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 !!
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).
Matrix Multiplication
Inverse of a matrix
Rank of a matrix
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) .
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 !!
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).
Matrix Multiplication
Inverse of a matrix
Rank of a matrix
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) .
Try to prove the same using inequalities involving rank of a matrix.
