reduction of matrix (nilpotence)

Question

I'm struggling doing this exercise:

Let A,B be two matrices in M_n(R) where N is nilpotent and NA = 0_n. We must show that the characteristic polynomial of A+N is exactly the one of A. Well, since I cannot solve it, I would like a hint please.

Though here is what I've tried so far: I have show that Tr(A+N) = Tr(A), I thought it was a good start but to go on I'd need also to prove det(A+N) = det(A) but so far I got det(A+N) = 1 which was not really convincing and then afterwards I'd need to show that the multiplity of the eigenvectors would be the same to finally get same characteristic polynomial, which I couldn't do so I gave up this idea. I also tried this : let k be the nilpotence index of N. Then I tried to compute (A+N)^k in order to get a clean closed form but I think I have done some mistakes, I found something like A^k + A^k * N, where actually I was expecting from this to get a similitude relation like P^-1(A+N)P=A so that since determinant is a similitude invariant I'd immediately get equality of characteristic polynomial.

I haven't mentionned but but I tried this since I know that N nilpotent <=> N trigonalizable <=> Sp(N) = {0}.

Though with all of this I could not conclude.

Thank you!

Answer 1

The relation $NA=\mathbf 0$ allows us to cleanly bionomial expand $\big(A+N\big)^n$ though what we get is akin to the identity for distinct $w,z\in \mathbb C$ that $\sum_{j=0}^n w^jz^{n-j} = \frac{w^{n+1}-z^{n+1}}{w-z}$

Base cases:
$\big(A+N\big)^2 = A^2 +AN +NA +N^2 = A^2 +AN +N^2$ $\big(A+N\big)^3 = \big(A+N\big)\big( A^2 +AN +N^2\big)=A^3+A^2N+AN^2+N^3=\sum_{j=0}^3 A^j N^{3-j}$

Inductive Case:
$\big(A+N\big)^{k+1}= \big(A+N\big)\big(A+N\big)^{k}=\big(A+N\big)\sum_{j=0}^k A^j N^{k-j}=\big(\sum_{j=0}^k A^{j+1} N^{k-j}\big) +A^0N^{k+1}$
$=\sum_{r=0}^{k+1} A^r N^{k+1-r}$

main argument:
For $k\in \big\{1,2,3 ,\cdots\big\}$ we have
$\text{trace}\Big(\big(A+N\big)^k\Big)= \sum_{j=0}^k \text{trace}\Big(A^j N^{k-j}\Big)= \text{trace}\Big(A^k\Big)+\sum_{j=0}^{k-1} \text{trace}\Big(A^j N^{k-j}\Big)= \text{trace}\Big(A^k\Big)$

since $N^k$ is nilpotent so $\text{trace}\Big(N^{k}\Big)=0$ and
for $0\lt j\lt k$ we have $\text{trace}\Big(A^j N^{k-j}\Big)=\text{trace}\Big(NA^j N^{k-j-1}\Big)=\text{trace}\Big(\mathbf 0\Big)=0$
by cyclic property of trace (alternatively: that matrix is nilpotent)

we conclude that $\big(A+N\big)$ and $A$ have the same characteristic polynomials, e.g. by Newton's Identities, since $\text{trace}\Big(\big(A+N\big)^k\Big)= \text{trace}\Big(A^k\Big)$ for all $k\in \mathbb N$

addendum
For an arbitrary field, I'd suggest a very simple invariant subspace argument. Let $V :=\mathbb F^n$ and use strong induction on $n$. There is nothing to do for the Base Case when $n=1$. For the inductive case, let $\dim \ker N = m \in \big\{1,2,\dots, n-1\big\}$ and note $m=n\implies N =\mathbf 0$ so there is nothing to do in that case. Then $\ker N$ is $A$-invariant because $w \in \ker N\implies N(Aw)=(NA)w=\mathbf 0w =\mathbf 0 \implies (Aw) \in \ker N$. Let $\mathbf B \in GL_n(\mathbb F)$ have its first $m$ columns be a basis for $\ker N$.
$N\mathbf B =\mathbf B \begin{bmatrix} \mathbf 0 & *\\ \mathbf 0 & N_{n-m} \end{bmatrix}$ and $A\mathbf B =\mathbf B \begin{bmatrix}A_m & *\\ \mathbf 0 & A_{n-m} \end{bmatrix}$
where we know $N_{n-m}$ is nilpotent and satisfies $N_{n-m}A_{n-m}=\mathbf 0$ so

$\det\big(A-\lambda I_n\big) = \det\big(A_m-\lambda I_m\big)\det\big(A_{n-m}-\lambda I_{n-m}\big)$
$=\det\big(A_m-\lambda I_m\big)\det\big(A_{n-m}+N_{n-m}-\lambda I_{n-m}\big)= \det\big(A + N -\lambda I_n\big)$
where the second equality holds by induction hypothesis

Answer 2

Note that if $A$ and $N$ are matrices such that $N$ is nilpotent and $NA = 0$, then the same is true for $PAP^{-1}$ and $PNP^{-1}$. Thus, may assume without loss of generality that $N^T$ is in Weyr canonical form. That is, we can assume that we have $$ N = \pmatrix{0_{k_1 \times k_1} \\ I_{k_2 \times k_1} & 0_{k_2\times k_2}\\ &\ddots & \ddots\\ &&I_{k_m\times k_{m-1}}&0_{k_m \times k_m}}, $$ with $k_1 \geq k_2 \geq \cdots \geq k_m$ and $I_{p \times q}$ being the block matrix $[I_{p} \ \ 0_{p,q-p}]$ where $I_p$ denotes the size-$p$ identity matrix.

With that, the fact that $NA = 0$ implies that $A$ has zero entries everywhere outside of its last $k_m$ rows. That is, we partition $A$ into the matrix $$ A = \pmatrix{0 & 0 \\A_{21} & A_{22}}, $$ where $A_{22}$ has shape $k_m \times k_m$.

Note that $A$ and $N$ are block-upper-triangular, and that they have the same diagonal blocks. The characteristic polynomial of a block-upper triangular matrix is the product of the characteristic polynomials of the blocks on the diagonal. Thus, $A$ and $A + N$ have the same characteristic polynomials.

In particular, if $p(x)$ denotes the characteristic polynomial of $A_{22}$, then both $A$ and $A + N$ have characteristic polynomials given by $x^{n - k_m} p(x)$.

Answer 3

Let $F$ be the underlying field (which is $\mathbb R$ in your case, but the proof here works over any field) and $x$ be an indeterminate. Denote by $F(x)$ be the field of fractions of $F[x]$. Since $\det(xI-A)$ is a monic polynomial in $x$, it is nonzero. Therefore $xI-A$ is invertible over $F(x)$. By Cayley-Hamilton theorem, $(xI-A)^{-1}$ is a polynomial in $xI-A$ with coefficients in $F(x)$. However, as $NA=0$, we have $N(xI-A)^k=x^kN$ for every integer $k\ge0$. It follows that $N(xI-A)^{-1}=r(x)N$ for some $r(x)\in F(x)$. Hence $N(xI-A)^{-1}$ is nilpotent and it is similar over $F(x)$ to a strictly upper triangular matrix. Therefore $\det\big(I-N(xI-A)^{-1}\big)=1$ and in turn $\det(xI-A-N)=\det(xI-A)$.