Jordan Block Eigenvalue Proof

Question

Let A be a near-Jordan block, that is, a matrix obtained from a Jordan block by possibly changing the first column. Prove that no two Jordan blocks in any Jordan canonical form for A have the same eigenvalue

Answer 1

Let $A \in M_n(\mathbb{F})$ be a perturbation of an $n \times n$ Jordan block with eigenvalue $\lambda \in \mathbb{F}$ so

$$ A = \begin{pmatrix} a_1 & 1 & 0 & \dots & 0 \\ a_2 & \lambda & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ a_{n-1} & \ddots & 0 & \lambda & 1 \\ a_n & 0 & 0 & 0 & \lambda \end{pmatrix}. $$

We can write $A = B + C$ where $B = \lambda I + N$ and

$$ N = \begin{pmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \ddots & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}, C = \begin{pmatrix} a_1 - \lambda & 0 & \dots & 0 \\ a_2 & 0 & \dots & 0 \\ \vdots & \vdots & \dots & \vdots \\ a_{n-1} & 0 & \dots & 0 \\ a_n & 0 & \dots & 0\end{pmatrix}. $$

Proving that any two blocks in the Jordan form of $A$ are associated to different eigenvalues is equivalent to proving that the minimal polynomial of $A$ and the characteristic polynomial of $A$ coincide. By Cayley-Hamilton, it is enough to prove that $p(A) \neq 0$ for all polynomials of degree $0 < k < n$.

For $B$, we know that the characteristic and the minimal polynomial coincide. Let us recall why. Denote by $(e_1, \dots, e_n)$ the standard basis vectors. We have $N^i e_n = e_{n-i}$. Now, if $p \in \mathbb{F}[X]$ is a polynomial of degree $k < n$, we claim that $p(B)e_n \in \operatorname{span} \{ e_n, \dots, e_{n-k} \}$ and $p(B)e_n = 0$ if and only if $p = 0$. To see this, write $p(x + \lambda) = g(x) + p(\lambda)$ where $g(x)$ is a polynomial of degree $k$ satisfying $g(0) = 0$. Writing $g(x) = \sum_{i=1}^k b_i X^i$, we have

$$ p(B) = p(N + \lambda I) = g(N) + p(\lambda)I = \sum_{i=1}^k b_i N^i + p(\lambda)I, \\ p(B)e_n = \sum_{i=1}^k b_i e_{n-i} + p(\lambda) e_n $$

which shows that $p(B)e_n \in \operatorname{span} \{ e_n, \dots, e_{n-k} \}$ and $p(B)e_n = 0$ if and only if $b_1 = \dots = b_k = 0$ and $p(\lambda) = 0$ which implies that $p = 0$.

Let us return to $A$. Note that $Ce_i = 0$ for $2 \leq i \leq n$. This implies by induction that $A^i e_n = B^i e_n$ for $0 \leq i < n$ as

$$ A^{i+1} e_n = A(A^i e_n) = A(B^i e_n) = (B + C)(B^i e_n) = B^{i+1} e_n + CB^i e_n $$

but since $B^i e_n \in \operatorname{span} \{ e_n, \dots, e_{n-i} \}$ and $i + 1 < n$ we have $CB^i e_n = 0$. Now, if $p(X) = \sum_{i=0}^k b_i X^i$ is a polynomial of degree $k < n$ that satisfies $p(A) = 0$, we have

$$ 0 = p(A)e_n = \sum_{i=0}^k b_i A^i e_n = \sum_{i=0}^k b_i B^i e_n = p(B) e_n $$

and by what we have shown before $p = 0$.

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Alternatively, following the suggestion of darij grinberg and using the notation above, note that $A = \lambda I + (C + N)$ and $C + N$ is a matrix that is similar to a companion matrix:

$$ C + N = \begin{pmatrix} a_1 - \lambda & 1 & 0 & \dots & 0 \\ a_2 & 0 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ a_{n-1} & \ddots & 0 & 0 & 1 \\ a_n & 0 & 0 & 0 & 0 \end{pmatrix} \sim \begin{pmatrix} 0 & 0 & \dots & 0 & a_n \\ 1 & 0 & \dots & 0 & a_{n-1} \\ 0 & 1 & \dots & 0 & a_{n-2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & (a_1 - \lambda) \end{pmatrix} $$

Now, we have $p(A) = 0$ if and only if $g(N + C) = 0$ where $g(x) = p(x + \lambda)$. Since $N + C$ is similar to a companion matrix, the minimal polynomial of $N + C$ has degree $n$ and so does the minimal polynomial of $A$. This method has the advantage of giving out explicitly the characteristic and minimal polynomial of $A$ which is

$$ X^n - \left( (a_1 - \lambda)(X - \lambda)^{n-1} + a_2(X - \lambda)^{n-2} + \dots + a_{n-1}(X - \lambda) + a_n \right). $$