How are vectors linearly independent?
So I have these $4$ vectors (they are column vectors, but I can't put them in the right form , somebody please help)
$v_1= (1,1,0,0)$
$v_2=(1,-1,0,0)$
$v_3=(0,2,0,0)$
$v_4=(0,2,1,0)$
The professor said that only $v_1,v_2,$ and $v_4$ are linearly independent.
He remarks by saying that that $v_3 = v_1 − v_2$; since $v_1$ and $v_2$ are clearly
independent and both have $0$ as their third coordinate while $v_4$ has a non-zero third coordinate, a set of three linearly independent vectors is clearly provided by ${v_1, v_2, v_4}$.
I did not understand this process. Can somebody please help me understand why only $3$ vectors out of $4$ are linearly independent? Thank you!
I know that vectors are L.I. if they have only the trivial solution , but I'm not sure how to apply this here.
Please explain as simply as you can too, if that is possible.
vector-spaces vectors
asked Nov 21 '18 at 12:35
GGGG
176
add a comment |
So I have these $4$ vectors (they are column vectors, but I can't put them in the right form , somebody please help)
$v_1= (1,1,0,0)$
$v_2=(1,-1,0,0)$
$v_3=(0,2,0,0)$
$v_4=(0,2,1,0)$
The professor said that only $v_1,v_2,$ and $v_4$ are linearly independent.
He remarks by saying that that $v_3 = v_1 − v_2$; since $v_1$ and $v_2$ are clearly
independent and both have $0$ as their third coordinate while $v_4$ has a non-zero third coordinate, a set of three linearly independent vectors is clearly provided by ${v_1, v_2, v_4}$.
I did not understand this process. Can somebody please help me understand why only $3$ vectors out of $4$ are linearly independent? Thank you!
I know that vectors are L.I. if they have only the trivial solution , but I'm not sure how to apply this here.
Please explain as simply as you can too, if that is possible.
vector-spaces vectors
asked Nov 21 '18 at 12:35
GGGG
176
1
Please review, and include, the definition you've learned for linear independence of vectors. $v_3$ is a linear combination of $v_1, v_2$, specifically, as your professor pointed out, $(0,2, 0,0) = (1,1,0,0) - (1, -1, 0, 0)$.
– amWhy
Nov 21 '18 at 12:37
How are linear combinations and indepenence connected ? Im sorry i dont understand
– GGGG
Nov 21 '18 at 12:42
A set of vectors is linearly independent if you cannot write one as a linear combination of the others
– Paul
Nov 21 '18 at 12:54
add a comment |
So I have these $4$ vectors (they are column vectors, but I can't put them in the right form , somebody please help)
$v_1= (1,1,0,0)$
$v_2=(1,-1,0,0)$
$v_3=(0,2,0,0)$
$v_4=(0,2,1,0)$
The professor said that only $v_1,v_2,$ and $v_4$ are linearly independent.
He remarks by saying that that $v_3 = v_1 − v_2$; since $v_1$ and $v_2$ are clearly
independent and both have $0$ as their third coordinate while $v_4$ has a non-zero third coordinate, a set of three linearly independent vectors is clearly provided by ${v_1, v_2, v_4}$.
I did not understand this process. Can somebody please help me understand why only $3$ vectors out of $4$ are linearly independent? Thank you!
I know that vectors are L.I. if they have only the trivial solution , but I'm not sure how to apply this here.
Please explain as simply as you can too, if that is possible.
vector-spaces vectors
asked Nov 21 '18 at 12:35
GGGG
176
So I have these $4$ vectors (they are column vectors, but I can't put them in the right form , somebody please help)
$v_1= (1,1,0,0)$
$v_2=(1,-1,0,0)$
$v_3=(0,2,0,0)$
$v_4=(0,2,1,0)$
The professor said that only $v_1,v_2,$ and $v_4$ are linearly independent.
He remarks by saying that that $v_3 = v_1 − v_2$; since $v_1$ and $v_2$ are clearly
independent and both have $0$ as their third coordinate while $v_4$ has a non-zero third coordinate, a set of three linearly independent vectors is clearly provided by ${v_1, v_2, v_4}$.
I did not understand this process. Can somebody please help me understand why only $3$ vectors out of $4$ are linearly independent? Thank you!
I know that vectors are L.I. if they have only the trivial solution , but I'm not sure how to apply this here.
