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.


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

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


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!

  • 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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