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\title{Lecture 3-27: Jordan decompositions}
\date{March 27, 2024}
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Last time we saw the Jordan canonical form of a square matrix $M$ over a field $k$ containing all eigenvalues of $M$ (in any extension field), which realizes $M$ as similar to a block diagonal matrix for which the (Jordan) blocks $B$ have equal entries along their main diagonals, 1s just above these diagonals, and 0s elsewhere.  In particular, we can write each $B$ as the sum of the diagonal matrix $B_D$ having the same diagonal entries as $B$ and a strictly upper triangular matrix $B_N$, whose nonzero entries are the 1s occurring in $B$.
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It is clear for each $B$ that the matrices $B_D$ and $B_N$ commute with each other.  Putting the blocks together and passing to similar matrices, we deduce that \alert{under the above hypothesis on $k$, any square matrix is similar to a sum $D+N$ of a diagonalizable matrix $D$ and a nilpotent matrix $N$ where $D$ and $N$ commute}.  Thus $N$ such that $N^m=0$ for some positive integer $m$.  But now it turns out that we can say even more about the matrices $D$ and $N$ in this sum.  To do this, we start over, writing the ambient vector space $V$ as the direct sum $\bigoplus\limits_{i=1}^m V_i$ of subspaces $V_i$ preserved by $M$, such that for each $i$ there is a scalar $\lambda_i\in k$ with $(M-\lambda_i I)^{n_i}$ acting by 0 for some integer $n_i$, with the $\lambda_i$ distinct.  
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Now use the Chinese Remainder Theorem to find a polynomial $p\in k[x]$ such that $p(x) \equiv \lambda_i\mod (x-\lambda_i)^{n_i}$ for each $i$ while $p(x)\equiv 0\mod x$; note that we can omit the last congruence if some $\lambda_i=0$, so that the moduli are always pairwise relatively prime.  Then the polynomial $p(M)$ acts as the scalar $\lambda_i$ on each $V_i$ and $p$ has constant term 0; moreover, $q(M)=M-p(M)$ acts nilpotently on each $V_i$ (and thus on all of $V$), with the constant term of $q$ also being 0.  The upshot is that the matrices $D=p(M),N=q(M)$ are diagonalizable and nilpotent, respectively.  Both commute with $M$ (being polynomials in $M$) and they commute with each other
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Moreover, $D+N$ is the {\sl only} decomposition of $M$ as the sum of commuting diagonalizable and nilpotent matrices.  Indeed, given such a decomposition $D'+N'$, we have $D-D' = N-N'$ and $D,D',N,N'$ commute pairwise, since all four matrices commute with $M$ and $D,N$ are polynomials in $N$.  Now the binomial theorem (valid for any pair of commuting elements in any ring) shows that $N-N'$ is also nilpotent.
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Likewise $D-D'$ is diagonalizable.  To see this, decompose the vector space $V$ as the sum of eigenspaces of $D$.  Since $D'$ commutes with $D$, it stabilizes all of its eigenspaces and must act diagonalizably on each one.  Hence there is a basis of $V$ which simultaneously diagonaiizes $D$ and $D'$; this same basis realizes $D-D'$ as diagonalizable.
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The upshot is that $D-D'=N-N'$ is both diagonalizable and nilpotent, whence this difference must be 0, as claimed.  We have proved
\begin{alertblock}{Jordan decomposition theorem}
Under the above hypothesis on $k$, any square matrix $M$ is uniquely the sum $D+N$ of commuting diagonalizable and nilpotent matrices.
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This basis-free version of the Jordan canonical form has many applications to Lie algebras, Lie groups, and algebraic groups.  The summands $D,N$ are called the \alert{semisimple} and \alert{nilpotent} parts of $M$.  This terminology is explained below.
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What about fields $k$ not containing the eigenvalues of a square matrix $M$?  It turns out that a weaker version of the Jordan decomposition (though not the Jordan canonical form) holds for most such fields.  Call a square matrix $D'$ {\sl potentially diagonalizable} if it is similar to a diagonal matrix when regarded as a matrix over some extension $k'$ of $k$.  Then we have
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\begin{alertblock}{Theorem}
If $k$ is perfect (so that it either has characteristic 0 or it has characteristic $p$ and is closed under taking $p$th roots) then any square matrix $M$ over $k$ can be uniquely written as the sum $D'+N$ of commuting potentially diagonalizable and nilpotent matrices.
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\noindent For the proof see the Wikipedia article \lq\lq Jordan-Chevalley decomposition" (but be warned that the argument there is garbled and needs fixing).
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We now define semisimple linear transformations.   Note first that given an irreducible polynomial $p(x)$ over any field $k$, the quotient ring $k[x]/(p(x))$ has no proper ideals.  Thus when regarded as a vector space $V$ over $k$, it has no proper subspaces stable under multiplication by $x$.  More generally, given a product $q=p_1(x)\ldots p_m(x)$of distinct irreducible polynomials $p_i(x)$, the quotient $k[x]/(q)$ is isomorphic to the direct sum of the quotients $V_i=k[x]/(p_i(x))$.  Then multiplication by $x$ acts on each $V_i$ without stabilizing any proper subspace of it.
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A linear transformation $T$ on a finite-dimensional vector space $V$ is called  \alert{semisimple} if $V$ can be written as the direct sum of subspaces $V_i$ stabilized by $T$ such that $T$ does not stabilize any proper subspace of $V_i$.  For example, a nontrivial rotation not by $\pi$ radians is semisimple as a linear transformation of $\mathbb R^2$.  Then it turns out that \alert{over a perfect field a transformation is semisimple if and only if it is potentially diagonalizable}.  In particular, if $k$ contains all eigenvalues of $T$, then $T$ is semisimple if and only if it is diagonalizable.  Any potentially diagonalizable linear transformation is semisimple, but the converse can fail for imperfect fields.
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The notion of semisimplicity actually has many different meanings in algebra, but all involve finite direct sums.  You learn more about some of the other meanings of this term later.
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