k*******r
发帖数: 16963 | 1 A, B are n*n matrices, if AB is invertible, how to prove A and B are
invertible? | hs
发帖数: 1549 | 2 看rank啊
【在 k*******r 的大作中提到】
: A, B are n*n matrices, if AB is invertible, how to prove A and B are
: invertible?
| k*******r
发帖数: 16963 | 3 How to prove it without rank, det etc.
Just use the basic definition of invertible matrix:
if AB=BA=I,A and B are invertible. | h***1
发帖数: 40 | 4 ABC=CAB=1 or you can read it as A(BC)=(CA)B=1 | k*******r
发帖数: 16963 | 5 but in order to prove A is invertible, A(BC)=1 is enough? how to prove (BC
)A=1? | f*********y
发帖数: 27 | 6 just look at determinant of AB, since A, B are n*n matrices, we have: det(AB
)=det(A)det(B), your book will contain this result. Hence if det(AB) is not
0, then det(A) and det(B) cannot be 0 either, thus A and B are invertible.
Q.E.D.
【在 k*******r 的大作中提到】
: A, B are n*n matrices, if AB is invertible, how to prove A and B are
: invertible?
| h***1
发帖数: 40 | | e****d
发帖数: 333 | 8 Totally agree.
If any one of A and B is singular, then AB is singular.
AB
not
【在 f*********y 的大作中提到】
: just look at determinant of AB, since A, B are n*n matrices, we have: det(AB
: )=det(A)det(B), your book will contain this result. Hence if det(AB) is not
: 0, then det(A) and det(B) cannot be 0 either, thus A and B are invertible.
: Q.E.D.
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