Question

how do I find the column space of A as a plane?

Answer 1

\[\left[\begin{matrix}2&4&6&4\\2&5&7&6\\2&3&5&2\end{matrix}\right]\]

Answer 2

isnt it the nulspace? memorys a little fuzzy

Answer 3

I got the nulspace of it

Answer 4

Solve the system \(\sf \color{blue}{Ax=0}\)

Answer 5

I think?

Answer 6

I did out v1, v2, v3 and v4 but don't get what I'm supposed to get from that

Answer 7

\[\left[\begin{matrix}2&4&6&4\\2&5&7&6\\2&3&5&2\end{matrix}\right]\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]=\left[\begin{matrix}b_1\\b_2\\b_3\end{matrix}\right]\]

Answer 8

x3(-1) + x4( 2)
x3(-1) + x4(-2)
x3( 1) + x4( 0)
x3( 0) + x4( 1)

that gives us 2 vectors to cross right? still fuzzy tho

Answer 9

I am not sure but are we not supposed to solve this using spans?

Answer 10

well in my notes they do...

Answer 11

I can't recall it. But you first find span of A then find out all the linear combination of span

Answer 12

my examples make no sense to me at all, how do you even find the span?

Answer 13

the column space would contain all the vectors that are expressed as columns in A, right?

Answer 14

http://www.wolframalpha.com/input/?i=column+space+%7B%7B2%2C4%2C6%2C4%7D%2C%7B2%2C5%2C7%2C6%7D%2C%7B2%2C3%2C5%2C2%7D%7D

this might help for some direction

Answer 15

the row reduced form tells us that the first 2 vectors are sufficient for a basis:

(2,2,2) and (4,5,3)

Answer 16

crossing vectors to define a normal, anchored at the origin is fine

Answer 17

2x -y -z = 0, is what i end up with if we are to define the plane that contains the column space

Answer 18

Just to show what's in my notes....(sorry terrible writing)

Answer 19

what did you mean by crossing vectors?

Answer 20

the rref form tells us the useless vectors, or rather that the last 2 columns can be formed as combinations of the first 2 vectors

Answer 21

crossing an operation that takes 2 vectors and finds a 3rd vector that is perp to both of them

Answer 22

I think we have to find Ax=b
where b = [x y z] (Transpose this)
Now if we do Augument A and b and put it in reduce row form we should get the answer

Answer 23

Yes. That's what I was thinking ^

Answer 24

the matrix given is formed from the span of its columns

a basis is an efficient span that only defines independent vectors.

Answer 25

the rref gives us:

1 0 1 -2
0 1 1 2
0 0 0 0
^^^^^
these 2 columns depend on the first 2
(can be made from linear combinations of the first 2)

this tells us that A exists in the plane that uses: (2,2,2) and (4,5,3)

Answer 26

crossing vectors is related to taking a determinant:

x 1 4
y 1 5
z 1 3

x = 1(3) - 1(5) = -2
y = -(1(3)-1(4)) = 1
z = 1(5)-1(4) = 1

Answer 27

I get y+z-2x=0

Answer 28

yep

Answer 29

or in standard form :) 2x -y-z=0

Answer 30

Yeah I both the approaches work.

Answer 31

Oh okay thanks guys!!! step by step helped a lot too btw

Answer 32

Ohh so you understood how to solve this?

Answer 33

you can watch the last 10 min of the video if any doubts https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/null_column_space/v/visualizing-a-column-space-as-a-plane-in-r3

Answer 34

Little wishy washy but I get the idea of it. 'll def check that out haha thank you