:Actually, having a unique left inverse does imply bijectivity, unless the domain of " f " is a singleton . J . 14 : 37, 7 May 2009 ( UTC)
12.
Then it suffices to show " g " ?, which is now over " B ", has a left homotopy inverse over " B " since that would imply that ? has such a left inverse.
13.
The left inverse can be used to determine the least norm solution of Ax = b, which is also the least squares formula for regression and is given by x = ( A ^ TA ) ^ {-1 } A ^ Tb.
14.
Let me try to summarize : Since the matrix A has a left inverse, then the transformation it represents ( T \ colon x \ mapsto Ax ) in injective-injectiveness is a necessary condition for having a left inverse in the category of sets and mappings, so we don't need to appeal to vector spaces for that.
15.
Let me try to summarize : Since the matrix A has a left inverse, then the transformation it represents ( T \ colon x \ mapsto Ax ) in injective-injectiveness is a necessary condition for having a left inverse in the category of sets and mappings, so we don't need to appeal to vector spaces for that.
16.
On the other hand, if m > n, then A'can have a left inverse ( depending on the rank of A ), and if it does, then AA', while not being an identity matrix, is a square matrix for which B is an eigenvector with a corresponding eigenvalue of 1; in other words : A'AB = B.
17.
The mapping T is injective, but we don't know that it's " surjective ", so B ( or B's columns-we can deal with them one at a time ) might not be in the range of T . If B is in the range of T, then A', being a left inverse for A, will map B back to its preimage, namely A'B . In that case, we're all good, and even though A'may not be a right inverse for A, we know that AA'will be a square matrix that maps B to B . On the other hand, if B is not in the range of T, then there is no solution.
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