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.

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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