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# ISI MStat PSB 2007 Problem 2 | Rank of a matrix

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 2 based on Rank of a matrix. Let's give it a try !!

## Problem- ISI MStat PSB 2007 Problem 2

Let $A$ and $B$ be $n \times n$ real matrices such that $A^{2}=A$ and $B^{2}=B$
Suppose that $I-(A+B)$ is invertible. Show that rank(A)=rank(B).

### Prerequisites

Matrix Multiplication

Inverse of a matrix

Rank of a matrix

## Solution :

Here it is given that $I-(A+B)$ is invertible which implies it's a non-singular matrix .

Now observe that ,$A(I-(A+B))=A-A^2-AB= -AB$ as $A^2=A$

Again , $B(I-(A+B))=B-BA-B^2=-BA$ as $B^2=B$ .

Now we know that for non-singular matrix M and another matrix N , $rank(MN)=rank(N)$ . We will use it to get that

$rank(A)=rank(A(I-(A+B)))=rank(-AB)=rank(AB)$ and $rank(B)=rank(B(I-(A+B)))=rank(-BA)=rank(BA)$ .

And it's also known that $rank(AB)=rank(BA)$ . Hence $rank(A)=rank(B)$ (Proved) .

## Food For Thought

Try to prove the same using inequalities involving rank of a matrix.

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