Linear independence of three vectors

[Salmone]

If I've got three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ and $\vec{a}$, $\vec{b}$ are linearly independent and $\vec{c}$ is linearly independent from $\vec{a}$, is $\vec{c}$ also linearly independent from $\vec{b}$?

 

[PeroK]

If I've got three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ and $\vec{a}$, $\vec{b}$ are linearly independent and $\vec{c}$ is linearly independent from $\vec{a}$, is $\vec{c}$ also linearly independent from $\vec{b}$?

You mean assuming $\vec c \ne \vec b$?

 

[Salmone]

You mean assuming $\vec c \ne \vec b$?

Yes.

 

[PeroK]

Yes.

Under what circumstances are two vectors linearly independent? There's a fairly simple criterion.

 

[Salmone]

Under what circumstances are two vectors linearly independent? There's a fairly simple criterion.

If the linear combination of the two is equal to $0$ only when both coefficients are equal to $0$.

 

[FactChecker]

Yes.

Can you use the $\vec c = \vec b$ counterexample to think up a whole family of other counterexamples?

 

[PeroK]

If the linear combination of the two is equal to $0$ only when both coefficients are equal to $0$.

Okay, but for only two vectors that implies something simple.

 

[Mark44]

If I've got three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ and $\vec{a}$, $\vec{b}$ are linearly independent and $\vec{c}$ is linearly independent from $\vec{a}$, is $\vec{c}$ also linearly independent from $\vec{b}$?

Under what circumstances are two vectors linearly independent? There's a fairly simple criterion.

If the linear combination of the two is equal to 0 only when both coefficients are equal to 0.

What does it mean geometrically if two vectors are linearly independent? I believe @PeroK is trying to get you to think in this direction.

 

[Salmone]

What does it mean geometrically if two vectors are linearly independent? I believe @PeroK is trying to get you to think in this direction.

They are orthogonal?

 

[PeroK]

That they are orthogonal?

There's no such thing as "orthogonal" until you define an inner product on your vector space.

The answer is that $\vec a$ and $\vec b$ are linearly independent as long as one is not a scalar multiple of the other. The only vectors that are linearly dependent with $\vec a$ are vectors of the form $\lambda \vec a$ for some scalar $\lambda$.

 

[mcastillo356]

Hi,@Salmone , a set of vectors $\{\vec{u_1},\vec{u_2},\ldots{\vec{u_{k}}}\}$ is linearly independent $\Longleftrightarrow{\lambda_{1}\vec{u_1}+\lambda_{2}\vec{u_2}+\ldots{+\lambda_{k}\vec{u_{k}}}=\vec{0}}$, then $\lambda_1=\lambda_2=\ldots{=\lambda_{k}=0}$,...Hmm, I think I'm late,

 

[Salmone]

There's no such thing as "orthogonal" until you define an inner product on your vector space.

The answer is that $\vec a$ and $\vec b$ are linearly independent as long as one is not a scalar multiple of the other. The only vectors that are linearly dependent with $\vec a$ are vectors of the form $\lambda \vec a$ for some scalar $\lambda$.

Okay, so can we say the same for more than three vectors. For example vectors $\vec{a},\vec{b},\vec{c}$ are linearly independent and a fourth vector $\vec{d}$ is linearly independent to $\vec{a}$ then it is linearly independent to $\vec{b}$ and $\vec{c}$ if $\vec{d} \neq \vec{b} \neq \vec{c}$

 

[PeroK]

Okay, so can we say the same for more than three vectors. For example vectors $\vec{a},\vec{b},\vec{c}$ are linearly independent and a fourth vector $\vec{d}$ is linearly independent to $\vec{a}$ then it is linearly independent to $\vec{b}$ and $\vec{c}$ if $\vec{d} \neq \vec{b} \neq \vec{c}$

It doesn't matter how many vectors you have. Linear independence between any pair of vectors is a simple matter or not being mutiples of each other.

It's only when you get to three or more vectors that linear independence becomes non-trivial.

 

[Salmone]

It doesn't matter how many vectors you have. Linear independence between any pair of vectors is a simple matter or not being mutiples of each other.

