Question

Linear algebra, a bit of help with a proof

Answer 1

Question: http://puu.sh/mIJgS/f39b777edf.png the 0 is what kind of gets me, but my idea if we say let A = that set {0, v1,...,vn} and Ax=0 corresponds to a system of p equations and n unknowns. So if n>p then there are more variables than equations so there is a free variable, then Ax=0 has nontrivial solution, and then columns of A are linearly dependent. Also is the set in R^n contains the zero vector, then the set is linearly dependent.

Answer 2

@ganeshie8 @dan815 @Kainui

Answer 3

a vector {v1,v2...vn} is linearly independent if a1*v1+a2*v2+an*vn=0

Answer 4

so in ur case linear dependent if the vector is not equal to 0

Answer 5

okay so basically its saying if the solution to Ax=0, contains only the trivial solution of vector x=0

Answer 6

okay so basically its saying if the solution to Ax=0, contains only the trivial solution of vector x=0, then there would be n lin dependant rows or colums for a n by n matrix A

Answer 7

wait so is that what your question is asking?

Answer 8

The way you stated it, doesn't sound the same to me

Answer 9

show that if Ax=0, only has x=0 as trivial solution, where x is a vector, then the rows or colums of A are lin independant

Answer 10

that is the question

Answer 11

What?

Answer 12

It's not talking about linearly independence it's asking me to proof it's linearly DEPENDENT

Answer 13

oh so your question says Ax=0, has a non trivial solution

Answer 14

Yes

Answer 15

wait so thats all there is to your question just that snippet you have there?

Answer 16

if u have a 0 vector in your set, then it has to be dependant

Answer 17

Thats the question, thats it

Answer 18

u can also show this by putting the vectors together in a matrix
and Ax=0 will have multiple non trivial solutions in the form
<k,0,0,0,0,0,0...,>
where k can be anything, if the first row of your A matridx is the 0 vector

Answer 19

Then the way I'm going about it is, let \[A = \left\{ 0, v_1, v_2,..., v_n \right\}\] where A is p x n, and the equation Ax=0 corresponds to a system of p equations and n unknowns, so if n>p then there are more variables than equations so there is a free variable, and we know it will have a nontrivial solution so it's linearly dependant. And also if the set contains the zero vector it's linearly dependant.

Answer 20

and as this dude stated earlier
"a vector {v1,v2...vn} is linearly independent if a1*v1+a2*v2+an*vn=0"
this is the lin independant statement

Answer 21

yeah this will do :

1*0 + 0*v1 + 0*v2 + .... + 0*vn = 0

Answer 22

Doesn't that prove it's linearly independent though..?

Answer 23

no u have infinite solution

Answer 24

"u can also show this by putting the vectors together in a matrix and Ax=0 will have multiple non trivial solutions in the form <k,0,0,0,0,0,0...,> where k can be anything, if the first row of your A matridx is the 0 vector"

Answer 25

that is a nontrivial linear combination yielding 0

therefore the vectors {0, v1, v2, ... , vn} are linearly "dependent"

Answer 26

finding one non trivial solution is sufficient to prove that the vectors are dependent

Answer 27

Right, I must've read that as something else, it makes sense now

Answer 28

Notice how this relates to the value of the determinant of a matrix

Answer 29

the determinant is 0 when any one line (row or column) contains all 0's

Answer 30

Right

Answer 31

when the determinant of a matrix is NOT 0, the equation Ax = 0 has "only one trivial solution"

what is it ?

Answer 32

Linearly indep

Answer 33

im asking about the trivial solution..

Answer 34

im asking about the trivial solution to Ax = 0 when the determinant of A is NOT 0

Answer 35

x + 2y = 0
x - y = 0


the only trivial solution is x = 0, y = 0

Answer 36

Wait what are we talking about here?

Answer 37

the corresponding matrix is \(\begin{bmatrix}1&2\\1&-1\end{bmatrix}\)

the column vectors of this matrix are \(v_1=\begin{pmatrix} 1\\1\end{pmatrix}\) and \(v_2=\begin{pmatrix} 2\\-1\end{pmatrix}\) are linearly independent, meaning, the only solution to \(x_1v_1+x_2v_2\) is \(x_1=x_2=0\)

Answer 38

Yes, I know that

Answer 39

If there exists a nontrivial solution, then the vectors are linearly "dependent"

so finding any one non trivial solution is sufficient

Answer 40

actually that is the definition of "linearly dependent vectors"

Answer 41

Yeah I know, I misread what you had written at first, I thought you had 0 + 0*v1 + 0*v2 + .... + 0*vn = 0 which is lin. indep. as it's all trivial solutions?

Answer 42

Oh crap the 0 vector

Answer 43

yeah in simple terms :

for a nontrivial solution you take "one" zero vector, and "none" other vectors

Answer 44

That's perfect, thank you!

Answer 45

Going back to the determinant, what you were getting across is if I sort of imagine the determinant of a 2x2 matrix as giving the area, if the 2 column vectors are the same then we will have a 0 area, something like that right haha?

Answer 46

Should've said volume and 3x3 that makes more sense but any case I think that is what you were trying to get scross

Answer 47

\(\color{blue}{\text{Originally Posted by}}\) @rational
actually that is the definition of "linearly dependent vectors"
\(\color{blue}{\text{End of Quote}}\)

Ah so that's what I explained eh haha. This was way simpler than I was thinking, thank you! :)