How to prove that a set of functions are linearly dependent? Is $S$ linearly dependent on $\textsf V = \mathcal{F}(\Bbb R,\Bbb R)$ and $S = \{t, e^t ,\sin(t)\}$.
How to prove that a set of functions are linearly dependent?
 A: Here is a another way to skin the cat:
Suppose $\alpha_1 t + \alpha_2 e^t + \alpha_3 \sin t = 0$ for all $t$.
If we differentiate twice and set $t = 0$, we get $\alpha_2 e^{\pi \over 2} = 0$ and so $\alpha_2 = 0$.
If we differentiate twice and set $t = -{\pi \over 2}$, we get $\alpha_3 = 0$.
Finally, set $t=1$ to get $\alpha_1 = 0$.
Hence $S$ is linearly independent.
A: Suppose we have some scalars $a_0,a_1,a_2$ in $\Bbb R$ such that
$$a_0t+a_1e^t+a_2\sin t =0 \tag{1}$$
for all real number $t$. Making $t=0$ this gives us $a_1=0$. Returning to $(1)$, we have
$$a_0t+a_2\sin t =0 \tag{2}$$
Now, make $t=\pi$ and then $a_0\pi=0$, which means $a_0=0$.
A: Hint. Write one as a linear combination of the others. That is, look for real numbers $a,b,c$ not simultaneously vanishing, so that we may write $$at+be^t+c\sin t=0.$$
A: HINT.- A good way to try this problem is to consider that the elements of $V$ are $f_1,f_2,f_3$ where $f_1(t)=t,f_2(t)=e^t,f_3(t)=\sin(t)$ so you have to show that if the linear combination
$$a_1f_1+a_2f_2+a_3f_3=0$$ where les $f_i$ are three vectors and les $a_i$ are scalars in your vectorial space, then $a_1=a_2=a_3=0$ (Obviously for each $t$ you have
$$(a_1f_1+a_2f_2+a_3f_3)(t)=a_1f_1(t)+a_2f_2(t)+a_3f_3(t)=0)$$
A: Also you may find if
$$\begin{vmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t)\\ f_1''(t) & f_2''(t) & f_3''(t)\end{vmatrix}= 0,$$ then the three functions are linearly dependent (LD) otherwise they are linearly independent (LI).
