Question

Linear algebra,

consider the following elements of R^3 x = (1 2 1), y = (1 3 2) , z = (1 1 0) , w = (3 8 5)

Show that Span{x,y} = Span{z,w}

Answer 1

what grade do you study ? @Christos

Answer 2

this question is looking like you study in master or in bachelor.

Answer 3

bachelor

Answer 4

yeah. I also covered this subject in my last semester " linear Algebra and Analytical Geometer". But our teacher was so bad, so I could not understand this whole chapter.
But I passed my this subject with A grade.

Answer 5

In other words, assume a vector \(v \in \mathbb R^3\) that is an element of the span of \(x,y\). We show that it belongs to the span of \(w, z\) too.

Answer 6

Then show it backwards.

Answer 7

What does Span{} = Span{} specifically mean

Answer 8

Im not sure what I am supposed to show exactly

Answer 9

It just means that x, y span over the same set as w, z. Are you familiar with the word span?

Answer 10

I would say that it means the vector space that can be created from a and b is the vector space that can be created from d and c

Answer 11

I know what "spans" mean not Span(something)

Answer 12

or am I worng in assuming its a vector space... so much ambiguity ll

Answer 13

Yeah, it's the same thing.

span(a1, a2, a3, ..., an) is the set of vectors that can be expressed as a linear combination of a1, a2, a3, ..., an.

Answer 14

yes yes

Answer 15

so how would you tackle this ? If I see one then I can solve anything same :D

Answer 16

I can tell you what I thought and you tell me if I am right but.... I have unorthodox ways

Answer 17

Oh, let's look at your thoughts first then. :D

Answer 18

taken from Dr strang

Answer 19

well

Answer 20

I would find the rref of the matrix of those collumn vectors

Answer 21

and that is gonna give me a zero row

Answer 22

and now my six sense tells me that they are related

Answer 23

cause of dependence

Answer 24

@Christos you should find the book of linear algebra and analytical algebra by Urs. Shaikh. This will help you a lot

Answer 25

Ah, I kinda see where you're coming from. If you want to do this purely from a basic point of view, then here's how you go about it:

assume that vector \(v\) is an element of span(x, y). then\[v = \alpha x + \beta y = \alpha (1, 2, 1) + \beta (1, 3, 2)\]we have to show that there exist \(\alpha_0, \beta_0\) such that \(v =\alpha_0 z+\beta_0w \). To find these, let's begin:\[v = (\alpha + \beta, 2\alpha + 3\beta, \alpha + 2 \beta) \]\[v=(\alpha_0 + 3 \beta_0, \alpha_0 + 8 \beta_0, 5\beta_0 )\]In other words,\[\beta_0 = \frac{\alpha + 2\beta}{5}\]\[\alpha_0 = 2\alpha + 3\beta - 8\left(\frac{\alpha+2\beta}{5}\right)\]So we have proven the existence of \(\alpha_0, \beta_0\).

Answer 26

see now, I didnt understand what you just did

Answer 27

from "let's begin: " to "in other words," I lose you

Answer 28

Basically, I started by assuming that \(v\) is an arbitrary vector in the span of \(x\) and \(y\). Then \(v\) is expressible as \(\alpha x + \beta y\). I later proved that this vector also belongs to the span of \(z \) and \(w\) by finding \(\alpha_0, \beta_0\) such that \(v = \alpha_0 w + \beta_0 z\).

Answer 29

;p

Answer 30

how did we find a_0 b_0

Answer 31

beta_0 *

Answer 32

read the solution once again

Answer 33

I dont understand what this is (α+β,2α+3β,α+2β)

Answer 34

Like how is this derived from what

Answer 35

those :

v=(α+β,2α+3β,α+2β)
v=(α0+3β0,α0+8β0,5β0)

Answer 36

Err, OK. \[v = \alpha x + \beta y = \alpha(1, 2, 1) + \beta(1, 3, 2)\]\[v = \alpha_0 z + \beta_0 w = \alpha_0 (1, 1, 0) + \beta_0 (3, 8, 5)\]

Answer 37

Then I solved for \(\alpha_0\) and \(\beta_0\) in terms of \(\alpha\) and \(\beta\).

Answer 38

ok so those are equations of the linear sytem

Answer 39

I really think there has to be some other way that I am missing

Answer 40

ill research more about it

Answer 41

Kinda. If you want to see it this way, then here it is:\[\alpha(1, 1, 0) + \beta(3, 8, 5)\]\[= \alpha(1, 1, 0) + 2\beta(1, 3, 2) + \beta(1, 2, 1)\]\[= 2\alpha (1, 2, 1) - \alpha (1, 3, 2) + 2 \beta(1, 3, 2) + \beta(1, 2, 1)\]\[=(2\alpha + \beta) (1, 2,1) + (2\beta - \alpha) (1, 3, 2)\]

Answer 42

Do you see what I did there? I started with a linear combination of w, z and ended up with a linear combination of x, y.