How to find kernel and image? I have doubts on how to find the kernel and image for this linear transformation $T:\mathbb{C(R)}\rightarrow\mathbb{C(R})$ defined by: $T(f(x))=\frac{f(x)+f(-x)}{2}$, where $\mathbb{C(R)}$ represents the set of continuous functions in $\mathbb{R}$. I'm lost. Any hints?
 A: For the kernel, you want $T(f(x))=0$, so $$\frac{f(x)+f(-x)}2=0$$ so $$f(x)=-f(-x).$$ This is the exact definition of an odd function.
The image is the set of even functions, because $T(f(-x))=T(f(x))$ by a simple calculation. And reciprocally, if you take an even function, you can always put it in this form.
A: For example
$$f\in\ker T\iff f(x)=-f(-x)\iff f\;\;\text{is an odd function}$$
Now continue on this path and you'll find out the image is pretty simple, too.
A: For $\ker(T)$ it suffices to apply the definition: the condition $T(f)=0$ means that $x\mapsto\frac{f(x)+f(-x)}2$ is the zero function, which means that $f(-x)=-f(x)$ for all $x\in\Bbb R$ (i.e., $f$ is an an odd function).
A direct computation shows that this linear transformation$~T$ satisfies $T^2=T$, so it is a projection. For a projection, being in the image is the same as being ones own image: if $f=T(g)$ for any function $g$ then one also has $f=T(T(g))=T(f)$. So the condition for $f$ being in the image of $T$ is that $0=T(f)-f=\left(x\mapsto\frac{f(x)+f(-x)}2-f(x)\right)=\left(x\mapsto\frac{-f(x)+f(-x)}2\right)$, in other words that $f(-x)=f(x)$ for all $x\in\Bbb R$ (i.e., $f$ is an an even function).
One may add, and this is true for any projection, that the whole space is the sum of the kernel and the image of $T$, since every $f$ satisfies $f=T(f)+(I-T)(f)$, where the first term is in the image of $T$ and the second term is (because of $T^2=T$) in the kernel of$~T$. Moreover this sum is direct (the decomposition of $f$ is unique), since a function $g$ cannot be simultaneously in the image of $T$ (so $T(g)=g$) and in the kernel of $T$ (so $T(g)=0$) unless $g=0$.
