intentionally put distinct variable names a,b,c, and d. This is to ensure that the possibility of a different number at each index is open. I derived ‘bc’ from the first equation and substituted it into its instance in the last equation to obtain a solution for ‘a’.

Since 0 cannot be divided by 0, I could not divide 0 by either term unless it was a non-zero term. Thus, I had two possibilities, to which I called case A and B.

I solved the four equations in case A by making substitutions into the 4 main equations. Later tested the solution with b=1.

As you can see, I could not use the elimination method in an advantageous manner for this case.

I couldn’t get a unique solution in either case. That is because there are many possible square matrices that are idempotent. However, I don’t feel comfortable to intuit that every 2 X 2 idempotent matrix has one of only two possible numbers as its first and last elements.

Others’ take on it

My classmate Sabari Sreekumar did manage to use elimination for the ‘bc’ term for the general case.
I took it a step further and defined the last element in terms of the other elements

So given any 2 X 2 idempotent matrix and its first three elements, you can find the last element unequivocally with this formula.


I wonder if multiples of matrices that satisfy either case are also idempotent. Perhaps I will see if I can prove that in another post.

In the next lecture, professor Venkata Ratnam suggested using the sure-shot approach of a zero matrix. And I was like “Why didn’t I think of that”?