basis and Jordan canonial form Let $P_3$ be the real vector space of polynomials of degree at most 3. Let $T:P_3\rightarrow P_3$ be defined by $T(p)=p''+p$ for all $p\in P_3$. Find a basis $X$ for $P_3$ such that the matrix of $T$ with respect to $X$ is in Jordan canonical form.
Can someone give me clue?
 A: You need to find $det([T] -\lambda I)=0$ to determine the eigenvalues. Any basis for $P_3$ will do, for example $\beta=\{ 1, x, x^2,x^3 \}$ for which $T(a+bx+cx^2+dx^3) =  2c+6dx+a+bx+cx^2+d^3 = a+2c+(b+6d)x+cx^2+dx^3$ hence $$[T] = \left[ \begin{array}{cccc} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$
Therefore, the eigenvalue is $\lambda = 1$ repeated four times. You can calculate eigenvectors and generalized eigenvectors for this matrix then translate back to polynomials in view of the basis $\beta$. Alternatively, calculate the eigenvectors directly in polynomial notation.
We're looking for as many solutions of $T(f(x))=f''(x)+f(x) = f(x)$ as possible. Clearly that gives $f''(x)=0$ which allows solutions $f(x) = ax+b$. This gives us two eigenvectors. Therefore, we need to find two generalized eigenvectors. 
We want them in a chain, $T(f(x))-f(x) = 1$ or $T(f(x))-f(x) = x$. (now I'm guessing, there is no reason given my current calculations that either $1$ or $x$ is attached to a chain). Continuing (with reckless abandon)
$$ f''(x) = 1 \qquad \text{or} \qquad f''(x) = x$$
hence $f(x) = \frac{1}{2}x^2$ or $f(x) = \frac{1}{6}x^3$. I believe, we can use $\gamma=\{ 1, x^2/2, x, x^3/6 \}$ as a Jordan basis. Finally, let's check:
$ T(a+bx^2/2+cx+dx^3/6)=$ 
$$= b+dx+a+bx^2/2+cx+dx^3/6 = (a+b)+bx^2/2+(c+d)x+dx^3/6. $$
it follows that $[T]_{\gamma,\gamma}$ is in Jordan form.
