The Jordan-Chevalley decomposition says that given a linear operator $L$, you can decompose it as $L = S + N$, where $S$ is diagonalizable and $N$ is nilpotent.

My textbook (Linear Algebra by Peterson) has a corollary of the Jordan-Chevalley that says that given $p(t) = (t-\lambda)^n$, the associated companion matrix $C_p$ is similar to a Jordan block (matrix with $\lambda$ on diagonal and $1$'s on superdiagonal).

So if I apply the JC decomposition to $C_p$, I get $C_p = \lambda I_n + (C_p - \lambda I_n) $. So I need to show that $C_p - \lambda I_n$ is similar to the matrix with $1$'s on the superdiagonal. I don't see how to do this without going into the Frobenius Canonical form and showing that the characteristic polynomial and minimal polynomial of $C_p$ are the same. (I.e. Using theorem that says if two operators have same minimal polynomial, then they are similar)

Is there an easy way to do this?


The eigenvalues of the companion matrix $C_p$ are the roots of $p(t)$. Thus we can write $C_p = S \Lambda S^{-1}$ where $\Lambda = \lambda I_n$. The eigenvalues of $C_p - \lambda I_n$ are therefore all zero, which means that $C_p - \lambda I_n$ is nilpotent. The Jordan canonical form of a nilpotent matrix, which is obtained through a similarity transformation, is a matrix with 1's on the superdiagonal and zero everywhere else. Hence, $C_p - \lambda I_n$ is similar to a matrix with 1's on the superdiagonal.

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    $\begingroup$ This is wrong. It is absurd to say that $C_p = S \lambda I_n S^{-1}$, since this would mean that $C_p =\lambda I_n$. It is true that the eigenvalues of $C_p-\lambda I_n$ are all zero (equivalently the eigenvalues of $C_p$ are all $\lambda$), but this is because the minimal polynomial of $C_p$ is $(X-\lambda)^n$ (it is also its characteristic polynomial, but that requires a bit more work, or applying Cayley-Hamilton). $\endgroup$ – Marc van Leeuwen Jan 31 '13 at 11:31
  • $\begingroup$ Moreover it is also untrue that the Jordan canonical form of a nilpotent matrix is a matrix with $1$'s on the superdiagonal and zero everywhere else, in other words a single Jordan block. Take the zero matrix of size $n>1$ for a counterexample, or any matrix of size $n\geq3$ with a single non-zero entry which is off the main diagonal. $\endgroup$ – Marc van Leeuwen Apr 4 '13 at 8:29

I cannot see why Peterson would want to invoke such a difficult result for such a simple conclusion. And you don't need the Frobenius Canonical form for showing that the characteristic polynomial and minimal polynomial of $C_p$ are the same (see this question and this one), nor indeed do you need the characteristic polynomial at all.

The companion matrix $C_p$ is defined so that $p(C_p)$, but no nonzero polynomial in $C_p$ of lower degree, annihilates the first basis vector $e_1$ (and as a consequence every vector). So the minimal polynomial of $C_p$ is $p=(X-\lambda)^n$, the Jordan normal form $J$ of $C_p$ has only entries $\lambda$ on the diagonal, and $(J-\lambda I)^{n-1}\neq0$, and the latter can only happen if $J$ consists of a single Jordan block (of size $n$).

In fact it is not hard to conjugate $C_p$ to its Jordan normal form explicitly. Define vectors $b_1,\ldots,b_n$ by $b_{n-i}=(C_P-\lambda I_n)^i\cdot e_1$, then $C_p\cdot b_1=\lambda b_1$ (because $(C_p-\lambda I_n)^n=p(C_p)$ annihilates $e_1$) and $C_p\cdot b_i=b_{i-1}+\lambda b_i$ for $i\geq2$, in other words change of basis to the basis $(b_1,\ldots,b_n)$ transforms $C_p$ into a Jordan block of size $n$ for $\lambda$.


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