Can we continously map coefficients of characteristic polynomials to matrices with specified endpoints?

Question

Let $A_1, A_2 \in \mathcal M_n(\mathbb C)$ be two fixed matrices with characteristic polynomials \begin{align*} &p_{A_1}(t) = t^n + \alpha_{n-1} t^{n-1} + \dots + \alpha_0, \\ &p_{A_2}(t) = t^{n} + \beta_{n-1}t^{n-1} + \dots + \beta_0. \end{align*} There is a bijection $\pi$ between $\mathbb C^n$ and the set of monic polynomials in degree $n$, i.e., $\pi(\alpha_{n-1}, \dots, \alpha_0) = t^n + \alpha_{n-1} t^{n-1} + \dots + \alpha_0$. Let $\gamma: [0,1] \to \mathbb C^n$ be a continuous path with $\gamma(0) = (\alpha_{n-1} , \dots, \alpha_0)$ and $\gamma(1) = (\beta_{n-1}, \dots, \beta_0)$. Does there exist a continuous function $f: [0,1] \to \mathcal M_n(\mathbb C)$ with $f(0) = A_1, f(1)=A_2$ such that $p_{f(s)} (t) = \pi(\gamma(s))$ for every $s \in [0,1]$ where $p_{f(s)}(t)$ denotes the characteristic polynomial of $f(s)$?

Answer 1

Yes, using Companion Matrices.

Let $\pi_k$ be the projection map from $\mathbb{C}^n$ to $\mathbb{C}$ that projects to the $k^{\textrm{th}}$ coordinate.

Consider the following path in the space $\mathcal{M}_{n}(\mathbb{C})$:

$$f(s) = \begin{bmatrix} 0 & 0 & \dots & 0 & -\pi_{n}(\gamma (s)) \\ 1 & 0 & \dots & 0 & -\pi_{n-1}(\gamma (s)) \\ 0 & 1 & \dots & 0 & -\pi_{n-2}(\gamma (s)) \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -\pi_{1}(\gamma (s)) \end{bmatrix}$$

Then each f(s) is a companion matrix for $\pi(\gamma (s))$ and so has that polynomial as its characteristic polynomial. This function is continuous since every entry of the matrix is either constant or the negative of one of the components of the continuous path $\gamma$.

Edit: After thinking about this some more, in order to get a path from arbitrary $A_1$ to $A_2$, we'll need to use the Jordan Normal Form Theorem instead. By that theorem, there exists invertible matrices $S_1$ and $S_2$ in $GL(n,\mathbb{C})$ such that:

$$A_1 = S_1 \begin{bmatrix} \lambda_{1} & \epsilon_{1} & 0 & \dots & 0 & 0 \\ 0 & \lambda_{2} & \epsilon_{2} & \dots & 0 & 0 \\ 0 & 0 & \lambda_{3} & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \dots & \lambda_{n-1} & \epsilon_{n-1} \\ 0 & 0 & 0 & \dots & 0 & \lambda_n \end{bmatrix} S_1^{-1}$$

where the $\lambda_i$ are the eigenvalues, the $\epsilon_{i} \in \{0,1\}$, and a similar form exists for $S_2$.

Since $GL(n,\mathbb{C})$ is path-connected, there is a continuous path $S(s)$ from $S_1$ to $S_2$ that stays in $GL(n,\mathbb{C}$).

Now, for every $s \in [0,1]$, use the Fundamental Theorem of Algebra to factor the polynomial $$\pi(\gamma(s)) = \prod_{i=1}^{n} (t-\lambda_i(s)).$$

Now, the roots of a polynomial vary continuously in terms of their coefficients. One way to make this precise is this paper by Naulin and Pabst. In particular, by always using a lexicographical ordering of $\mathbb{C}$ when writing our roots, we get $n$ continuous functions $\lambda_i(s)$ that maintain that order. We now adopt the convention that whenever writing Jordan normal forms, our eigenvalues will also be ordered lexicographically. (This convention can easily be used in picking $S_1$ and $S_2$ above.)

Next, we need to deal with the $\epsilon_i$'s. Look at the Jordan normal forms for $A_1$ and $A_2$ (as always with eigenvalues ordered lexicographically). Let $\epsilon_{j,i}$ be the $(i,i+1)$ entry of $A_j$. Then define the following continuous paths:

$$E_i(s) = \begin{cases} s, & \textrm{if } \epsilon_{1,i} = 0 \textrm{ and } \epsilon_{2,i} = 1 \\ 1-s, & \textrm{if } \epsilon_{1,i} = 1 \textrm{ and } \epsilon_{2,i} = 0 \\ 1, & \textrm{if } \epsilon_{1,i} = \epsilon_{2,i} = 1 \\ 0, & \textrm{if } \epsilon_{1,i} = \epsilon_{2,i} = 0 \end{cases}$$

Finally (!), we're ready to put these all together to get our path:

$$f(s) = S(s) \begin{bmatrix} \lambda_{1}(s) & E_1(s) & 0 & \dots & 0 & 0 \\ 0 & \lambda_{2}(s) & E_2(s) & \dots & 0 & 0 \\ 0 & 0 & \lambda_{3}(s) & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \dots & \lambda_{n-1}(s) & E_{n-1}(s) \\ 0 & 0 & 0 & \dots & 0 & \lambda_n(s) \end{bmatrix} S(s)^{-1}$$

This is a continuous path in $\mathcal{M}_n(\mathbb{C})$, $f(0) = A_1$, $f(1) = A_2$, and the characterisitic polynomial of $f(s)$ is (by similarity and properities of upper triangular matrices) $$\prod_{i=1}^{n} (t-\lambda_i(s)) = \pi(\gamma(s)).$$

Answer 2

First of all, let's note that this question is subtle, because the answer is no over $\mathbb{R}$! The matrices $$\begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix} \ \mbox{and}\ \begin{bmatrix} 0&-1 \\ 1&0 \end{bmatrix}$$ both have characteristic polynomial $\lambda^2+1$, but there is no path from one to the other through matrices with characteristic polynomial $\lambda^2+1$. To see this, note that a upper triangular matrix cannot have characteristic polynomial $\lambda^2+1$, so it is not possible to continuously change from lower left entry negative to lower left entry positive while keeping the same characteristic polynomial.

However, the original question was over $\mathbb{C}$, and there the answer is yes. As in Charlie Cunninham's answer, for a degree $n$ polynomial $p$, let $C(p)$ be the companion matrix. We can write our original matrices as $A_1 = S_1 C(p_1) S_1^{-1}$ and $A_2 = S_2 C(p_2) S_2^{-1}$ for some invertible $S_1$ and $S_2$. The group $GL_n(\mathbb{C})$ is path connected (here is the first citation I could find) so we can build a path $S(t)$ from $S_1$ to $S_2$. Then $S(t) C(p_{\gamma(t)}) S(t)^{-1}$ is your desired path.