2
$\begingroup$

Edit: Since these are pretty small assignments each and all of the same topic, I've decided to post them into one thread. I hope that's ok. Thank you.

Question

I have the following assignment:

Assignment 1 a)

Let $V$ be a vector space over a field $\mathbb{K}$ and $v_i \in \mathbb{K}$ for $i \in \mathbb{N}$. Prove or falsify the following statements about linear independency:

  1. If $(v_1,v_2)$ are linear independent, $(v_1+v_2,v_2)$ also will be linear independent.

  2. If $v_1$ is $\in \text{span}\{v_2, v_3\}$, then $v_3 \in \text{span}\{v_1,v_2\}$

  3. Let $v = \begin{pmatrix} a \\ b \end{pmatrix}$, $w = \begin{pmatrix} c \\ d \end{pmatrix} \in \mathbb{K}^2$. $(v,w)$ are linear independent if and only if $ad-bc \ne 0$

Assignment 1 b)

Are the following vectors in the $\mathbb{R}$-vector space of continuous functions $\mathbb{C^0(\mathbb{R},\mathbb{R}})$ linear independent?

$$v_1 = \sin, ~~~ v_2 = \cos ~~~ v_3 = \exp$$

Can you please check whether the following solutions I provided are correct? Thank you in advance!


My approach

Assigmnent 1 a) 1.

That statement is true. Proof:

Linear independency of $v_i$ vectors means:

$$0 = \sum \lambda_i v_i \Rightarrow \lambda_1 = ... = \lambda_i = 0 $$

Applying this to that problem we get for $v_1$ and $v_2$:

$$0 = \lambda_1 v_1 + \lambda_2 v_2 \Rightarrow \lambda_1 = \lambda_2 = 0 =: \alpha$$

and thus $\alpha(v_1 + v_2) = 0$ Therefore we get for $v_1 + v_2$ and $v_2$:

$$0 = \lambda_1 (v_1 + v_2) + \lambda_2 v_2 \Rightarrow 0 = \alpha (v_1 + v_2) + \lambda_2 v_2 \Rightarrow \alpha = \lambda_1 = \lambda_2 = 0$$


Assigmnent 1 a) 2.

This statement is false. Proof:

Let $v_1 \in \text{span}\{v_2\}$ and $v_3 \notin \text{span}\{v_2\}$. Therefore we get that there is a $\lambda \in \mathbb{R}$ for which we get $\lambda\cdot v_2 = v_1$. We further get there are $\lambda_1 = 0$ and $\lambda_2 \in \mathbb{R}$ for which we get $\lambda_1 \cdot v_3 + \lambda_2\cdot v_2 = v_1$ and thus $v_1 \in \text{span}\{v_2,v_3\}$.

Now because of $v_1 \in \text{span}\{v_2\}$ we can reduce $\text{span}\{v_1, v_2\}$ to $\text{span}\{v_2\}$. Thus $v_3 \in \text{span}\{v_1,v_2\}$ would be a contradiction to $v_3 \notin \text{span}\{v_2\}$.


Assigmnent 1 a) 3.

This statement is false. Proof:

We already know that $e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, $e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \in \mathbb{K}^2$ are linear independent. Replace $v$ by $e_1$ and $w$ by $e_2$ and we get

$$ab-bc = 1\cdot0 - 0\cdot1 = 0$$

contradiction.

Edit: I've posted a different version of that proof here


$$$$

Assignment 1 b)

This statement is true. Proof:

Linear independency of $v_i$ vectors means:

$$0 = \sum \lambda_i v_i \Rightarrow \lambda_1 = ... = \lambda_i = 0 $$

Therefore we have:

$$0 = \lambda_1 \sin(x) + \lambda_2 \cos(x) + \lambda_3 \exp(x) \Rightarrow \lambda_1 = \lambda_2 = \lambda_3 = 0$$

because already $\sin(x)$ and $\cos(x)$ can't be both $0$ for the same $x$.


$$$$

Thank you very much for your help!

