Let A be a square matrix. Assume that the following statement is true: If B is an invertible matrix then rank BA = rank A. (a) Show that if B is an invertible matrix then rank AB = rank A. Hint: What is rank(AB)" ? (b) Show that for any invertible matrix P, rank(P-'AP) = rank A. (c) If P is invertible then nullity P-'AP = nullity A.

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varshamittal029 varshamittal029 · Answer 1 · 2022-12-18T09:27:08-05:00

We proved the expression rank(AB)=rank(B) and rank(BA)=rank(B),

where A and B are the two matrices of same order.

Given, A is a square matrix.

If A is invertible and a n×n square matrix of full rank.

Since both AB and BA exist, where B is also a n×n square matrix.

rank(B)=rank(BT)

On using the rank–nullity theorem, we get

nullity of a matrix be n minus the rank of the matrix.

So let U be a matrix whose columns are a basis for the nullspace of AB, so that ABU = 0.

Pre-multiplying both sides by A−1 , we obtain BU=0,

As B and AB have the same rank.

Similarly, let V be a matrix whose columns are a basis for the null space of ATBT, so that ATBTV=0 .

Pre-multiplying both sides by (A−1)T , we get

BTV=0.

Since rank(BT)=rank(B) and rank(ATBT)=rank((BA)T)=rank(BA)

we get rank(AB)=rank(B) and rank(BA)=rank(B).

So, rank(AB)=rank(B) and rank(BA)=rank(B).

Hence, rank(AB)=rank(B) and rank(BA)=rank(B).

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