It's only when you get to three or more vectors that linear independence becomes non-trivial.

So it's yes? If a vector is linearly independent on another vector and this last one is LI to another vector and so on, they are all LI?

 

[PeroK]

So it's yes? If a vector is linearly independent on another vector and this last one is LI to another vector and so on, they are all LI?

Definitely not.

 

[Salmone]

Definitely not.

So I can't understand your previous answer "It doesn't matter how many vectors you have."

 

[Mark44]

Hi,@Salmone , a set of vectors $\{\vec{u_1},\vec{u_2},\ldots{\vec{u_{k}}}\}$ is linearly independent $\Longleftrightarrow{\lambda_{1}\vec{u_1}+\lambda_{2}\vec{u_2}+\ldots{+\lambda_{k}\vec{u_{k}}}=\vec{0}}$, then $\lambda_1=\lambda_2=\ldots{=\lambda_{k}=0}$,...Hmm, I think I'm late,

That's not quite right. That set of vectors is linearly independent iff the only solution to the equation $c_1x_1 + c_2x_2 + \dots + c_nx_n = 0$ is $c_1 = c_2 = \dots = c_n = 0$

Consider the vectors x, 2x, and 3x. I notice that for the equation $c_1x + c_2*2x + c_3*3x = 0$, that $c_1 = c_2 = c_3 = 0$ is a solution. Can you conclude that these vectors are linearly independent?

 

[Mark44]

So it's yes? If a vector is linearly independent on another vector and this last one is LI to another vector and so on, they are all LI?

No. Draw a picture to see why the above isn't true.

What @PeroK was saying is that it's easy to determine whether two vectors are dependent. If they are, they either point the same direction or in opposite directions. I.e., they are either parallel or antiparallel (point in opposite directions). That's what "scalar multiples" of one another means.

If you have three or more vectors it's harder to tell whether they are linearly dependent or linearly independent.

 

[Salmone]

No. Draw a picture to see why the above isn't true.

What @PeroK was saying is that it's easy to determine whether two vectors are dependent. If they are, they either point the same direction or in opposite directions. I.e., they are either parallel or antiparallel (point in opposite directions). That's what "scalar multiples" of one another means.

If you have three or more vectors it's harder to tell whether they are linearly dependent or linearly independent.

Ok and so what is the answer to my first question? Now I'm way more confused.

 

[Mark44]

Ok and so what is the answer to my first question? Now I'm way more confused.

We're not going to just tell you the answer, but we'll help you arrive at it. If you draw a picture of three vectors, you'll probably get the answer in short order.

 

[Salmone]

We're not going to just tell you the answer, but we'll help you arrive at it. If you draw a picture of three vectors, you'll probably get the answer in short order.

I drew the three vectors and have no idea what I am supposed to figure out from this.

 

[FactChecker]

You have said that if $\vec c = \vec b$, then it is not true. In fact, if $\vec c$ is a multiple of $\vec b$, then it is not true (if this is not clear to you, then you should prove it). In fact, what can you say if $\vec c$ has even a tiny component that is a multiple of $\vec b$? So what should the answer to the problem be?

 

[Salmone]

You have said that if $\vec c = \vec b$, then it is not true. In fact, if $\vec c$ is a multiple of $\vec b$, then it is not true (if this is not clear to you, then you should prove it). In fact, what can you say if $\vec c$ has even a tiny component that is a multiple of $\vec b$? So what should the answer to the problem be?

If $\vec{c}$ has a tiny component that is a multiple of $\vec{b}$ they are linearly indipendent.

 

[Salmone]

We're not going to just tell you the answer, but we'll help you arrive at it. If you draw a picture of three vectors, you'll probably get the answer in short order.

What I can imagine is that: if $\vec{a}$ and $\vec{b}$ are linearly independent then they are not a multiple of each other and if $\vec{c}$ is linearly independent with $\vec{a}$ the same can be said of these two BUT, since $\vec{c}$ and $\vec{b}$ are different they cannot even be multiples of each other so the answer should be yes, right?

 

[Mark44]

Due to an error on one poster's part, I have deleted that post as well as another post that pointed out the error.