$\endgroup$
2
  • 1
    $\begingroup$ Regards 1 a) 3. The condition is $a\color{red} d-bc\ne0$. $\endgroup$ Jan 30 '14 at 11:57
  • $\begingroup$ Oh right - I somehow did't notice that. Will go over that again thanks $\endgroup$ Jan 30 '14 at 11:59
3
$\begingroup$

I disagree with you proof of (1.a.1), although the statement is indeed correct. You used:

$$\lambda_1v_1+\lambda_2v_2 = \alpha(v_1+v_2) + \lambda_2v_2 = 0$$

where $\alpha = 0$, which is just a leap of logic! I think the correct way to go around it is the following:

We want to prove that if $(v_1,v_2)$ are linearly independent, then $(v_1+v_2,v_2)$ is linearly independent. So, let's have:

$$\lambda_1(v_1+v_2)+\lambda_2v_2 = 0$$ $$\lambda_1v_1+(\lambda_1+\lambda_2)v_2 = 0$$

As $(v_1,v_2)$ are linearly independent, we must have:

$$\lambda_1 = 0$$ $$\lambda_1+\lambda_2=0$$

Hence, $\lambda_1=0$ and $\lambda_2=0$, that is $(v_1+v_2,v_2)$ is linearly independent.

$\endgroup$
3
  • $\begingroup$ Hmm I think you are right. Your proof is much more detailed. Still I have a question: I did the $\alpha$-thing because I thought it won't be possible to use the same $\lambda 's$ both equations you get. I started my thoughts based on solving the equations $\lambda_1 v_1 + \lambda_2 v_2 = 0$ and $\lambda_3 (v_1 + v_2) + \lambda_4 v_2 = 0$. Is this distinction redundant? Thank you very much! $\endgroup$ Jan 30 '14 at 12:36
  • 1
    $\begingroup$ It's not that we are using the same or different $\lambda$'s! It's just that whenever you get to a point where you've written $\lambda_1 v_1+\lambda_2 v_2 = 0$, you can pull the fact that $(v_1,v_2)$ are linearly independent and go to $\lambda_1=\lambda_2=0$, no matter what symbols are there! First stuff times $v_1$ plus second stuff times $v_2$ equals zero only if both stuff are zero.You have to go from the thing you need to prove (that the same happen for $(v_1+v_2,v_2)$) and get to a point where you can use the information you have $\endgroup$
    – lsoranco
    Jan 30 '14 at 12:43
  • $\begingroup$ You are right. Thank you! :) $\endgroup$ Jan 30 '14 at 12:44
2
$\begingroup$

1.a)2 Is false. Consider the vector space $\mathbb{R}^3$ and let $v_1=0$ and $v_2 \perp v_3$.

1.b) Is true, they are linearly independent over the field $\mathbb{R}$, but not over the field $\mathbb{C}$. Your proof however is wrong.

Consider $a, b, c \in \mathbb{R}$ such that for all $t \in \mathbb{R}$ we have: $$a sin(t) + b cos(t) + c e^t =0$$

Setting $t=0$ we get $b + c=0$, hence $b=-c$.

Next we let $t \rightarrow \infty$ to obtain $c=0$ (because both $sin$ and $cos$ are bounded but $lim_{t \rightarrow \infty}e^t=\infty$).

We now have $b=c=0$ from which it immediately also follows that $a=0$, hence the given functions are linearly independent over the field $\mathbb{R}$.

Further explanation: For fixed $a, b, c$ we have for any $t \in \mathbb{R}$ that $|a sin(t) + b cos(t)| \leq |a sin(t)| + |b cos(t)| \leq |a| + |b|$ which is bounded because $a, b, \in \mathbb{R}$. However if $c \neq 0$ the term $c e^t$ is unbounded (and also $\neq 0$) which means that for $t$ large enough $a sin(t) + b cos(t)$ and $c e^t$ cannot cancel each other out.

$\endgroup$
1
  • $\begingroup$ Hey, thanks, your posts makes sense to me. Still would you please explain further how you obtain c by letting t approach $\infty$? I don't really understand that. Thank you! $\endgroup$ Jan 30 '14 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.