Question

Basis for the Column space of the matrix A. and Basis for the Null space.

A = [1 2 1 -1]
[1 2 -1 1]

RREF = [1 2 1 0]
[0 0 0 1]

so is the column space = [1] , [-1] is this correct?
[1] , [ 1]

how do I get the nullspace? I know its when the matrix is multiplied with a vector = the 0 vector

Answer 1

you rref is incorrect

Answer 2

ok. one sec

Answer 3

https://www.wolframalpha.com/input/?i=rref%7B+%7B1%2C2%2C1%2C-1%7D+%2C+%7B1%2C2%2C-1%2C1%7D+%7D

Answer 4

RREF = [1 2 0 0]
[0 0 1 -1]

so then the col(A) = [1] , [1] ??
[1] , [-1]

Answer 5

correct,

for the null space:
it's
[1 2 0 0 ] [x] = [0]
[0 0 0 1 ] [y] = [0]

yes?

Answer 6

yes

Answer 7

oops correction

Answer 8

would it be just [-2 1 0 0]

Answer 9

null space:
1x + 1y + 0z + 0a = 0
0x + 0y + 0z + 1a = 0

Answer 10

so we ge x + y = 0
and a = 0

Answer 11

the matrix multiplied with the vector {-2, 1, 0 ,0} = 0

is there another vector that works?

Answer 12

so x = -y

[x] [-y] [-y] [0] [-1] [0]
[y] = [y ] = [y] [0] y[1] z[0]
[z] [z] [0] + [z] = [0] + [1]
[a] [0] [0] [0] [0] [0]

Answer 13

crap mistake again -.- should have been
1x + 2y + 0z + 0a = 0
0x + 0y + 1z - 1a = 0

Answer 14

again, solving for x we get, x = -y
and solving for z we get, z = a

so
(x,y,z,a) = (-y, y, a, a)

so we have
(-y,y,0,0) + (0,0, a,a) = y(-1,1,0,0) + a (0,0,1,1)

so the basis for the null space is (-1,1,0,0) and (0,0,1,1)

Answer 15

OMG! brain fart again, x = -2y? -.-

I feel really stupid again.

Answer 16

sall good dude I got the two vectors x2(-2,1,0,0) and x4(0,0,1,1) both = the 0 vector when multiplied with the original matrix

Answer 17

yep, Don't make that stupid mistake like i did on the test. -2 is 1!!! and be sure to have your eyes check before the test XDDD

Answer 18

-2 is *not* 1. That's it. I'm done for today! -.-.

Answer 19

lol thanks for the help :)