Welcome to our community

Be a part of something great, join today!

Are the vectors linearly independent?

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
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 - - -

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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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)