Linear independence in $V =$ the space of functions from $\mathbb R$ to $\mathbb R$ Problem 8.25 in Schaum's 3000 Solved Problems in Linear Algebra states:

Let $V$ be the vector space of functions from $\mathbb R$ into $\mathbb R$. Show that $\{ e^{2t}, t^2, t\} $ is a set of linearly independent vectors.

The answer is supplied but I'm curious if my alternate proof is valid. Suppose $\alpha e^{2t} + \beta t^2 + \gamma t = 0$. When $t = 0$, this becomes $\alpha  + 0 + 0 = 0$, so $\alpha = 0$. When $t = 1$ and given that $\alpha = 0$, this becomes $\beta + \gamma = 0$, so $\gamma = -\beta.$ Finally, combining all of that with the equation obtained when $t = 2$, we have $4 \beta = 2 \beta $ and thus $\beta = \gamma = 0$. Thus $ \alpha = \beta = \gamma = 0$, which means that the vectors are linearly independent. $\square$
 A: Consider the subspace $C^\infty(\mathbb{R})$. Then $T=\frac{d}{dx}$ is a linear operator. $x$ is an eigenvector with $\lambda=0$, $x^2$ is a generalized eigenvector with eigenvalue $\lambda=0$ and $e^{2x}$ is an eigenvector with eigenvalue $\lambda=0$. Thus, linear independent
A: Yes, your argument is fine.
I would have used $t=-1$ for the last point, and that way you get $\beta+\gamma=0$ and $\beta-\gamma=0$.
A: The approach works just by the definition of independence. Start with the supposition that for all real number  $t$ we have $\alpha e^{2t}+\beta t^{2}+\gamma t=0$ $(*)$ and then we try to see if the above assumption implies that $\alpha=\beta=\gamma=0$ in order to see the independence. Now, since for all $t$ we have $(*)$, then in particular for $t^{*}$ the equation $(*)$ have to hold and in this way you search the scalars $\alpha, \beta,\gamma$. If all the scalars they are zeros then the set must be linearly independent.
Alternatively, since we have differentiable functions then the Wronskian give $\det\begin{bmatrix}e^{2t}&t^{2}&t\\2e^{2t}&2t&1\\2^{2}e^{2t}&2&0 \end{bmatrix}=-2e^{2t}(2t^{2}-2t+1)\not=0$ for all real number $t$. Thus, the set give is linearly independent.
