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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?

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2 Answers 2

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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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). –  Marc van Leeuwen Jan 31 '13 at 11:31
    
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. –  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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