Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For each subset of each vector space, justify whether their vectors are linearly independent or dependent:

$$\{f,g,h\}, f(x)=e^{2x} , g(x)=x^2 , h(x)=x,\mbox{ in }F(\mathbb R,\mathbb R).$$

share|improve this question
use latex and wrap the $ \LaTeX $ between $'s – experimentX Dec 26 '12 at 19:42

1 Answer

Assuming you meant $f(x) = e^{2x}, g(x) = x^2, h(x) = x$, you need to show that if there are constants $\alpha_f, \alpha_g, \alpha_h$ such that $\alpha_f f + \alpha_g g + \alpha_h h = 0$ (that is, equal to the the function $t \mapsto 0$), then these constants must be zero.

Some tricks for this sort of problem are evaluating $\alpha_f f + \alpha_g g + \alpha_h h$ at various points, or differentiating an appropriate number of times and evaluating.

Let $\phi(x) = \alpha_f f(x) + \alpha_g g(x) + \alpha_h h(x)$. We have $\phi(x) = 0$ for all $x$. So, $\phi(0) = \alpha_f = 0$. So we know that $\alpha_f = 0$. We note that since $\phi$ is contstant, $\phi' = 0$, and so $\phi'(0) = \alpha_g g'(0) + \alpha_h h'(0) = \alpha_h = 0$. So we know that $\alpha_h = 0$. Finally, since $\phi(1) = \alpha_g g(1) = \alpha_g = 0$, we have $\alpha_g = 0$.

Hence $f,g,h$ are linearly independent.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.