Are the proofs I made correct? 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:

*

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


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


*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 $x$.

$$$$
Thank you very much for your help!
 A: I disagree with you proof of (1.a.1), although the statement is indeed correct. You used:
$$\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$$
Hence, $\lambda_1=0$ and $\lambda_2=0$, that is $(v_1+v_2,v_2)$ is linearly independent.
A: 1.a)2 Is false. Consider the vector space $\mathbb{R}^3$ and let $v_1=0$ and $v_2 \perp v_3$.
1.b) Is true, they are linearly independent over the field $\mathbb{R}$, but not over the field $\mathbb{C}$. Your proof however is wrong.
Consider $a, b, c \in \mathbb{R}$ such that for all $t \in \mathbb{R}$ we have:
$$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.