 

[Mark44]

I drew the three vectors and have no idea what I am supposed to figure out from this.

Can you post the picture you drew?

 

[vela]

What I can imagine is that: if $\vec{a}$ and $\vec{b}$ are linearly independent then they are not a multiple of each other and if $\vec{c}$ is linearly independent with $\vec{a}$ the same can be said of these two BUT, since $\vec{c}$ and $\vec{b}$ are different they cannot even be multiples of each other so the answer should be yes, right?

Your argument is similar to:

1. 5 is not a multiple of 2
2. 5 is not a multiple of 4
3. 4 isn't equal to 2; therefore, 4 is not a multiple of 2.

See the problem?

 

[mcastillo356]

Hi, PF, this is hardwork!(for me, of course)
$\{(0,0,1),(0,0,2),(0,0,3)\}$ is dependent, though could be multyplied by $\lambda_1=\lambda_2=\lambda_3=0$ to be $\vec{0}$
Thanks, @Mark44 !
Where I wrote "then", it is $\iff$
Quite sure

 

[Salmone]

Your argument is similar to:

1. 5 is not a multiple of 2
2. 5 is not a multiple of 4
3. 4 isn't equal to 2; therefore, 4 is not a multiple of 2.

See the problem?

No, $5*2/5=2$ so they are the same vector, $2/5$ is a scalar. If $\vec{a}$ doesn't the same direction of $\vec{b}$ and $\vec{c}$ doesn't have the same direction of $\vec{a}$ and $\vec{c} \neq \vec{b}$, how can $\vec{c}$ and $\vec{b}$ have the same direction?

 

[Salmone]

Can you post the picture you drew?

It seems to me that three vectors like these are linearly independent

 

[PeroK]

View attachment 301867
It seems to me that three vectors like these are linearly independent

Note that for three vectors linear independence is a test of all three vectors. It cannot be reduced to a test for each pair of vectors.

In your diagram, any of the three vectors can be expressed as a sum of the other two. So, they are linearly dependent. You can have at most two linearly independent vectors in $\mathbb R^2$.

It's not a question of linear independence of any two vectors. It's a question of all three vectors taken together.

 

[Salmone]

Note that for three vectors linear independence is a test of all three vectors. It cannot be reduced to a test for each pair of vectors.

In your diagram, any of the three vectors can be expressed as a sum of the other two. So, they are linearly dependent. You can have at most two linearly independent vectors in $\mathbb R^2$.

It's not a question of linear independence of any two vectors. It's a question of all three vectors taken together.

Ok so in general the answer to my first question is no, if I have 3 or more vectors I can't check their linearly independence just looking at 2 vectors per time, I need to check the whole independence.

 

[PeroK]

Ok so in general the answer to my first question is no, if I have 3 or more vectors I can't check their linearly independence just looking at 2 vectors per time, I need to check the whole independence.

Precisely.

Note that unless two vectors are "in the same direction", i.e. "essentially the same vector", then they are always linearly independent.

 

[Salmone]

Precisely.

Note that unless two vectors are "in the same direction", i.e. "essentially the same vector", then they are always linearly independent.

So the same can be said about a basis, a basis is the set made of the maximum number of linearly independent vectors, let's say $B={\vec{a},\vec{b},\vec{c}}$. This means that no vectors linearly independent with the vectors of the set can be left out of the set, but a single vector $\vec{d}$, for example linearly independent only with $\vec{a}$, doesn't need to be in the set since the linear independence must be among all the vectors in the set and for sure the vectors of the vectorial space outside the basis will be "not on the same direction" of the single vectors inside the basis, otherwise the whole vectorial space would consist of only the vectors of the basis. Hope the question is clear.

 

[Delta2]

They are linearly independent in any possible pair, but not as triplet.

How do I know that without trying to applying the definition of linear independence and see where i get?

Well, there are some theorems governing linear independence. One of those theorems say that if the dimension of a vector space is n, then any $N>n$ vectors of the vector space are linearly dependent. Now also another theorem says that the dimension of the vector space that has all the vectors of an euclidean plane is 2. Here we have 3>2 vectors of the plane so I guess you understand how the theorems apply here.