Question

Suppose AD=MxM identity Matrix. Show that any vector b in space R Mthe equation Ax=b where b and x are vectors.

Answer 1

i can not wait for this response. i am sitting on the edge of my seat

Answer 2

i'm not seeing a question here

Answer 3

show that any vector in \(\mathbb R^m\) satisfies \(A\vec x=\vec b\) if \[AB_{m\times m}=I\]?

Answer 4

its a proof. Show that any vector b in \[\mathbb{R}^m\] the equation ax=b

Answer 5

yes

Answer 6

or you have \[AD=I_{m\times m}\]correct?

Answer 7

yes that is correct

Answer 8

and there is a hint saying think about the eq ADb=b

Answer 9

oh ok that may help

Answer 10

show that any vector in Rm satisfies Ax⃗ =b⃗ if
ABm×m=I
this is correct

Answer 11

ok i understand the problem i think but i'm rusty lol

Answer 12

This is what i had
ADb=b
ADAx=b
IAx=b
Ax=b
but i used the fact that Ax=b which is what i needed to prove in step 2

Answer 13

we also know that AD=Ad1+Ad2....Adn where di is the columns of D. from the definition of matrix multiplication

Answer 14

and if that equals the identity does that imply that Ad1=e1 and so forth?

Answer 15

where e1= equals the idenity matrix of column 1

Answer 16

i suppose it would yeah

Answer 17

what about something like\[AB\vec b=\vec b\\\vec b=B^{-1}A^{-1}\vec b=\vec x\\A\vec b=\vec x\]i don't know if that is less tautological than yours though

Answer 18

I think i got it. To show that all Ax=b for all b we need to show that Ax is an onto function. To show some thing is onto we need to show that the columns of A span Rm which is done by showing AD=I

Answer 19

o okay yours makes sense as well

Answer 20

hm you are going beyond what i remember from LA well lol, but i think it works if you just assert that \(\vec b\) can be \(\vec x\) since the inverse of the identity is the identity

Answer 21

in the first part of the proof i mean

Answer 22

maybe take out the x though, hm still seems tautological

Answer 23

alright sounds good

Answer 24

no it's all wrong lol i need to rethink it

Answer 25

haha alright

Answer 26

yeah way too rusty on my LA.... dang
maybe @AccessDenied or @thomaster can help?

Answer 27

remove my medal :P

Answer 28

ill leave it thanks for the help

Answer 29

hehe thanks for the reminder i need to review this stuff

Answer 30

oh dude i think i got it

Answer 31

alright sounds good haha

Answer 32

dang it false alarm lol

Answer 33

its fine i am sure that it will be okay.

Answer 34

i think that it is enough to say that\[AB=I_{m\times m}\implies B=A^{-1}\]so you have\[A^{-1}AA^{-1}\vec b=A^{-1}\vec b=\vec x\]which must be in \(\mathbb R^m\) because \(A\) is an \(m\times n\) matrix, and since \(A\) is invertible, for every \(\vec x\) in \(\mathbb R^m\) there must be a unique \(\vec b\) such that\[A\vec x=\vec b\]

Answer 35

yep that works. because you showed that A is onto and 1 to 1. Thanks!

Answer 36

yw :)