Question

determine whether the following are linearly indepent or not
alpha=(1,-1,0) ;
beta=(1,3,-1) ;
gamma=(5,3,-1)

Answer 1

the first one is alpha =(1,-1,0)

Answer 2

Are these vectors or points?

Answer 3

alpha, beta, gamma usually denote angles.

Answer 4

vectors

Answer 5

A vector is linearly dependent iff one can be expressed as a multiply of another given vector, for instance

\[ \Large \vec{a}=c\vec{b}, \text{where c is scalar} \] are two linearly dependent vectors.

Answer 6

You maybe also hear or read about redundant vectors, which means that one vector just doesn't carry any additional informations.

Answer 7

do you know how to do elimination?

Answer 8

yes

Answer 9

If you get a row with all zeros, then the 3 vectors are dependent.
if you get 3 pivots they are independent

Answer 10

but a vector is linearl independent if all the scalars are equal to 0. but i don't get that.

Answer 11

ie let x,y,z be scalars ,
then by using methods (e.g gramer 's etc) to solve ,i must get everything as zero .

Answer 12

I believe you could setup a \( 3 \times 3 \)-Matrix and then check it with the determinant method, if you get determinant = 0 then they are linearly dependent.

Answer 13

but if I am not completely mislead by my intuition, then this vectors are linearly independent anyway.

Answer 14

think of it as solving
\[\left[\begin{matrix}1 & -1 & 0\\ 1 & 3 & -1 \\ 5 &3 & -1\end{matrix}\right]\left[\begin{matrix}a \\ b\\c\end{matrix}\right]=\left[\begin{matrix}0 \\ 0\\0\end{matrix}\right]\]

Answer 15

@Spacelimbus if my det =0 my scalars will be undefined because anything divide by zero is undefined

Answer 16

elimination is the easiest way to solve, and if you get 3 pivots, the c must be 0
and this will force b to be 0, and then a to be 0 (when you backsolve)
so 3 pivots means 3 independent vectors.

Answer 17

I maybe shouldn't have said determinant method, the determinant isn't what you divide by.

\[\left(\begin{matrix}2 \\ 3\end{matrix}\right) \text{and} \left(\begin{matrix}4 \\ 6\end{matrix}\right)\] are linearly dependent because their determinant is 0.

Answer 18

\[\det \left[\begin{matrix}2 & 4 \\ 3 & 6\end{matrix}\right]=0\]

Answer 19

yes, in the textbook they are saying it linearly dependebd since we can find infinitely many solution . i don't how do they got this answer.

Answer 20

@Spacelimbus are using my points or yours

Answer 21

they would be dependent if gamma were (5,3,-2)

Answer 22

linearly dependent?

\[1\left[\begin{matrix}3 & -1 \\ 3 & -1\end{matrix}\right]+1\left[\begin{matrix}1 & -1 \\ 5 & -1\end{matrix}\right]+0\left[\begin{matrix}1 & 3 \\ 5 & 3\end{matrix}\right]\]

Answer 23

not sure yet if I am missing something.

Answer 24

they are saying one such solution is 3(alpha)+2(beta)-gamma=(0,0,0)

Answer 25

that means gamma must be (5,3,-2)
so maybe a typo in the question as you start with gamma= (5,3 ,-1)

Answer 26

yes @phi, then my method would work too.

\[1\left[\begin{matrix}3 & -1 \\ 3 & -1\end{matrix}\right]+1\left[\begin{matrix}1 & -1 \\ 5 & -2\end{matrix}\right]+0\left[\begin{matrix}1 & 3 \\ 5 & 3\end{matrix}\right]\]

Answer 27

determinant is zero, therefore linearly dependent.

Answer 28

\[  1\det\left[\begin{matrix}3 & -1 \\ 3 & -2\end{matrix}\right]+1\det\left[\begin{matrix}1 & -1 \\ 5 & -2\end{matrix}\right]+0\det\left[\begin{matrix}1 & 3 \\ 5 & 3\end{matrix}\right]\]

Correction*

Answer 29

@Spacelimbus wat if my det is not zero but my scalars a all zero wat will happen

Answer 30

they are saying one such solution is 3(alpha)+2(beta)-gamma=(0,0,0)

if we arrange the vectors as columns in a matrix,

\[\left[\begin{matrix}1 & 1 & 5 & | & 0\\ -1 &3 & 3&|&0\\0& -1&-2&|&0\end{matrix}\right]\]
now do elimination to reduced row echelon form

Answer 31

vector times scalar when scalar is zero will also equal to zero. But I don't see why I would want to include scalar identities for that, if I read that correctly, all we want to check is if or if not the given vectors are linearly dependent, For the method via a Matrix.

