# Linear Algebra- Jordan Normal Form Ok before anyone says Ive posted 3 questions in one post, I would like to state this is given as a single whole question and I'm sure parts before help the parts after. This is a question I'm not able to do I reckon due to my weakness with partial derivatives etc. so it would be great if someone can show me how to deduce the answers. My attempts have been rather weak and this is the only question I couldn't do on my question sheet and it was a rather early question so I'm quite sure its just down to my weakness in p.d.e.etc.

(a) Suppose that the operator has a nonzero eigenvalue $\lambda$.

Then there is a nonzero polynomial $f\in P_2$ such that $$D(f)=\lambda f.$$

Now, consider the total degree of the above. For polynomials of positive total degree, the operator $D$ reduces at least $1$ degree of $f$. However, the right hand side has the same total degree as $f$. Thus, $f$ cannot have positive total degree, hence $f$ is of total degree $0$, which are constant polynomials.

We see that $D(f)=0$ for all constant polynomials. Therefore, we must have $\lambda=0$. This yields that $D$ has the only eigenvalue $0$.

Let $I$ be the identity mapping of $P_2$. Consider $$(D-0I)(f)=0.$$

From the matrix representation of $D$ on the basis $\{x^2y^2,x^2y,x^2,xy^2,xy,x,y^2,y,1\}$, we see that the space of $f$ satisfying $D(f)=0$ has dimension $3$. Also, $1$, $x-y$, and $(x-y)^2$ satisfy $D(f)=0$.

Thus, $\textrm{Null}(D)$ has the basis $\{1,(x-y), (x-y)^2\}$.

The least positive integer $k$ satisfying $D^k=0$ is $k=5$. (check $x^2y^2$) This means that the Jordan Normal Form(JNF) should contain a $5\times 5$ block.

The remaining blocks can be found from considering $$D(x^2y-xy^2)=x^2-y^2, \textrm{ } D(x^2-y^2)=2x-2y, \textrm{ }D(x-y)= 0.$$

Therefore, there is one $3\times 3$ block, and one $1\times 1$ block in JNF of $D$.

c.f) The JNF of $D$: $$J_2=\pmatrix{{0}&{1}&{}&{}&{}&{}&{}&{}&{}\\{}&{0}&{1}&{}&{}&{}&{}&{}&{}\\{}&{}&{0}&{1}&{}&{}&{}&{}&{}\\{}&{}&{}&{0}&{1}&{}&{}&{}&{}\\{}&{}&{}&{}&{0}&{}&{}&{}&{}\\{}&{}&{}&{}&{}&{0}&{1}&{}&{}\\{}&{}&{}&{}&{}&{}&{0}&{1}&{}\\{}&{}&{}&{}&{}&{}&{}&{0}&{}\\{}&{}&{}&{}&{}&{}&{}&{}&{0}}$$ with the basis for $P_2$ consisting of generalized eigenvectors is: $$B_2=\{D^i(x^2y^2) \textrm{ for } i=0,1,2,3,4\}\cup \{D^i(x^2y-xy^2) \textrm{ for }i=0,1,2\} \cup \{(x-y)^2\}$$ The general case can be dealt with by induction on $n$.

Let $J_1$ be the JNF for $D:P_1\rightarrow P_1$, and $B_1=\{xy,D(xy), D^2(xy), x-y\}$. Then the JNF $J_n$ for $D:P_n\rightarrow P_n$ satisfies $$J_{n}=J(0,2n+1)\oplus J_{n-1} , \textrm{ for n\geq 2}$$ with the basis $B_n$ for $P_n$ consisting of generalized eigenvectors $$B_n=\{D^i(x^ny^n) \textrm{ for } i=0, \cdots, 2n\} \cup (x-y)B_{n-1}, \textrm{ for n\geq 2}.$$