Question

True or False? If Ax=b has infinitely many solutions then so does Ax=0? And if Ax=b is inconsistent then Ax=0 has only trivial solution

Answer 1

Could someone help explain?

Answer 2

are we dealing with numbers or matrices? that capital A makes me think of a matrix

Answer 3

Yes it is with matrix

Answer 4

my gut says matrix

Answer 5

ah ok, so the first one is true. if Ax=b has more than one solution, then that means Ax= 0 must have some non-trivial solution. For example, if the vector u is a solution to Ax = b, and if a vector v is a solution to Ax = 0, then the vector (u + v) is also a solution to Ax = b because:
A(u+v) = Au+Av = b+0 = b

Answer 6

also any multiple of v will work along with u as well:
A(u+cv) = Au+A(cv) = b+cAv = b + c*0 = b + 0 = b
so any scalar multiple of v + u will be a solution.

Answer 7

Okay whoa that actually makes sense would have never thought up of that proof and gotten b

Answer 8

For the second one....im thinking this:
Since the solution Ax = b is inconsistent, that means the vector b isnt in the range of the matrix A. That means the range of A doesnt cover the whole vector space, because if it did, there would be a solution. Now, lets say we are dealing with n-dimensional vectors (the space would be R^n). The dimension of the range of A has to be less than n (because the range doesnt cover the whole space). That means the dimension of the Null Space (all the vectors that satify Ax = 0) has to be at least dimension 1. And this means that Ax = 0 has some non trivial solution.

Answer 9

In short: false <.<

Answer 10

Kinda confusing but I think I can get it if I read more about it

Answer 11

im trying to think of a shorter way to say it >.<

Answer 12

well, if Ax = b is inconsistent, then you must have gotten an all zero row when you started row reducing A. That right there shows that Ax = 0 must have a non-trivial solution in a couple of different ways. Do you know about determinants?

Answer 13

Yeah

Answer 14

Alright, if the determinant of a matrix is 0, then automatically Ax = 0 has non-trivial solutions. So back from the beginning (lol):
Ax = b is inconsistent means when we row reduce A we get a row with all zeros.
Getting a row with all zeros means the determinant of A = 0.
The determinant of A = 0 means Ax = 0 has non-trivial solutions.
Done and Done (answer is false)

Answer 15

i like my longer proof, it makes me sound smarter <.< lolol jk

Answer 16

OH I get it now!!

Answer 17

that make sense I think man wish I could give you more medals

Answer 18

thanks for your help!

Answer 19

you get a medal for that comment lol

Answer 20

Haha I have a midterm tomorrow, linear algebra is hard to understand sometimes especially subspaces

Answer 21

But enough of that thanks for your help

Answer 22

no prob :)