Sage-Tips

Is AB invertible if B is not invertible?

It is easy to show using determinants: det(AB) = det(A)det(B)= 0, so AB is not invertible if either A or B are not invertible.

How do you prove that AB is not invertible?

The proof is to compute the determinant of every elementary row operation matrix, E, and then use the previous theorem. det(AB) = det(A) det(B). Proof: If A is not invertible, then AB is not invertible, then the theorem holds, because 0 = det(AB) = det(A) det(B)=0.

Does AB BA for invertible matrices?

Definition: A matrix A is invertible if there exists a matrix B such that AB = BA = In. In this case, the matrix B is said to be the inverse of A. But this means AB is n × n and BA is m × m, so in order for these two matrices to be equal, we must have m = n. This proves that A is square.

Is I-a⁢b invertible if I-B⁢a?

I-B⁢Ais invertible. Exchanging the roles of Aand Bwe can prove the ”if” part. So I-A⁢Bis invertible if and only if I-B⁢Ais invertible. 2. Let us first recall that a linear mapbetween vector spaces is invertible if and only if its kernel keris the zero vector(see this page (http://planetmath.org/KernelOfALinearTransformation)).

How do you prove that a matrix is invertible?

If B is nonsingular, you can write y = B -1 x, and y = 0 if and only if x = 0. Now think about Ax and ABy. An indirect way to prove this is to first show that a square matrix is invertible if and only if its determinant is not 0. Then if AB is invertible, det (AB) is not 0.

Is (ab) -1 singular or invertible?

Proof: If AB is defined and (AB) -1 exists, then there are four possibilities: A and B are both invertible, A is invertible and B is singular, A is singular and B is invertible, or A and B are both singular. Case 2: Let A be an invertible matrix and B be a singular matrix, let AB be defined, and let (AB) -1 exist.

Is b invertible if a is n by N?

Hence they span and since A is n by n they form a basis for R n so A is invertible. With its inverse present you can immediately get B invertible too. Or if you assume B is singular you can find some nonzero matrix C such that BC is the zero matrix which means ABC is the zero matrix which is impossible if C is nonzero and AB is invertible.