5

Let, $C[0, 1]$ be the real vector space of all continuous real valued functions on $[0, 1]$, and let $T$ be the linear operator on $C[0, 1]$ given by $$(Tf)(x) =\int_{0}^{1}\sin(x + y)f(y) dy,\quad x\in[0, 1].$$ Then what is the dimension of the range space of $T\;?$

My attempt:

I know that $C[0, 1]$ is a vector space of infinite dimension. But I don't know what is basis of $C[0, 1]$, that's why could not find the matrix representation. I've read this problem in MSE. But I'm unable to solve the problem. Any hints will be apprciated.

Empty
  • 13,012
SAHEB PAL
  • 1,299
  • 11
  • 19

1 Answers1

6

Note that $$\sin(x+y) = \sin(x) \cos(y) + \sin(y) \cos(x).$$ Using this, we see that for any $f \in C[0,1]$, $$(Tf)(x) = \left(\int^1_0 \cos(y) f(y) dy \right) \sin(x) + \left(\int^1_0 \sin(y) f(y) dy \right) \cos(x).$$ Thus the image of any continuous function under $T$ will be a linear combination of two functions: $\sin(x)$ and $\cos(x)$. This shows that the range of $T$ has dimension $2$.

User8128
  • 15,485
  • 1
  • 18
  • 31
  • Your approach seems to me correct, but can you explain how could you say "This shows that the range of T has dimension 2". – SAHEB PAL Mar 07 '18 at 17:48
  • The dimension of a space is the number of elements in a basis for the space. This shows that ${\sin(x), \cos(x)}$ is a basis for the range of $T$. – User8128 Mar 07 '18 at 20:32
  • @User8128 To be precise it shows that ${\sin(x),\cos(x)}$ spans the range (i.e. that $\dim \mathop{\text{range}}(T) \le 2$). It is of course trivial to show that both $\sin(x)$ and $\cos(x)$ are in the range (and that they are linearly independent) so it is indeed a basis. – Jamie Radcliffe Nov 30 '22 at 03:41