Answer 32

I have to disappear for a moment, but @REMAINDER, did you check already if the given points above in your question are correct?

Answer 33

@Spacelimbus ok

Answer 34

we would get
\[\left[\begin{matrix}1 & 0 & 3 & | & 0\\ 0 &1 & 2&|&0\\0& 0&0&|&0\end{matrix}\right]\] first we see there are 2 pivots (not 3) so linearly dependent
second the 3rd column tells us that 3 * alpha (the 1st vector) + 2* beta will equal the 3rd vector gamma

Answer 35

yes i also got that wat do u mean beta will equall the 3rd vector gamma

Answer 36

the third column of the rref matrix (3,2,0) tells us that 3*alpha + 2*beta= gamma

Answer 37

@beketso check this

Answer 38

@Spacelimbus wat about this one\[\left\{ -1+x-x ^{2};2+x+x^{2};1-x+x^{2};1+x\right\}\]

Answer 39

Interesting notation, are these supposed to be functions in dimensions? first one would be the \(i\) component, second one the \(j\) component, third one the \(k\) and last one the \(m\) component?

Answer 40

yes

Answer 41

I would say linearly depending, because the \(i\) component added to the \(k\) component gives you

\[ 0+ 0 + 0 \]

Answer 42

It looks like they are asking if these polynomials are independent, and as space points out the 3rd polynomial is -1* first polynomial
so this set is dependent.
we can still use elimination to show the other 3 are independent.

Answer 43

otherwise, I would maybe consider using the Wronski Determinant.

Answer 44

oh by the way @phi in the first question at the last row wat if the last pivot was having a 1 ,wat i'm i gonna say to my answer.

Answer 45

If you get 3 pivots (using the original gamma = (5,3,-1) that would happen)
then the 3 vectors are independent.
If I were answering this, I would answer it both ways (w the correct gamma), and ask for extra credit!

Answer 46

@experimentX check this

Answer 47

huh ... where are we?

Answer 48

try to determinate the rank of that matrix or try to reduce it to reduced row echelon form ... try pluggin into wolf

Answer 49

http://tinyurl.com/9os48og

Answer 50

let me try one another thing

Answer 51

walf it gives me the inverse

Answer 52

of what? these vectors are linearly independent

Answer 53

of the original matrix

Answer 54

yep ...

Answer 55

http://www.wolframalpha.com/input/?i=solve+a+%281%2C3%2C-1%29+%2B+b+%285%2C3%2C-1%29+%2B+c+%281%2C-1%2C0%29+%3D+0

Answer 56

the answer from the textbook is that it is linearly dependent

Answer 57

well ... there must be some typo ... check this ... did I make any error input
http://www.wolframalpha.com/input/?i=Determinant%5B%7B%7B1%2C-1%2C0%7D%2C%7B1%2C3%2C-1%7D%2C%7B5%2C3%2C-1%7D%7D%5D

Answer 58

@beketso check this

Answer 59

@REMAINDER it is clear, based on the textbook answer, that the question as posed has a typo. They give gamma as (5,3,-1) and they meant (5,3,-2)

(math books often have typos, so this is not surprising)

Answer 60

@phi i agree with u it is that wolf confused me

Answer 61

@REMAINDER ,remember,vectors are linearly dependent if they have a non-trivial linear relation.based on that i say ur given vectors are linearly dependent.

Answer 62

Here is wolf's answer using the correct gamma
http://www.wolframalpha.com/input/?i=rref%5Btranspose%7B%7B1%2C-1%2C0%7D%2C%7B1%2C3%2C-1%7D%2C%7B5%2C3%2C-2%7D%7D%5D

Answer 63

and its determinant is 0 (indicating the matrix is non-invertible, and so has linearly dependent columns (or rows))

Answer 64

so @phi .frankly,that means if the rows/columns are linearly independent then the vectors are also linearly dependent?