Are the vectors linearly independent?

In summary: Nod)If we substitute them in the original equation, we get $(\alpha_1+\alpha_2)\vec v + (\alpha_1 + \alpha_2)\vec w = 0$.Now since $\alpha_1 + \alpha_2 \neq 0$, this contradicts the fact that $\vec v$ and $\vec w$ are linearly independent. In summary, we have proven that $\vec{v}$ and $\vec{v}+\vec{w}$ are linearly independent by contradiction.
  • #1
mathmari
Gold Member
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Hey! :eek:

We have that the vectrs $\vec{v},\vec{w}, \vec{u}$ are linearly independent.

I want to check if the pairs
  • $\vec{v}, \vec{v}+\vec{w}$
  • $\vec{v}+\vec{u}$, $\vec{w}+\vec{u}$
  • $\vec{v}+\vec{w}$, $\vec{v}-\vec{w}$
are linearly indeendent or not.

Since $\vec{v}, \vec{w}, \vec{u}$ are linearly independet it holds that $\lambda_1\vec{v}+\lambda_2\vec{w}+\lambda_3\vec{u}=0 \Rightarrow \lambda_1=\lambda_2=\lambda_3=0$ ($\star$). We have the following:
  • $\vec{v}, \vec{v}+\vec{w}$ :

    $\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0 \Rightarrow (\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$

    How can we continue here? (Wondering)

    $$$$
  • $\vec{v}+\vec{u}$, $\vec{w}+\vec{u}$ :

    $\alpha_1(\vec{v}+\vec{u})+\alpha_2(\vec{w}+\vec{u})=0 \Rightarrow \alpha_1\vec{v}+(\alpha_1+\alpha_2)\vec{u}+\alpha_2\vec{w}=0$

    From ($\star$) it folows that $\alpha_1=\alpha_1+\alpha_2=\alpha_2=0\Rightarrow \alpha_1=\alpha_2=0$ and so this means that the vectors $\vec{v}+\vec{u}$ and $\vec{w}+\vec{u}$ are linearly independent.

    $$$$
  • $\vec{v}+\vec{w}$, $\vec{v}-\vec{w}$ :

    $\alpha_1(\vec{v}+\vec{w})+\alpha_2(\vec{v}-\vec{w})=0\Rightarrow (\alpha_1+\alpha_2)\vec{v}+(\alpha_1-\alpha_2)\vec{w}=0$

    How can we continue here?
 
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  • #2
mathmari said:
Since $\vec{v}, \vec{w}, \vec{u}$ are linearly independent it holds that $\lambda_1\vec{v}+\lambda_2\vec{w}+\lambda_3\vec{u}=0 \Rightarrow \lambda_1=\lambda_2=\lambda_3=0$ ($\star$).

We have the following:
  • $\vec{v}, \vec{v}+\vec{w}$ :

    $\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0 \Rightarrow (\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$

    How can we continue here?

Hey mathmari!

Let's try with a proof by contradiction. (Thinking)

Suppose they are not linearly independent. Then there must be $\alpha_1,\alpha_2$ such that $\alpha_1\ne 0$ and/or $\alpha_2\ne 0$.
Let $\lambda_1 = \alpha_1+\alpha_2$ and $\lambda_2=\alpha_2$.
What if we substitute them in your expression for independence of $\vec{v}, \vec{w}, \vec{u}$?
Can we get a contradiction? (Wondering)
 
Last edited:
  • #3
Klaas van Aarsen said:
Hey mathmari!

Let's try with a proof by contradiction. (Thinking)

Suppose they are not linearly independent. Then there must be $\alpha_1,\alpha_2$ such that $\alpha_1+\alpha_2\ne 0$ and/or $\alpha_2\ne 0$.
Let $\lambda_1 = \alpha_1+\alpha_2$ and $\lambda_2=\alpha_2$.
What if we substitute them in your expression for independence of $\vec{v}, \vec{w}, \vec{u}$?
Can we get a contradiction? (Wondering)

That would mean that $\lambda_1$ and/or $\lambda_2$ is non-zero, which is a contradiction, correct? (Wondering)
 
  • #4
mathmari said:
That would mean that $\lambda_1$ and/or $\lambda_2$ is non-zero, which is a contradiction, correct?

Yep. (Nod)

Btw, I made a mistake before. It should be $\alpha_1\ne 0$ and/or $\alpha_2\ne 0$. (Blush)
 
  • #5
Klaas van Aarsen said:
Let $\lambda_1 = \alpha_1+\alpha_2$ and $\lambda_2=\alpha_2$.
I got stuck right now. Why can we just take these $\lambda$'s ? (Wondering)
 
  • #6
mathmari said:
Hey! :eek:

We have that the vectrs $\vec{v},\vec{w}, \vec{u}$ are linearly independent.

I want to check if the pairs
  • $\vec{v}, \vec{v}+\vec{w}$
  • $\vec{v}+\vec{u}$, $\vec{w}+\vec{u}$
  • $\vec{v}+\vec{w}$, $\vec{v}-\vec{w}$
are linearly indeendent or not.

Since $\vec{v}, \vec{w}, \vec{u}$ are linearly independet it holds that $\lambda_1\vec{v}+\lambda_2\vec{w}+\lambda_3\vec{u}=0 \Rightarrow \lambda_1=\lambda_2=\lambda_3=0$ ($\star$). We have the following:
  • $\vec{v}, \vec{v}+\vec{w}$ :

    $\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0 \Rightarrow (\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$
How can we continue here? (Wondering)
Didn't you just say that the fact that $\vec{v}$ and $\vec{w}$ are independent requires that $\alpha_1+ \alpha_2= 0$ and $\alpha_2= 0$?

