linearly independent in Hilbert Space

Question

Please help me to solve the linearly independent of functions in Hilbert Space

how i can show that the functions $\sin(t)$ and $\cos(t)$ are linearly independent in Hilbert Space (L^2[0,pi])?

Which Hilbert space are we talking about? For example in $\Bbb{R}^1$ they are not (irrespective of the value of $t$). In $L^2([0,2\pi])$ they are, because they are orthogonal and non-zero. — Jyrki Lahtonen, Dec 23 '13 at 14:00

score 4 · Answer 1 · answered Dec 23 '13 at 14:02

4

Hint

Let $\alpha,\beta\in\mathbb R$ such that $$\alpha \cos(t)+\beta\sin(t)=0\;\forall t\in\mathbb R$$ now choose a particular value of $t$ to show that $\alpha=\beta=0$ and conclude.

answered Dec 23 '13 at 14:02

score 0 · Accepted Answer · answered Dec 23 '13 at 14:10

0

Hint: You're in a Hilbert space, so you have the scalar product $\langle\cdot\vert\cdot\rangle$. Nonzero orthogonal vectors are always linearly independent. Can you check that $\int_0^{2\pi}\sin(t)\overline{\cos(t)}\,dt=0$?

answered Dec 23 '13 at 14:10

tomasz

35,474

GEdgar · Answer 3 · 2013-12-23T15:00:59.453

0

Another method would be the Wronskian.

added

The Wronskian for a linearly dependent list of functions is identically zero. Compute the Wronskian for $\sin x, \cos x$, $$ \left|\begin{matrix}\sin x & \cos x\\ \cos x & -\sin x\end{matrix}\right| = -\sin^2 x - \cos^2 x = -1 $$ to see that it is not identically zero.

edited Dec 23 '13 at 15:00

answered Dec 23 '13 at 14:18

GEdgar

111,679

Dear GEdgar: I'm positive this is the kernel of a great hint, but as written it's pretty vague and doesn't really rise beyond the content of a comment. If you find the time, could you beef it up with another sentence alluding to the relevant property? Thanks! – rschwieb Dec 23 '13 at 14:47

linearly independent in Hilbert Space

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