Please explain as simply as you can too, if that is possible.
vector-spaces vectors
vector-spaces vectors
asked Nov 21 '18 at 12:35
GGGG
176
asked Nov 21 '18 at 12:35
GGGG
176
edited Nov 21 '18 at 12:38
asked Nov 21 '18 at 12:35
GGGG
176
asked Nov 21 '18 at 12:35
GGGG
176
asked Nov 21 '18 at 12:35
GGGG
176
176
1
Please review, and include, the definition you've learned for linear independence of vectors. $v_3$ is a linear combination of $v_1, v_2$, specifically, as your professor pointed out, $(0,2, 0,0) = (1,1,0,0) - (1, -1, 0, 0)$.
– amWhy
Nov 21 '18 at 12:37
How are linear combinations and indepenence connected ? Im sorry i dont understand
– GGGG
Nov 21 '18 at 12:42
A set of vectors is linearly independent if you cannot write one as a linear combination of the others
– Paul
Nov 21 '18 at 12:54
add a comment |
1
Please review, and include, the definition you've learned for linear independence of vectors. $v_3$ is a linear combination of $v_1, v_2$, specifically, as your professor pointed out, $(0,2, 0,0) = (1,1,0,0) - (1, -1, 0, 0)$.
– amWhy
Nov 21 '18 at 12:37
How are linear combinations and indepenence connected ? Im sorry i dont understand
– GGGG
Nov 21 '18 at 12:42
A set of vectors is linearly independent if you cannot write one as a linear combination of the others
– Paul
Nov 21 '18 at 12:54
1
1
Please review, and include, the definition you've learned for linear independence of vectors. $v_3$ is a linear combination of $v_1, v_2$, specifically, as your professor pointed out, $(0,2, 0,0) = (1,1,0,0) - (1, -1, 0, 0)$.
– amWhy
Nov 21 '18 at 12:37
Please review, and include, the definition you've learned for linear independence of vectors. $v_3$ is a linear combination of $v_1, v_2$, specifically, as your professor pointed out, $(0,2, 0,0) = (1,1,0,0) - (1, -1, 0, 0)$.
– amWhy
Nov 21 '18 at 12:37
How are linear combinations and indepenence connected ? Im sorry i dont understand
– GGGG
Nov 21 '18 at 12:42
How are linear combinations and indepenence connected ? Im sorry i dont understand
– GGGG
Nov 21 '18 at 12:42
A set of vectors is linearly independent if you cannot write one as a linear combination of the others
– Paul
Nov 21 '18 at 12:54
A set of vectors is linearly independent if you cannot write one as a linear combination of the others
– Paul
Nov 21 '18 at 12:54
add a comment |
1 Answer
1
active
oldest
votes
(*Since you are a new user I shall try to explain. But is best if you show in this forum what you have done so that we can guide you.)
Linear Independence: $v_1,v_2,cdots,v_n$ are linearly independent if the only solution to the equation $a_1v_1+a_2v_2+cdots+a_nv_n=0$ is $a_1=a_2=cdots=a_n=0$.
So here we have $a_1v_1+a_2v_2+cdots+a_nv_n=0 implies a_1begin{bmatrix}1\1\0\0end{bmatrix}+a_2begin{bmatrix}1\-1\0\0end{bmatrix}+a_3begin{bmatrix}0\2\0\0end{bmatrix}+a_4begin{bmatrix}0\2\1\0end{bmatrix}=begin{bmatrix}0\0\0\0end{bmatrix}$ which gives us four equations $$a_1+a_2=0\a_1-a_2+2a_3+2a_4=0\a_3=0.$$ Using $a_3=0,a_1=-a_2$ in the second equation gives you $a_3=a_4$. So we have got a non-zero solution $(-a_2,a_2,0,a_2)$ for $a_1v_1+a_2v_2+cdots+a_4v_4=0 $. Thus the four vectors ${a_1,a_2,a_3,a_4}$ are liearly dependent.
You may try to show linear independence or dependence with any set of $3$ vectors from the above $4$ vectors to understand further.
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
But on the answers it states that only 3 out the 4 vectors are linearly independent? How come here you say that all 4 are linearly dependent?
– GGGG
Nov 21 '18 at 13:21
@GGGG: Here all four vectors taken together as a set are linearly dependent. But any three of them taken as a set can be linearly dependent. For this take any three from the four vectors you have mentioned and do the repeat the above procedure. If you end up with $a_1=a_2=cdots=a_n=0$ being the only solution, then those set of vectors are linearly independent.
– Yadati Kiran
Nov 21 '18 at 13:26
Im sorry , so i have to repeat this procedure ? I didn't understand. How can i choose the three vectors? Isn't there a simpler solution ?
