Linear independence of basis vectors $\def\braket#1#2{\langle#1|#2\rangle}\def\bra#1{\langle#1}\def\ket#1{#1\rangle}$ Consider the vector space of functions $r(t)$ that are solutions of the differential equation $\frac{d^2r}{dt^2}-6\frac{dr}{dt}+9r=0$. Given two vectors $|\ket{x_1}=e^{3t}$ and $|\ket{x_2}=te^{3t}$, I must show that they form a basis in this space. To do this, I must show that they are independent. I am not sure if I should write $\alpha|\ket{x_1}+\beta|\ket{x_2}=\alpha e ^{3t}+\beta te^{3t}=0$ and show that $\alpha$ and $\beta$ are zero. Or, say $\alpha\begin{bmatrix}e^{3t}\\\0\end{bmatrix}+\beta\begin{bmatrix}0\\\ te^{3t}\end{bmatrix}=0$, which would give me two equations, $\alpha|\ket{x_1}=\alpha e^{3t}=0$, and $\beta|\ket{x_2}=\beta te^{3t}=0$ (a lot easier to show $\alpha$ and $\beta$ must be zero). Or, is my notation all messed up? Or, are they the same thing? Thank you

Another approach would be to calculate the Wronskian.

The Wronskian of a finite family $f_1,\ldots, f_n$ of functions, which are $(n – 1)$ times differentiable on an interval $I$ is defined as the determinant $$W(f_1,\ldots,f_n)(x)= \begin{vmatrix} f_1(x) & \cdots & f_n(x) \\\ f_1'(x) & \cdots & f_n'(x) \\\ \vdots & \vdots & \vdots \\\ f_1^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{vmatrix}, \ x \in I.$$

If $W(f_1,\ldots,f_n)(x) \
eq 0$ for some $x \in I$, then the set $\\{f_1,\ldots,f_n\\}$ is linearly independent.

In your case, we would have that

$$W(e^{3t},te^{3t})= \begin{vmatrix} e^{3t} & te^{3t} \\\ 3e^{3t} & e^{3t} + 3te^{3t} \end{vmatrix}=e^{6t} > 0 \ \ \ \forall \ t \in \mathbb{R}.$$ Hence the set $\\{e^{3t},te^{3t}\\}$ is linearly independent.