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bio website
location St. Petersburg, Russia
age 32
visits member for 2 years, 8 months
seen Sep 30 at 16:58
mail: mlvl.jr@gmail.com

Nov
25
comment Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations
Huh, that's nice! :)
Feb
18
comment Looking for insightful explanation as to why right inverse equals left inverse for square invertible matrices
@JackSchmidt You'd probably be interested in math.stackexchange.com/questions/110336/…
Feb
17
comment Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations
@Arturo That's what I was thinking to try, thanks! (Once again you help me :) )
Feb
17
comment Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations
But how do I deal with the left inverse? (Is there a simmetry I have overlooked?)
Feb
17
comment Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations
@Arturo Thanks for keeping the comment.
Feb
17
comment Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations
Whatever row operations we apply to A to get it into reduced row-echelon form, if the same row operations are applied to I, the resulting matrix will have no zero rows. Then, if and only if A is row-equivalent to I, there will be a single set of solutions (i.e. B's column values) for every equation A_<j>*B=I_<j>, where A_ and I_ are transformed A and I and <j> denotes j column. If there's a zero row i in A_, there will be at least one such j that I_<j> has a non-zero element in row i, and thus the system of equations has no solution.
Feb
17
comment Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations
Or is it really that exotic?
Feb
17
comment Proving that a right (or left) inverse of a square matrix is unique using only basic matrix operations
The idea I've got is that since elementary transforms of I will never give a zero row, any matrix A which is not row-equivalent to I, will produce an unsolvable set of equations for at least one column of B in A*B = I. Hence, for that matrix equation to hold, A must be row-equvalent to I and thus there's only one B sufficing the equation.
Feb
14
comment General rules to keep an eye on division by zero when dealing with a system of equations
@Nuxonic, thanks, I know of the pivoting method(s?) existence, though have not yet looked at it.
Feb
14
comment General rules to keep an eye on division by zero when dealing with a system of equations
Ok, do we never loose solutions utilizing the divid-by-possibly-zero-and-check-later method?