– GGGG
Nov 21 '18 at 13:27
You may try this. It might take time the first time but it is simple enough though. You will also intuitively find other ways to solve once you have understood this. Take ${v_1,v_2,v_3}$ for the first try and follow the above procedure. Then take any other set of three vectors say ${v_1,v_2,v_4}$ and see what happens.
– Yadati Kiran
Nov 21 '18 at 13:29
The problem is that i don't understand how only 3 out of 4 can be independent. ISn't there another way to do this?
– GGGG
Nov 21 '18 at 13:32
|
show 2 more comments
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(*Since you are a new user I shall try to explain. But is best if you show in this forum what you have done so that we can guide you.)
Linear Independence: $v_1,v_2,cdots,v_n$ are linearly independent if the only solution to the equation $a_1v_1+a_2v_2+cdots+a_nv_n=0$ is $a_1=a_2=cdots=a_n=0$.
So here we have $a_1v_1+a_2v_2+cdots+a_nv_n=0 implies a_1begin{bmatrix}1\1\0\0end{bmatrix}+a_2begin{bmatrix}1\-1\0\0end{bmatrix}+a_3begin{bmatrix}0\2\0\0end{bmatrix}+a_4begin{bmatrix}0\2\1\0end{bmatrix}=begin{bmatrix}0\0\0\0end{bmatrix}$ which gives us four equations $$a_1+a_2=0\a_1-a_2+2a_3+2a_4=0\a_3=0.$$ Using $a_3=0,a_1=-a_2$ in the second equation gives you $a_3=a_4$. So we have got a non-zero solution $(-a_2,a_2,0,a_2)$ for $a_1v_1+a_2v_2+cdots+a_4v_4=0 $. Thus the four vectors ${a_1,a_2,a_3,a_4}$ are liearly dependent.
You may try to show linear independence or dependence with any set of $3$ vectors from the above $4$ vectors to understand further.
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
But on the answers it states that only 3 out the 4 vectors are linearly independent? How come here you say that all 4 are linearly dependent?
– GGGG
Nov 21 '18 at 13:21
@GGGG: Here all four vectors taken together as a set are linearly dependent. But any three of them taken as a set can be linearly dependent. For this take any three from the four vectors you have mentioned and do the repeat the above procedure. If you end up with $a_1=a_2=cdots=a_n=0$ being the only solution, then those set of vectors are linearly independent.
– Yadati Kiran
Nov 21 '18 at 13:26
Im sorry , so i have to repeat this procedure ? I didn't understand. How can i choose the three vectors? Isn't there a simpler solution ?
– GGGG
Nov 21 '18 at 13:27
You may try this. It might take time the first time but it is simple enough though. You will also intuitively find other ways to solve once you have understood this. Take ${v_1,v_2,v_3}$ for the first try and follow the above procedure. Then take any other set of three vectors say ${v_1,v_2,v_4}$ and see what happens.
– Yadati Kiran
Nov 21 '18 at 13:29
The problem is that i don't understand how only 3 out of 4 can be independent. ISn't there another way to do this?
– GGGG
Nov 21 '18 at 13:32
|
show 2 more comments
(*Since you are a new user I shall try to explain. But is best if you show in this forum what you have done so that we can guide you.)
Linear Independence: $v_1,v_2,cdots,v_n$ are linearly independent if the only solution to the equation $a_1v_1+a_2v_2+cdots+a_nv_n=0$ is $a_1=a_2=cdots=a_n=0$.
So here we have $a_1v_1+a_2v_2+cdots+a_nv_n=0 implies a_1begin{bmatrix}1\1\0\0end{bmatrix}+a_2begin{bmatrix}1\-1\0\0end{bmatrix}+a_3begin{bmatrix}0\2\0\0end{bmatrix}+a_4begin{bmatrix}0\2\1\0end{bmatrix}=begin{bmatrix}0\0\0\0end{bmatrix}$ which gives us four equations $$a_1+a_2=0\a_1-a_2+2a_3+2a_4=0\a_3=0.$$ Using $a_3=0,a_1=-a_2$ in the second equation gives you $a_3=a_4$. So we have got a non-zero solution $(-a_2,a_2,0,a_2)$ for $a_1v_1+a_2v_2+cdots+a_4v_4=0 $. Thus the four vectors ${a_1,a_2,a_3,a_4}$ are liearly dependent.
You may try to show linear independence or dependence with any set of $3$ vectors from the above $4$ vectors to understand further.
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
But on the answers it states that only 3 out the 4 vectors are linearly independent? How come here you say that all 4 are linearly dependent?
