I am trying to determine whether the vector functions
x1 = col(e^x, 2e^x, 3e^x) x2 = col(2e^2x, 4e^2x, 6e^2x) x3 = col(3e^3x, 6e^3x, 9e^3x)
are linearly independent over all real numbers x.
My instinct is that they are, but when I pose the question to WolframAlpha, it tells me otherwise, and I think it could be because it is interpreting x as a constant rather than a variable?
I know that when we use x = 0, we obtain 3 equations which effectively tell us only c1 + 2c2 + 3c3 = 0.
However, can we deduce nonzero constants from this fact that also work for all other x? Again, I think not, but Wolfram tells me otherwise.
Set the linear relation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3= 0$ with
$\mathbf{v}_1 = [e^x \quad 2e^{x} \quad 3e^{x}]^T$, $\quad$ $\mathbf{v}_2 = [2e^{2x} \quad 4e^{2x} \quad 6e^{2x}]^T$, $\quad$ $\mathbf{v}_3 = [3e^{3x} \quad 6e^{3x} \quad 9e^{3x}]^T$.
You will find that the only solution to the linear relation above is the trivial solution, $c_1 = c_2 = c_3 = 0$.
Hence, the three vector functions $\mathbf{v}_1, \mathbf{v}_2, $ and $ \mathbf{v}_3$ are linearly independent, provided that $x \neq 0$. If $x$ were zero, one can find by inspection that $\mathbf{v}_1 + \mathbf{v}_2 - \mathbf{v}_3 = 0$, hence resulting in a nontrivial solution.
[Edit] My apologies, I should have thought about the condition when $x = 0$.
BHP