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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.

## Subscribe to Cheenta at Youtube

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.

## Subscribe to Cheenta at Youtube

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