– GGGG
Nov 21 '18 at 13:21
@GGGG: Here all four vectors taken together as a set are linearly dependent. But any three of them taken as a set can be linearly dependent. For this take any three from the four vectors you have mentioned and do the repeat the above procedure. If you end up with $a_1=a_2=cdots=a_n=0$ being the only solution, then those set of vectors are linearly independent.
– Yadati Kiran
Nov 21 '18 at 13:26
Im sorry , so i have to repeat this procedure ? I didn't understand. How can i choose the three vectors? Isn't there a simpler solution ?
– GGGG
Nov 21 '18 at 13:27
You may try this. It might take time the first time but it is simple enough though. You will also intuitively find other ways to solve once you have understood this. Take ${v_1,v_2,v_3}$ for the first try and follow the above procedure. Then take any other set of three vectors say ${v_1,v_2,v_4}$ and see what happens.
– Yadati Kiran
Nov 21 '18 at 13:29
The problem is that i don't understand how only 3 out of 4 can be independent. ISn't there another way to do this?
– GGGG
Nov 21 '18 at 13:32
|
show 2 more comments
(*Since you are a new user I shall try to explain. But is best if you show in this forum what you have done so that we can guide you.)
Linear Independence: $v_1,v_2,cdots,v_n$ are linearly independent if the only solution to the equation $a_1v_1+a_2v_2+cdots+a_nv_n=0$ is $a_1=a_2=cdots=a_n=0$.
So here we have $a_1v_1+a_2v_2+cdots+a_nv_n=0 implies a_1begin{bmatrix}1\1\0\0end{bmatrix}+a_2begin{bmatrix}1\-1\0\0end{bmatrix}+a_3begin{bmatrix}0\2\0\0end{bmatrix}+a_4begin{bmatrix}0\2\1\0end{bmatrix}=begin{bmatrix}0\0\0\0end{bmatrix}$ which gives us four equations $$a_1+a_2=0\a_1-a_2+2a_3+2a_4=0\a_3=0.$$ Using $a_3=0,a_1=-a_2$ in the second equation gives you $a_3=a_4$. So we have got a non-zero solution $(-a_2,a_2,0,a_2)$ for $a_1v_1+a_2v_2+cdots+a_4v_4=0 $. Thus the four vectors ${a_1,a_2,a_3,a_4}$ are liearly dependent.
You may try to show linear independence or dependence with any set of $3$ vectors from the above $4$ vectors to understand further.
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
(*Since you are a new user I shall try to explain. But is best if you show in this forum what you have done so that we can guide you.)
Linear Independence: $v_1,v_2,cdots,v_n$ are linearly independent if the only solution to the equation $a_1v_1+a_2v_2+cdots+a_nv_n=0$ is $a_1=a_2=cdots=a_n=0$.
So here we have $a_1v_1+a_2v_2+cdots+a_nv_n=0 implies a_1begin{bmatrix}1\1\0\0end{bmatrix}+a_2begin{bmatrix}1\-1\0\0end{bmatrix}+a_3begin{bmatrix}0\2\0\0end{bmatrix}+a_4begin{bmatrix}0\2\1\0end{bmatrix}=begin{bmatrix}0\0\0\0end{bmatrix}$ which gives us four equations $$a_1+a_2=0\a_1-a_2+2a_3+2a_4=0\a_3=0.$$ Using $a_3=0,a_1=-a_2$ in the second equation gives you $a_3=a_4$. So we have got a non-zero solution $(-a_2,a_2,0,a_2)$ for $a_1v_1+a_2v_2+cdots+a_4v_4=0 $. Thus the four vectors ${a_1,a_2,a_3,a_4}$ are liearly dependent.
You may try to show linear independence or dependence with any set of $3$ vectors from the above $4$ vectors to understand further.
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
answered Nov 21 '18 at 13:00
Yadati Kiran
1,693619
1,693619
But on the answers it states that only 3 out the 4 vectors are linearly independent? How come here you say that all 4 are linearly dependent?
– GGGG
Nov 21 '18 at 13:21
@GGGG: Here all four vectors taken together as a set are linearly dependent. But any three of them taken as a set can be linearly dependent. For this take any three from the four vectors you have mentioned and do the repeat the above procedure. If you end up with $a_1=a_2=cdots=a_n=0$ being the only solution, then those set of vectors are linearly independent.
– Yadati Kiran
Nov 21 '18 at 13:26
Im sorry , so i have to repeat this procedure ? I didn't understand. How can i choose the three vectors? Isn't there a simpler solution ?
