Find a basis for the null space of A.

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- KonradZuse

Find a basis for the null space of A.

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- KonradZuse

My book says that the null space is the solution space of Ax = b when b = 0. The example in the book shows this...

- KonradZuse

##### 1 Attachment

- KonradZuse

Do I just solve it and that's all...?

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## More answers

- KonradZuse

I'm also a bit confused how they got the column matrix's filled out... Maybe that's my issue...

- TuringTest

pretty much, yeah
just solve it and rewrite each variable as a new vector, i.e. one for r, one for s, and one for t

- KonradZuse

hmm okay maybe this is easier than I'm thinking let me try it. My example is
A = (Matrix(3, 4, {(1, 1) = 1, (1, 2) = 4, (1, 3) = 5, (1, 4) = 2, (2, 1) = 2, (2, 2) = 1, (2, 3) = 3, (2, 4) = 0, (3, 1) = -1, (3, 2) = 3, (3, 3) = 2, (3, 4) = 2}));
print(`output redirected...`); # input placeholder

- KonradZuse

woah... LOL

- KonradZuse

1 4 5 2
2 1 3 0
-1 3 2 2

- KonradZuse

now do I have to augment it with the 0 column-vector?

- TuringTest

go ahead and row reduce it as much as you can
brb, gotta go to get dinner

- KonradZuse

the book does 2 eq's which is really annoying....

- KonradZuse

how they would both be the same, so it seems like it's not an augmented matrix....

- TuringTest

row reduce, what do you get?

- KonradZuse

OOOOOOOOOOOOOOOo There's a Null Space/Nullity thing on Maple hehehe. I guess it isn't augmented then :D

- TuringTest

no, it's not

- KonradZuse

1 0 1 -2/7
0 1 1 4/7
0 0 0 0

- KonradZuse

now I actually found something that says the num space is the subspace of vectors x satisfying A. x = 0

- KonradZuse

it shows the null space as
2/7
-4/7
0
1
then another column matrix next to it.
-1
-1
1
0..

- KonradZuse

Is there a matrix thing in the equations..... This is stupid writing it out like this :P.

- KonradZuse

oh I found it.... :)

- TuringTest

setting up matrices with latex is a pain, you are doing it the easy way

- KonradZuse

HJHAHAH

- KonradZuse

\[\left(\begin{matrix}2/7 \\-4/7\\ 0\\ 1\end{matrix}\right), \left(\begin{matrix}-1 \\-1\\ 1\\ 0\end{matrix}\right)\]

- TuringTest

yes, you get that from
1 0 1 -2/7
x1+r-2/7s=0->x1=-r+2/7s
0 1 1 4/7
x2+r+4/7s=0->x2=-r-4/7s
collect the r's in one vector and the s's in another

- KonradZuse

my eyes they burn.... :)

- KonradZuse

you're so smart TT :Pe

- TuringTest

first row of your matrix reduced:
1 0 1 -2/7
this is the same as the equation
x1+r-2/7s=0
which leads to
x1=-r+2/7s
and thanks, but I need to get better at computer stuff now, so maybe I'll need your help
sorry, back in 15

- KonradZuse

yeah yeah yeah!

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