# The Unapologetic Mathematician

## The Jordan-Chevalley Decomposition (proof)

We now give the proof of the Jordan-Chevalley decomposition. We let $x$ have distinct eigenvalues $\{a_i\}_{i=1}^k$ with multiplicities $\{m_i\}_{i=1}^k$, so the characteristic polynomial of $x$ is

$\displaystyle\prod\limits_{i=1}^k(T-a_i)^{m_i}$

We set $V_i=\mathrm{Ker}\left((x-a_iI)^{m_i}\right)$ so that $V$ is the direct sum of these subspaces, each of which is fixed by $x$.

On the subspace $V_i$, $x$ has the characteristic polynomial $(T-a_i)^{m_i}$. What we want is a single polynomial $p(T)$ such that

\displaystyle\begin{aligned}p(T)&\equiv a_i\mod (T-a_i)^{m_i}\\p(T)&\equiv0\mod T\end{aligned}

That is, $p(T)$ has no constant term, and for each $i$ there is some $k_i(T)$ such that

$\displaystyle p(T)=(T-a_i)^{m_i}k_i(T)+a_i$

Thus, if we evaluate $p(x)$ on the $V_i$ block we get $a_i$.

To do this, we will make use of a result that usually comes up in number theory called the Chinese remainder theorem. Unfortunately, I didn’t have the foresight to cover number theory before Lie algebras, so I’ll just give the statement: any system of congruences — like the one above — where the moduli are relatively prime — as they are above, unless $0$ is an eigenvalue in which case just leave out the last congruence since we don’t need it — has a common solution, which is unique modulo the product of the separate moduli. For example, the system

\displaystyle\begin{aligned}x&\equiv2\mod3\\x&\equiv3\mod4\\x&\equiv1\mod5\end{aligned}

has the solution $11$, which is unique modulo $3\cdot4\cdot5=60$. This is pretty straightforward to understand for integers, but it works as stated over any principal ideal domain — like $\mathbb{F}[T]$ — and, suitably generalized, over any commutative ring.

So anyway, such a $p$ exists, and it’s the $p$ we need to get the semisimple part of $x$. Indeed, on any block $V_i$ $x_s=p(x)$ differs from $x$ by stripping off any off-diagonal elements. Then we can just set $q(T)=T-p(T)$ and find $x_n=q(x)$. Any two polynomials in $x$ must commute — indeed we can simply calculate

\displaystyle\begin{aligned}x_sx_n&=p(x)q(x)\\&=q(x)p(x)\\&=x_nx_s\end{aligned}

Finally, if $x:B\to A$ then so must any polynomial in $x$, so the last assertion of the decomposition holds.

The only thing left is the uniqueness of the decomposition. Let’s say that $x=s+n$ is a different decomposition into a semisimple and a nilpotent part which commute with each other. Then we have $x_s-s=n-x_n$, and all four of these endomorphisms commute with each other. But the left-hand side is semisimple — diagonalizable — but the right hand side is nilpotent, which means its only possible eigenvalue is zero. Thus $s=x_s$ and $n=x_n$.

August 28, 2012 Posted by | Algebra, Linear Algebra | 1 Comment

## The Jordan-Chevalley Decomposition

We recall that any linear endomorphism of a finite-dimensional vector space over an algebraically closed field can be put into Jordan normal form: we can find a basis such that its matrix is the sum of blocks that look like

$\displaystyle\begin{pmatrix}\lambda&1&&&{0}\\&\lambda&1&&\\&&\ddots&\ddots&\\&&&\lambda&1\\{0}&&&&\lambda\end{pmatrix}$

where $\lambda$ is some eigenvalue of the transformation. We want a slightly more abstract version of this, and it hinges on the idea that matrices in Jordan normal form have an obvious diagonal part, and a bunch of entries just above the diagonal. This off-diagonal part is all in the upper-triangle, so it is nilpotent; the diagonalizable part we call “semisimple”. And what makes this particular decomposition special is that the two parts commute. Indeed, the block-diagonal form means we can carry out the multiplication block-by-block, and in each block one factor is a constant multiple of the identity, which clearly commutes with everything.

More generally, we will have the Jordan-Chevalley decomposition of an endomorphism: any $x\in\mathrm{End}(V)$ can be written uniquely as the sum $x=x_s+x_n$, where $x_s$ is semisimple — diagonalizable — and $x_n$ is nilpotent, and where $x_s$ and $x_n$ commute with each other.

Further, we will find that there are polynomials $p(T)$ and $q(T)$ — each of which with no constant term — such that $p(x)=x_s$ and $q(x)=x_n$. And thus we will find that any endomorphism that commutes with $x$ with also commute with both $x_s$ and $x_n$.

Finally, if $A\subseteq B\subseteq V$ is any pair of subspaces such that $x:B\to A$ then the same is true of both $x_s$ and $x_n$.

We will prove these next time, but let’s see that this is actually true of the Jordan normal form. The first part we’ve covered.

For the second, set aside the assertion about $p$ and $q$; any endomorphism commuting with $x$ either multiplies each block by a constant or shuffles similar blocks, and both of these operations commute with both $x_n$ and $x_n$.

For the last part, we may as well assume that $B=V$, since otherwise we can just restrict to $x\vert_B\in\mathrm{End}(B)$. If $\mathrm{Im}(x)\subseteq A$ then the Jordan normal form shows us that any complementary subspace to $A$ must be spanned by blocks with eigenvalue $0$. In particular, it can only touch the last row of any such block. But none of these rows are in the range of either the diagonal or off-diagonal portions of the matrix.

August 28, 2012 Posted by | Algebra, Linear Algebra | 3 Comments