– GGGG
Nov 21 '18 at 13:27
You may try this. It might take time the first time but it is simple enough though. You will also intuitively find other ways to solve once you have understood this. Take ${v_1,v_2,v_3}$ for the first try and follow the above procedure. Then take any other set of three vectors say ${v_1,v_2,v_4}$ and see what happens.
– Yadati Kiran
Nov 21 '18 at 13:29
The problem is that i don't understand how only 3 out of 4 can be independent. ISn't there another way to do this?
– GGGG
Nov 21 '18 at 13:32
|
show 2 more comments
But on the answers it states that only 3 out the 4 vectors are linearly independent? How come here you say that all 4 are linearly dependent?
– GGGG
Nov 21 '18 at 13:21
@GGGG: Here all four vectors taken together as a set are linearly dependent. But any three of them taken as a set can be linearly dependent. For this take any three from the four vectors you have mentioned and do the repeat the above procedure. If you end up with $a_1=a_2=cdots=a_n=0$ being the only solution, then those set of vectors are linearly independent.
– Yadati Kiran
Nov 21 '18 at 13:26
Im sorry , so i have to repeat this procedure ? I didn't understand. How can i choose the three vectors? Isn't there a simpler solution ?
– GGGG
Nov 21 '18 at 13:27
You may try this. It might take time the first time but it is simple enough though. You will also intuitively find other ways to solve once you have understood this. Take ${v_1,v_2,v_3}$ for the first try and follow the above procedure. Then take any other set of three vectors say ${v_1,v_2,v_4}$ and see what happens.
– Yadati Kiran
Nov 21 '18 at 13:29
The problem is that i don't understand how only 3 out of 4 can be independent. ISn't there another way to do this?
– GGGG
Nov 21 '18 at 13:32
But on the answers it states that only 3 out the 4 vectors are linearly independent? How come here you say that all 4 are linearly dependent?
– GGGG
Nov 21 '18 at 13:21
But on the answers it states that only 3 out the 4 vectors are linearly independent? How come here you say that all 4 are linearly dependent?
– GGGG
Nov 21 '18 at 13:21
@GGGG: Here all four vectors taken together as a set are linearly dependent. But any three of them taken as a set can be linearly dependent. For this take any three from the four vectors you have mentioned and do the repeat the above procedure. If you end up with $a_1=a_2=cdots=a_n=0$ being the only solution, then those set of vectors are linearly independent.
– Yadati Kiran
Nov 21 '18 at 13:26
@GGGG: Here all four vectors taken together as a set are linearly dependent. But any three of them taken as a set can be linearly dependent. For this take any three from the four vectors you have mentioned and do the repeat the above procedure. If you end up with $a_1=a_2=cdots=a_n=0$ being the only solution, then those set of vectors are linearly independent.
– Yadati Kiran
Nov 21 '18 at 13:26
Im sorry , so i have to repeat this procedure ? I didn't understand. How can i choose the three vectors? Isn't there a simpler solution ?
– GGGG
Nov 21 '18 at 13:27
Im sorry , so i have to repeat this procedure ? I didn't understand. How can i choose the three vectors? Isn't there a simpler solution ?
– GGGG
Nov 21 '18 at 13:27
You may try this. It might take time the first time but it is simple enough though. You will also intuitively find other ways to solve once you have understood this. Take ${v_1,v_2,v_3}$ for the first try and follow the above procedure. Then take any other set of three vectors say ${v_1,v_2,v_4}$ and see what happens.
– Yadati Kiran
Nov 21 '18 at 13:29
You may try this. It might take time the first time but it is simple enough though. You will also intuitively find other ways to solve once you have understood this. Take ${v_1,v_2,v_3}$ for the first try and follow the above procedure. Then take any other set of three vectors say ${v_1,v_2,v_4}$ and see what happens.
– Yadati Kiran
Nov 21 '18 at 13:29
The problem is that i don't understand how only 3 out of 4 can be independent. ISn't there another way to do this?
– GGGG
Nov 21 '18 at 13:32
The problem is that i don't understand how only 3 out of 4 can be independent. ISn't there another way to do this?
– GGGG
Nov 21 '18 at 13:32
|
show 2 more comments
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1
Please review, and include, the definition you've learned for linear independence of vectors. $v_3$ is a linear combination of $v_1, v_2$, specifically, as your professor pointed out, $(0,2, 0,0) = (1,1,0,0) - (1, -1, 0, 0)$.
– amWhy
Nov 21 '18 at 12:37
How are linear combinations and indepenence connected ? Im sorry i dont understand
– GGGG
Nov 21 '18 at 12:42
A set of vectors is linearly independent if you cannot write one as a linear combination of the others
– Paul
Nov 21 '18 at 12:54