# Answer to Question #23765 in Linear Algebra for Matthew Lind

Question #23765

Let A and B be M by N matrices, P an invertible M by M matrix, and Q an invertible N by N matrix, such that B = PAQ, that is, the matrices A and B are equivalent. Show that the rank of B is the same as the rank of A. (Show that A and AQ have the same rank).

Expert's answer

Every invertible matrix is afinite product of some elementary matrices, that are in one to one

correspondence with elementary row and column operations of matrix. Thus, as it

is known, every elementary row/column operation preserves rank of matrix. So, matrix

Q=E1*E2*...*En, where each Ei is elementary matrix, AQ=(...((A*E1)*E2)*...)*En

and in every bracket rank is equal rank of matrix A.

So rank(A)=rank(AQ).

As rank(A^T)=rank(A), then as P^T is also invertible,

rank(PAQ)=rank(Q^T * A^T * P^T)=rank(Q^T * A^T)=rank(AQ)=rank(A) and we are

done.

correspondence with elementary row and column operations of matrix. Thus, as it

is known, every elementary row/column operation preserves rank of matrix. So, matrix

Q=E1*E2*...*En, where each Ei is elementary matrix, AQ=(...((A*E1)*E2)*...)*En

and in every bracket rank is equal rank of matrix A.

So rank(A)=rank(AQ).

As rank(A^T)=rank(A), then as P^T is also invertible,

rank(PAQ)=rank(Q^T * A^T * P^T)=rank(Q^T * A^T)=rank(AQ)=rank(A) and we are

done.

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