Show that the set of all functions described by $y(t) = c_1\cos\omega t + c_2\sin\omega t$ is a vector space. ($c_1,c_2$ are constants and $\omega$ is fixed).

I am having real issues imagining multiple functions as a vector space. I don't understand because it is just one function so how is there a "set of all functions?"

  • $\begingroup$ There's lots of functions, actually; $c_1$ and $c_2$ are free variables. $\endgroup$ – user61527 Oct 21 '13 at 3:57
  • 1
    $\begingroup$ What is $\omega$ in this case? (It turns out not to matter, but for context's sake, it's worth mentioning.) $\endgroup$ – Cameron Buie Oct 21 '13 at 3:57
  • $\begingroup$ Take two functions, $y_1(t), y_2(t)$. Show that $k \cdot y_1 + y_2$ has the form given. $\endgroup$ – The Chaz 2.0 Oct 21 '13 at 4:02

Every vector space has three properties:

  • The space must contain a zero element relevant to the other elements in the set. For example, the set of 2-element vectors with real-valued elements is $\ R^2 $ which contains the zero vector $\ <0,0> $.

  • Closure under vector addition must exist for the space in question.
    For two elements in the space, $\ a,b $, there must be some element
    $\ a + b$ also in the space.

  • Closure under scalar multiplication must exist for the space in
    question. For an element $a$ in the space and some scalar $c$ in the set of real numbers $R$, there must be an element $ca$ in the space.

No doubt you'd agree that $cos(ωt)$ is a function and $sin(ωt)$ is also a function. Then, we can see that $y(t) = c_1cos(ωt) + c_2sin(ωt)$ is a linear combination of the two functions $cos(ωt)$ and $sin(ωt)$, which we can regard separately.

Functions are the basic elements of the function space we're concerned with (much as vectors are the basic elements of vector spaces), so we know that there must be some way to combine elements of the space to form other elements as with the vector spaces we're more familiar with. We know this as closure under vector addition, or in this case function addition. For example, if you imagine two polynomial equations of order $n$, say $\ A$ = $c_1x^n + c_2x^{n-1} + ...$ and $\ B$ = $r_1x^n + r_2x^{n-1} + ...$, you could add these together to get $\ A + B$ = $(c_1 + r_1)x^n + (c_2 + r_2)x^{n-1} + ... $

We also know that closure under scalar multiplication exists, since for some scalar $c$ in $R$ and a polynomial function $ f(x) = x^n + x^{n-1} + ... $, we could multiply $\ cA$ = $cx^n + cx^{n-1} + ...$ to get another function.

Finally, a zero element exists in the function space: we'd certainly say that $ f(x) = 0 $ is a function, right?

If you look at your initial function $y(t) = c_1cos(ωt) + c_2sin(ωt)$, you'll see that it is closed under scalar multiplication: for some $r$ in the reals $R$, and a function $x(t) = cos(ωt) + sin(ωt)$ in the function space $y(t)$, we can take $r*x(t) = rcos(ωt) + rsin(ωt)$ to also be in that space.

Similarly, two elements of the space $A = c_1cos(ωt) + c_2sin(ωt)$ and $B = d_1cos(ωt) + d_2sin(ωt)$ with $c,d$ in $R$, can be combined with function addition to yield $A+B = c_1cos(ωt) + c_2sin(ωt) + d_1cos(ωt) + d_2sin(ωt) $, or $ (c_1 + d_1)cos(ωt) + (c_2 + d_2)sin(ωt)$ which is of the same form as your initial $y(t) = c_1cos(ωt) + c_2sin(ωt)$.

Finally, we already know that $y(t) = 0$ is an element of the space $y(t) = c_1cos(ωt) + c_2sin(ωt)$ if we set $c_1,c_2 = 0$, so we arrive at the conclusion that $y(t) = c_1cos(ωt) + c_2sin(ωt)$ is indeed a vector space.

Hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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