Question

I have another linear algebra question :D

Answer 1

wait a sec gonna attach the file

Answer 2

Yikes, I'll have to review on this stuff, give me a while...

Answer 3

hehe ok :D i will go and i will be back in hr

Answer 4

Oh I was just about to take a stab at this, but let's see how it's supposed to be done...
I was thinking of using the theorem
Rank(A)+Nullity(A)=m

Answer 5

...and trying to prove that Nullity>0

Answer 6

Pippa and TuringTest are typing replies…

Answer 7

oh ya the theroem that Rank(A)+Nullity(A)=n

Answer 8

but like how wld u show that? can u kind of show me?

Answer 9

still Pippa and TuringTest are typing replies…

Answer 10

if the points are collinear, call them p1, p2, p3, then you can write one of them as the sum of the other three. I.e., there exist scalars a and b such that
p3 = ap1 + bp2

Now, that being the case, we need to show we can row eliminate the entire last row. In which case the rank must be less than 3.

Answer 11

What's not obvious to me right now is why you can choose such a and b so you eliminate the "1" in the bottom right corner.

Answer 12

The whole thing isnt obvious to me lol

Answer 13

I don't really see what needs to be shown
if p3 is a possible linear combination of p1 and p2 it can be eliminated because that's how elimination works

maybe: since
p3=ap1+bp2
then
p3-ap1-bp2=0

...Like I was saying I'm not good with rigor in proofs :/

Answer 14

why do you have to choose a particular a and b ?

I was thinking...
if x and y are dependent then Ax=0 must have some non-zero solution, so the Nullity(A)>0
hence m=3=Rank(A)+Nullity(A)
so Rank(A)=3-Nulltiy(A)<3

does that mean anything at all James?

Answer 15

lol ok i think i just have to diegt what u guys said

Answer 16

*digest

Answer 17

James is not present, and that's the only proof I have in mind...

Answer 18

Ok. So I have a proof by it's not elegant.

Answer 19

neither is mine, but I bet yours is better

Answer 20

If the points are collinear, the lie on the same "direction" vector from each other.

let r = p2 - p1 then
p2 = p1 + r
p3 = p1 + a.r
for some constant a.

So far so good?

Answer 21

I don't know where pippa went, but I wanna see where it's going if you feel like continuing

Answer 22

Well, that means that we can write
\[ x_3 = x_1 + a(x_2 - x_1) \] and
\[ y_3 = y_1 + a(y_2 - y1) \]

Answer 23

lol I am here but os is freezing on me

Answer 24

\[ y_1 \]

Now, explicitly write out the determinant of that matrix and use those identities for x3 and y3, and the whole thing turns out to be zero.

Answer 25

From that, it follows the nullity of the matrix is greater than zero and the rank must be less than 3.

Answer 26

kk i think i may have understood it Thanks guys :D

Answer 27

Pippa, I figured out the question we were both stumped on. Just message me somehow and I'll explain it :-)

Answer 28

Turing did u follow cuz I didnt

Answer 29

pipa help me!

Answer 30

I need to help myself :D

Answer 31

okay i

Answer 32

did you need help with that last proof pippa?

Answer 33

i didnt follow th ewhole thing seriously lol

Answer 34

y1
Now, explicitly write out the determinant of that matrix and use those identities for x3 and y3, and the whole thing turns out to be zero.
Like this part

Answer 35

you can't get zero, or you don't understand what to do?

Answer 36

I dont understand what to do

Answer 37

Sorry Os is very slow today

Answer 38

James pointed out that you can make a vector in the direction of the line that would contain those points by calling each point p, and defining a vector r=p2-p1
how are you so far?

Answer 39

That I understood

Answer 40

so then that relates point p1 and p2
and p3 must be from some addition of that vector r to a known point like p1 (since it is also co-linear):
p3=p1+ar

so p3 can be written as \[p_3=p_1+ar\]

Answer 41

ya i got that part

Answer 42

looking at each component individually:\[x_3=x_1+a(x_2-x_1)\]\[y_3=y_1+a(y_2-y_1)\]now sub this into the matrix....

Answer 43

ohhh i think i get it

Answer 44

If there is a particular part of that you don't get let me know

Answer 45

wait let me just read it over

Answer 46

ya ok

Answer 47

if\[A\vec x=\vec 0\]has any solution other that the trivial solution, then it has a zero determinant. If it has a zero determinant, it has a Null Space>0
if it has a Null Space greater than zero, the theorem

m=3=Rank(A)+NullSpace(A)
implies that
Rank(A)=3-Nullspace(A)

which, if the null space is greater than zero, means that the rank is less than 3

Answer 48

ohhhh I get it Thanks Turing. That was great It just took a while for it to click in my brain

Answer 49

I was working on this problem all evening and I cldnt figure it out THANKS :DDDDDD

Answer 50

http://tutorial.math.lamar.edu/Classes/LinAlg/FundamentalSubspaces.aspx
look at all the equivalent statements of the theorem at the bottom of this link
linear algebra uses that a lot...

Answer 51

...and you're welcome :D

Answer 52

Thanks seriously that was great

Answer 53

3 points are on the nonvertical line y=mx+b, then matrices:

x1 y1 1 x1 mx1+b 1

x2 y2 1 = x2 mx2+b 1

x3 y3 1 x3 mx3+b 1

Observe that second column is a linear combination of the first and third columns. Therefore this determinant is zero and the rank is less than 3.

On the other hand, if the rank of the matrix below is less than 3:

x1 y1 1

x2 y2 1

x3 y3 1

then this determinant is zero and it implies that the three points are collinear.

If the 3 points are collinear and on the same vertical line, then x1 = x2 = x3

and the matrix has two equal columns and the rank is therefore less than 3.

Answer 54

I asked my prof to give me the answer just to make sure and this is what he sent me

Answer 55

I guess this is what we were saying but in just a simpler manner

Answer 56

Yeah, exactly the same idea, but it used the point-slope formula for a line instead of the vector one that James used.

Answer 57

slope-intercept form rather*