Finding kernel and range Let $V = \operatorname{Span}[{\sin(x), \cos(x),\sin(2x), \cos(2x)}] ⊂ C^∞ (R)$. Show that the mapping $L : V → V$ defined by $L(y) = y + y''$ is a linear transformation, and show that $L$ really does map to $V$. 
Find the kernel and range of $L$, and find the matrix of $L$ with respect to the basis $B = [{\sin(x), \cos(x),\sin(2x), \cos(2x)}]$ of $V$.
To prove that it is a linear transformation I believe I am just making sure $(u+w)+(u+w)''= (u)+(u)''+(w)+(w)''$, right?
But then I am stuck on computing the kernel with respect to the basis. Once I find one of them I can find the other by using the rank-nullity theorem, correct? But which one should I try and find and how do I begin? 
 A: Because V is finite dimentional, you can compute the matrix representation of L with respect to basis $B={sin(x), cos(x), sin(2x), cos(2x)}$.
$$L(sin(x))=0$$
$$L(cos(x))=0$$
$$L(sin(2x))=sin(2x)-4sen(2x)=-3sen(2x)$$
$$L(cos(2x))=cos(2x)-4cos(2x)=-3cos(2x)$$
Then, calculating the coordinates, the matrix representation is
$$A=\begin{pmatrix}
0&0&0&0\\
0&0&0&0\\
0&0&-3&0\\
0&0&0&-3\\
\end{pmatrix}$$
It is easy to see image of A is generated by third and fourth column of A , $im(A)=span\{(0, 0, -3, 0), (0, 0, 0, -3)\}$, then $im(L)=span\{sin(2x), cos(2x)\}$. Also, it is straightforward to see that kernel of A is span by $\{(1, 0, 0, 0), (0, 1, 0, 0)\}$ which means $ker(L)=span\{sin(x), cos(x)\}$ because the kernel of this matrix is the coordinates of the kernel of L (and the same for the image).
A: It is clear that $y''+y=0$ forces $y=c_1\cos(t)+c_2\sin(t),$ namely, linear combinations of sine and cosine. That is to say, 
$$\ker(L)=\text{span}\{\cos(t),\sin(t)\}.$$
The image of this map can be computed as follows:
$$L(y)=y''+y=-4c_3\cos(2t)-4c_4\sin(2t)$$
so 
$$\text{Im}(L)=\text{span}\{\cos(2t),\sin(2t)\}.$$
To establish linearity, as was commented you must show that
$$L(\alpha v+w)=\alpha L(v)+ L(w)$$
for any $v,w\in V$ and $\alpha \in \Bbb{R}.$
You can (and should) show this directly. You should also recognize that the derivative operator, $\frac{d}{dt}$ is a linear operator, and so $\frac{d^2}{dt^2}$ is applying a linear map twice. Then $$L(v)=\left(\frac{d^2}{dt^2}+I\right)v$$ where $I$ denotes the identity map, is a sum of linear maps, and is therefore a linear map.
