# Thread: linearly dependent in Q, linearly independent in R

1. ## linearly dependent in Q, linearly independent in R

I need help with this one.

Find $\displaystyle \lambda \in \mathbb{R}$ such that $\displaystyle \{(\lambda,1,0),(1,\lambda,1),(0,1,\lambda)\}$ is linearly dependent in $\displaystyle \mathbb{Q}$ and linearly independent in $\displaystyle \mathbb{R}$.

Here's what I did:
$\displaystyle \alpha (\lambda, 1, 0)+\beta(1, \lambda, 1)+\gamma (0, 1, \lambda)=0$
Solving this gets me:
$\displaystyle \alpha \lambda+\beta=0$
$\displaystyle \alpha + \beta \lambda + \gamma=0$ and
$\displaystyle \beta + \gamma \lambda=0$.

This gets me:
$\displaystyle \lambda(\gamma - \alpha)=0$ which means that either $\displaystyle \lambda=0$ or $\displaystyle \gamma - \alpha=0$.

The first case gives me $\displaystyle \alpha, \beta, \gamma=0$, which means that the set is linearly independent both in $\displaystyle \mathbb{R}$ and $\displaystyle \mathbb{Q}$ so that is obviously not the $\displaystyle \lambda$ I'm looking for.

Now, the other case, $\displaystyle \gamma-\alpha=0$. This means that $\displaystyle \alpha=\gamma$. The three equations I had are now two equations:
$\displaystyle \alpha \lambda + \beta=0$ and
$\displaystyle 2\alpha + \beta \lambda=0$.

Solving that gets me:
$\displaystyle \alpha (2-\lambda^2)=0$.

Now, if $\displaystyle \lambda=\pm \sqrt{2} \in \mathbb{R}$, I get that $\displaystyle \alpha, \beta, \gamma$ are not necessarily zero, but if $\displaystyle \lambda \in \mathbb{Q}$, the equation $\displaystyle \alpha (2-\lambda^2)=0$ implies that $\displaystyle \alpha=0$, and so $\displaystyle \beta=0$ and $\displaystyle \gamma=0$.
Obviously, $\displaystyle \lambda=\pm \sqrt{2}$ is the one we're looking for.

Is this correct?
Is this the only possible $\displaystyle \lambda$?

Thank you!

2. Originally Posted by harriette
I need help with this one.

Find $\displaystyle \lambda \in \mathbb{R}$ such that $\displaystyle \{(\lambda,1,0),(1,\lambda,1),(0,1,\lambda)\}$ is linearly dependent in $\displaystyle \mathbb{Q}$ and linearly independent in $\displaystyle \mathbb{R}$.

Here's what I did:
$\displaystyle \alpha (\lambda, 1, 0)+\beta(1, \lambda, 1)+\gamma (0, 1, \lambda)=0$
Solving this gets me:
$\displaystyle \alpha \lambda+\beta=0$
$\displaystyle \alpha + \beta \lambda + \gamma=0$ and
$\displaystyle \beta + \gamma \lambda=0$.

This gets me:
$\displaystyle \lambda(\gamma - \alpha)=0$ which means that either $\displaystyle \lambda=0$ or $\displaystyle \gamma - \alpha=0$.

The first case gives me $\displaystyle \alpha, \beta, \gamma=0$, which means that the set is linearly independent both in $\displaystyle \mathbb{R}$ and $\displaystyle \mathbb{Q}$ so that is obviously not the $\displaystyle \lambda$ I'm looking for.

Now, the other case, $\displaystyle \gamma-\alpha=0$. This means that $\displaystyle \alpha=\gamma$. The three equations I had are now two equations:
$\displaystyle \alpha \lambda + \beta=0$ and
$\displaystyle 2\alpha + \beta \lambda=0$.

Solving that gets me:
$\displaystyle \alpha (2-\lambda^2)=0$.

Now, if $\displaystyle \lambda=\pm \sqrt{2} \in \mathbb{R}$, I get that $\displaystyle \alpha, \beta, \gamma$ are not necessarily zero, but if $\displaystyle \lambda \in \mathbb{Q}$, the equation $\displaystyle \alpha (2-\lambda^2)=0$ implies that $\displaystyle \alpha=0$, and so $\displaystyle \beta=0$ and $\displaystyle \gamma=0$.
Obviously, $\displaystyle \lambda=\pm \sqrt{2}$ is the one we're looking for.

Is this correct?
Is this the only possible $\displaystyle \lambda$?

Thank you!
Hmm...surely if $\displaystyle \lambda = 0$ then you just have $\displaystyle \beta=0$ and that $\displaystyle \alpha + \gamma=0$, not that $\displaystyle \alpha=0=\gamma$...however, you are right - you can discard this case as it is linearly dependent in both the reals and the rationals.

And yes, $\displaystyle \pm\sqrt{2}$ are your only two solutions, and your working looks correct. Further, $\displaystyle b=-a\lambda$. So clearly one of $\displaystyle a$ or $\displaystyle b$ is in $\displaystyle \mathbb{R}$ (as $\displaystyle \lambda \not\in \mathbb{Q}$). And so the set is linearly independent over $\displaystyle \mathbb{Q}$ but linearly dependent over $\displaystyle \mathbb{R}$ (which is the opposite of what you said, but I think you just wrote them the wrong way round...or I have written them the wrong way round...but I'm pretty sure I haven't!)

3. Originally Posted by Swlabr
Hmm...surely if $\displaystyle \lambda = 0$ then you just have $\displaystyle \beta=0$ and that $\displaystyle \alpha + \gamma=0$, not that $\displaystyle \alpha=0=\gamma$...however, you are right - you can discard this case as it is linearly dependent in both the reals and the rationals.
Yes, I see this now!
But how do I conclude that the set is lineraly dependent? Is it because $\displaystyle \alpha$ and $\displaystyle \gamma$ can be anything, as long as it's $\displaystyle \alpha=\gamma$? So they are not necessarily $\displaystyle 0$?

(which is the opposite of what you said, but I think you just wrote them the wrong way round...or I have written them the wrong way round...but I'm pretty sure I haven't!)
I mixed it up, of course.
Thank you!!!

4. Originally Posted by harriette
Yes, I see this now!
But how do I conclude that the set is lineraly dependent? Is it because $\displaystyle \alpha$ and $\displaystyle \gamma$ can be anything, as long as it's $\displaystyle \alpha=\gamma$? So they are not necessarily $\displaystyle 0$?
Watch - you missed a minus sign! It is linearly dependent as for any $\displaystyle \alpha, \gamma$ such that $\displaystyle \alpha=-\gamma$ you get zero. So, as always with a counter-example, plug in a number. So $\displaystyle \alpha=1$ and $\displaystyle \gamma=-1$ work.

5. Thank you once again!