How to find a basis for an eigenspace? Consider the real vector space $P_2(\mathbb{R})$ of real polynomials of grade $\leq 1$. Consider the inner product defined as
$$
\langle p,q \rangle = p(0)q(0)+p(1)q(1)
$$
and the linear operator
$$
L: P_2(\mathbb{R}) \rightarrow P_2(\mathbb{R})
$$
defined as
$$
L(\alpha + \beta X) = (8 \alpha + 2 \beta) + (\beta - 3 \alpha) X
$$
Then I have to determine all eigenvalues for $L$ as well as the basis for the eigenspaces but I am not sure about the part about eigenspaces. I have found the eigenvalues to be $2$ and $7$
I know that from my book that 
$$
E_L(\lambda) = N(A - \lambda I)
$$
where $I$ is the identity matrix. Normally I use this to find the eigenspaces and if I do so I get $(-2,1)^T$ is a basis for $E_L(7)$ and that $(-1/3,1)^T$ is a basis for $E_L(2)$. Does this mean now that I have to express these bases as polynomials?
So the desired bases are $p_1 = -2+x$ for $E_L(7)$ and $p_2 = -1/3 + x$ for $E_L(2)$. Have I understood this correctly?
Thanks for your help in advance. 
 A: Hint:
What you are after are the Eigenvectors of the matrix of the transformation,
$$\begin{pmatrix}8&&2\\-3&&1.\end{pmatrix}$$
A: $\newcommand{\span}{\textrm{span}\{}$
Yes you have. There is a simple way to verify this.  
Since $L(-2+x)=-14+7x=7(-2+x)$ and $L\left(-\frac{1}{3}+x\right)=-\frac{2}{3}+2=2\left(-\frac{1}{3}+x\right)$, $-2+x$ and $-\frac{1}{3}+x$ are eigenvalues of $L$ corresponding to distinct eigenvalues. As the two polynomials correspond to distinct eigenvalues they are necessarily linearly independent.  
Now, notice that since $L$ acts on a vector space of $\dim 2$, the eigenspace corresponding to $7$ and $2$ must be $\span 2+x\}$ and $\span -\frac{1}{3}+x\}$ respectively.  
If not, suppose for instance that $E_L(7)>\span 2+x\}$. Then, there would be a vector $v\in E_L(7)$ that is linearly independent of $2+x$. Given that the domain of $L$ is of $\dim 2$, we'll have that $E_L(7)=\span v,2+x\}$ equals the entire vector space. But this is clearly impossible, as $P_2(\mathbb R)$ contains, in particular, $-\frac{1}{3}+x$ which  doesn't belong to $E_L(7)$.
