Proving functions are linearly dependent

[brotherbobby]

TL;DR

We have set of functions $\varphi (t)$ continuous in the interval $[a,b]$. The set is a linear (vector) space with the usual definitions of addition and multiplication by real numbers. We denote this space by $C[a,b]$.

Statement of the problem : Prove that the following set of functions are linearly independent in the space $C[a,b]$ mentioned above : $\varphi_1(t) = \sin^2 t, \varphi_2(t) = \cos^2 t, \varphi_3(t) = t, \varphi_4(t) = 3 \; \text{and} \; \varphi_1(t) = e^t$

We can make the first three functions add up to zero in the following way : $\sin^2 t+\cos^2 t-\frac{1}{3}\times 3 = \varphi_1(t) + \varphi_2(t) - \frac{1}{3} \varphi_3(t) = 0$.

However, look at $\varphi_4(t) = t$ and $\varphi_5(t) = e^t$. How does one combine the two to add up to zero? I can't see a way out.

Any help would be appreciated.

 

[PeroK]

What about using the differentiability of these functions?

 

[fresh_42]

Summary:: We have set of functions $\varphi (t)$ continuous in the interval $[a,b]$. The set is a linear (vector) space with the usual definitions of addition and multiplication by real numbers. We denote this space by $C[a,b]$.

$\text{Statement of the problem :}$ Prove that the following set of functions are linearly independent in the space $C[a,b]$ mentioned above : $\varphi_1(t) = \sin^2 t, \varphi_2(t) = \cos^2 t, \varphi_3(t) = t, \varphi_4(t) = 3 \; \text{and} \; \varphi_1(t) = e^t$

We can make the first three functions add up to zero in the following way : $\sin^2 t+\cos^2 t-\frac{1}{3}\times 3 = \varphi_1(t) + \varphi_2(t) - \frac{1}{3} \varphi_3(t) = 0$.

However, look at $\varphi_4(t) = t$ and $\varphi_5(t) = e^t$. How does one combine the two to add up to zero? I can't see a way out.

Any help would be appreciated.

You say you want to show linear independence, but you gave a non trivial linear combination?
If $\vec{u},\vec{v}$ are linearly dependent, then $\{\,\vec{u},\vec{v},\vec{w}\,\}$ are automatically linearly dependent, too.

 

[PeroK]

You say you want to show linear independence, but you gave a non trivial linear combination?
If $\vec{u},\vec{v}$ are linearly dependent, then $\{\,\vec{u},\vec{v},\vec{w}\,\}$ are automatically, too, linearly dependent.

I assumed the question is to investigate the linear indepence of these functions.

 

[brotherbobby]

You say you want to show linear independence, but you gave a non trivial linear combination?
If $\vec{u},\vec{v}$ are linearly dependent, then $\{\,\vec{u},\vec{v},\vec{w}\,\}$ are automatically, too, linearly dependent.

Many thanks. Sorry I forgot this is a theorem. "Given any two linearly dependent functions, any larger set of vectors involving the two is also linearly dependent".

Many thanks and apologies.

 

[Infrared]

You don't need to cite any theorem: $\sin^2(t)+\cos^2(t)+0\cdot t-\frac{1}{3}3+0\cdot e^t$ is a nontrivial linear combination that is zero (and this example should make it clear why that theorem is true).