  • $\vec{v}+\vec{u}$, $\vec{w}+\vec{u}$ :

    $\alpha_1(\vec{v}+\vec{u})+\alpha_2(\vec{w}+\vec{u})=0 \Rightarrow \alpha_1\vec{v}+(\alpha_1+\alpha_2)\vec{u}+\alpha_2\vec{w}=0$

    From ($\star$) it folows that $\alpha_1=\alpha_1+\alpha_2=\alpha_2=0\Rightarrow \alpha_1=\alpha_2=0$ and so this means that the vectors $\vec{v}+\vec{u}$ and $\vec{w}+\vec{u}$ are linearly independent.

    $$$$
  • $\vec{v}+\vec{w}$, $\vec{v}-\vec{w}$ :

    $\alpha_1(\vec{v}+\vec{w})+\alpha_2(\vec{v}-\vec{w})=0\Rightarrow (\alpha_1+\alpha_2)\vec{v}+(\alpha_1-\alpha_2)\vec{w}=0$

    How can we continue here?
 
Last edited by a moderator:
  • #7
mathmari said:
I got stuck right now. Why can we just take these $\lambda$'s ?

We can.

It's just that we want to prove that $\vec v$ and $\vec v + \vec w$ are linearly independent.
To do so, we need to prove that $a_1\vec v + a_2 (\vec v + \vec w)=0 \implies a_1=a_2=0$.
So for the proof by contradiction we assume that $a_1\ne 0$ and/or $a_2\ne 0$.

Now we can pick those lambda's and continue... (Thinking)
 
  • #8
HallsofIvy said:
Didn't you just say that the fact that $\vec{v}$ and $\vec{w}$ are independent requires that $\alpha_1+ \alpha_2= 0$ and $\alpha_2= 0$?

Do you mean the following? (Wondering)

Let $\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0 \Rightarrow (\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$.

Since $\vec{v}$, $\vec{w}$ and $\vec{u}$ are linearly independent, then $\vec{v}$ and $\vec{w}$ are also linearly independent and this means that $\alpha_1+\alpha_2=\alpha_2=0 \Rightarrow \alpha_1=\alpha_2=0$.

- - - Updated - - -

Klaas van Aarsen said:
We can.

It's just that we want to prove that $\vec v$ and $\vec v + \vec w$ are linearly independent.
To do so, we need to prove that $a_1\vec v + a_2 (\vec v + \vec w)=0 \implies a_1=a_2=0$.
So for the proof by contradiction we assume that $a_1\ne 0$ and/or $a_2\ne 0$.

Now we can pick those lambda's and continue... (Thinking)

We suppose that $\vec{v}$ and $\vec{v}+\vec{w}$ are linearly dependent.

Then at $\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0$ we have that $\alpha_1\neq 0$ and/or $\alpha_2\neq 0$.

From the above equation we have that $(\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$.

Since $\vec{v}, \vec{w}, \vec{u}$ are linearly independet it holds that $\lambda_1\vec{v}+\lambda_2\vec{w}+\lambda_3\vec{u}=0 \Rightarrow \lambda_1=\lambda_2=\lambda_3=0$.

Do you mean that we define these $\lambda$'s ? (Wondering)
 
  • #9
mathmari said:
We suppose that $\vec{v}$ and $\vec{v}+\vec{w}$ are linearly dependent.

Then at $\alpha_1\vec{v}+\alpha_2(\vec{v}+\vec{w})=0$ we have that $\alpha_1\neq 0$ and/or $\alpha_2\neq 0$.

From the above equation we have that $(\alpha_1+\alpha_2)\vec{v}+\alpha_2\vec{w}=0$.

Since $\vec{v}, \vec{w}, \vec{u}$ are linearly independet it holds that $\lambda_1\vec{v}+\lambda_2\vec{w}+\lambda_3\vec{u}=0 \Rightarrow \lambda_1=\lambda_2=\lambda_3=0$.

Do you mean that we define these $\lambda$'s ?

Yep. (Nod)
 

FAQ: Are the vectors linearly independent?

What does it mean for vectors to be linearly independent?

Linear independence refers to the relationship between vectors in a vector space. If a set of vectors is linearly independent, it means that no vector in the set can be written as a linear combination of the other vectors in the set. In other words, each vector in the set adds something new to the span of the other vectors.

How can I determine if a set of vectors is linearly independent?

To determine if a set of vectors is linearly independent, you can use the following method:
1. Write the vectors as columns in a matrix.
2. Use elementary row operations to reduce the matrix to its reduced row echelon form.
3. If the reduced matrix has a pivot in every column, then the vectors are linearly independent. Otherwise, they are linearly dependent.

Can a set of two vectors be linearly independent?

Yes, a set of two vectors can be linearly independent if they are not scalar multiples of each other. In other words, one vector cannot be a multiple of the other.

What is the significance of linear independence in linear algebra?

Linear independence is an important concept in linear algebra because it allows us to determine the dimension of a vector space. If a set of vectors is linearly independent, then the number of vectors in the set is equal to the dimension of the vector space they span. Additionally, linear independence is necessary for certain operations, such as finding a basis or solving systems of linear equations.

Can a set of linearly dependent vectors be useful in any way?

Yes, a set of linearly dependent vectors can still be useful in certain situations. For example, they can be used to represent a system of equations that has infinitely many solutions. Additionally, linearly dependent vectors can be used to find the rank of a matrix or to solve for free variables in a system of equations.

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