# Monday August 26

**Definition (Jordan Decomposition):**

Let $X \in \mathrm{End}(V)$ for $V$ finite dimensional. Then,

(a) There exists a unique $X_s, X_n \in \mathrm{End}(V)$ such that $X = X_s + X_n$ where $X_s$ is semisimple, $X_n$ is nilpotent, and $[X_s, X_n] = 0$.

(b) There exists a $p(t), q(t) \in t \FF[t]$ such that $X_s = p(X), X_n = q(X)$.

(Polynomials with no constant term.)

*Proof of (a):* 
Assume $X_s = X_s + X_n = X_s' + X_n'$, so both have bracket zero. Assuming that (b) holds, we have $X_s = p(X)$, and so
$$
[X, X_s] = [X_s + X_n', X_s'] = [X_s', X_s'] + [X_s', X_n'] = 0 \implies
[p(X), X_S'] = 0 = [X_s, X_s']
$$

Using fact (c) from last time, then $X_s, X_s'$ can be diagonalized simultaneously, and so $X_s - X_s'$ is semisimple.

On the other hand, if $X_n', X_n$ are nilpotent, and since these commute, $X_n - X_n'$ is nilpotent. But then this is a Jordan decomposition of the zero map, i.e.
$$
0 = X - X = (X_s - X_s') + (X_n + X_n')
$$
where the first term is semisimple and the second is nilpotent. Then each term is both semisimple *and* nilpotent, so they must be zero, which is what we wanted to show.

*Proof of part (b):* 
Let $m(t) = \prod_{i=1}^r (t-\lambda_i)^{m_i}$ be the minimal polynomial of $X$, where each $m_i \geq 1$ and the $\lambda_i$ are distinct.
Then the primary composition of $V$ is given by
$$
V = \bigoplus_{i=1}^r V_i,\quad V_i = \ker(X - \lambda_i I_V) \neq 0, \quad X(V_i) \subseteq V_i
$$
**Claim:**
There exists a polynomial $p\in F[t]$ such that
\begin{align*}
p &= \lambda \mod (t-\lambda_i)^{m_i} \quad \forall i, \\
p &= 0 \mod t
.\end{align*}

The existence follows from the Chinese Remainder Theorem.

What is $p(x) \actson V_i$? This acts by scalar multiplication by $\lambda_i$ for all $i$. (Check).
Because of the restrictive conditions, $p(x)$ has no constant term.

![???](figures/2019-08-28-09:28.png)

So $p(X) = X_s$ is the semisimple part we want. Now just set $q(t) = t - p(t)$, then $X_n \coloneqq q(X) = X - X_s$ is nilpotent.

*Example:*
The Jordan Decomposition is invariant under taking adjoints.

If we have $X = X_s + X_n$, then $\ad_X \in \mathrm{End}(\mathrm{End}(V))$. It can be shown that $(\ad_X)_s + (\ad_X)_n = \ad(X_s) + \ad(X_n)$.

Let $e_{ii}$ be the elementary matrix with a 1 in the $i, j$ position. You can write $\ad_X$ as a $4\times 4$ matrix (see image).

![Image](figures/2019-08-28-09:39.png)
![Image](figures/2019-08-28-09:40.png)
![Image](figures/2019-08-28-09:40.png)

You can check that $(\ad_X)_S = 0, \ad(X_s) = 0$, and $(\ad X)_n$ is the Jordan form given above.

**Lemma:**

(a) $x \in \mathrm{End}(V) \implies \ad(x)_s = \ad(x_s)$ and $\ad(x)_n = \ad(x_n)$.

(b) If $A$ is a finite dimensional $\FF\dash$algebra, then $\delta \in \mathrm{Der}(A) \implies \delta_s, \delta_n \in \mathrm{Der}(A)$ as well.


*Proof of (a):*

Check that $\ad(x) = \ad(x_s) + \ad(x_n)$.
Then for $y\in \mathrm{End}(V)$, we have
\begin{align*}
(\ad(x))(y)
&= [x, y] \\
&= [x_s + x_n, y] \\
&= [x_s, y] + [x_n, y] \\
&- (\ad(x_s))(y) + (\ad(x_n))(y)
.\end{align*}

Using theorem 3.3, $x_n$ nilpotent $\implies$ $\ad(x_n)$ is also nilpotent. So write $x_s = \sum \lambda_i e_{ii}$ with the eigenvalues on the diagonal. Then $\ad x_s (e_{ij}) = (\lambda_i - \lambda_j)e_{ij}$ for all $i, j$. But then $\ad x_s$ is given by a matrix with $\lambda_i - \lambda_j$ in the $i,j$ position and zeros elsewhere. By the uniqueness of the Jordan decomposition, the statement follows.

*Proof of (b):*

Since $\delta \in \mathrm{Der}(A)$, the primary decomposition with respect to $\delta$ is given by
$$
A = \bigoplus_{\lambda \in F} A_\lambda \quad \text{where} ~A_\lambda = \theset{a\in A \suchthat (\delta - \lambda I)^k a = 0 ~\text{for some}~ k >> 0}.
$$
So $\delta_s \actson A_\lambda$ by scalar multiplication (by $\lambda$). Then for $\lambda, \mu \in \FF$, we have

![Image](figures/2019-08-28-09:54.png)

So $[A_x, A_y]  \subseteq A_{\lambda + \mu}$ for all $x, y \in A$. But then

![Image](figures/2019-08-28-09:56.png)

and so $\delta_s \in \mathrm{Der}(A)$, and $\delta_n = \delta - \delta_s \in \mathrm{Der}(A)$ as well.