By symmetric characteristic polynomial, I mean for example... the characteristic polynomial of the $3\times3$ identity matrix is:

$x^3 - 3x^2 + 3x - 1$

similarly for the $4\times4$ identity matrix it is:

$x^4 - 4x^3 + 6x^2 - 4x + 1$

The absolute values of the coefficients are symmetric about the center.

Is there some general property of matrices that leads to this symmetry in the characteristic polynomial in cases other than the identity matrices or a scalar times identity? For example, is there something interesting we can say about a $4\times4$ matrix that has this characteristic polynomial?

$x^4 - 14x^3 + 26x^2 - 14x + 1$

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    $\begingroup$ If $\lambda$ is an eigenvalue, so is $\frac{1}{\lambda}$. $\endgroup$ – André Nicolas Jul 2 '16 at 5:27
  • $\begingroup$ @AndréNicolas, thanks this is what I was looking for. $\endgroup$ – Ameet Sharma Jul 2 '16 at 5:33
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Jul 2 '16 at 5:35

Those are examples of palindromic or anti-palindromic polynomials (I will call them pal and a-pal respectively [and (a)pal, for one that is either] in this answer). The identity matrix of order $n$ has characteristic polynomial $(x - 1)^n$, whose expansion has binomial coefficients, which are symmetric. This makes it obvious why its characteristic polynomial is (a)pal.

Now more generally, a polynomial is (a)pal iff all its roots are multiplicatively symmetric about $1$. That is, for each root $\lambda$ of a polynomial that is (a)pal, $\dfrac 1 \lambda$ is also a root.

One obvious and simple implication of this is that zero can never be a root, and any matrix $A$ with (a)pal characteristic polynomial is therefore non-singular. Since the eigenvalues of the inverse matrix $A^{-1}$ are exactly the reciprocals of the eigenvalues of $A$, this also implies that $A$ and $A^{-1}$ have the same spectrum.

We can do better than this.

A square matrix with entries from an algebraically closed field has (a)pal characteristic polynomial if and only if it is similar to its inverse.

Proof: Let $A$ be a matrix with (a)pal characteristic polynomial, and let $$J = P^{-1}AP = \operatorname{diag}(J_1, \ldots, J_p)$$ be its Jordan normal form. Each block $J_k$ is of the form \begin{equation*} J_k = \begin{bmatrix} \lambda_k & 1 & 0 & \cdots & 0\\ 0 & \lambda_k & 1 & \cdots & 0\\ 0 & 0 & \lambda_k & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & \lambda_k \end{bmatrix}. \end{equation*} We know that $A^{-1}$ has Jordan normal form $J^{-1} = P^{-1}A^{-1}P$, where $J^{-1} = \operatorname{diag}(J_1^{-1}, \ldots, J_p^{-1})$, and each block $J_k^{-1}$ is of the form \begin{equation*} J_k^{-1} = \begin{bmatrix} \frac 1 {\lambda_k} & 1 & 0 & \cdots & 0\\ 0 & \frac 1 {\lambda_k} & 1 & \cdots & 0\\ 0 & 0 & \frac 1 {\lambda_k} & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & \frac 1 {\lambda_k} \end{bmatrix}, \end{equation*} and has the same size as $J_k$. However, as the characteristic polynomial of $A$ is (a)pal, $J_k^{-1} = J_l$, for some $l$ (and hence $J_l^{-1} = J_k$). Thus, $J$ and $J^{-1}$ differ only by a rearrangement of the blocks $J_1, \ldots, J_p$, and are therefore similar. Specifically, there exists a permutation matrix $Q$ such that $J^{-1} = QJQ^{-1}$ (see note below). Then \begin{equation*} A^{-1} = PJ^{-1}P^{-1} = P(QJQ^{-1})P^{-1} = (PQP^{-1})(PJP^{-1})(PQP^{-1})^{-1} = RAR^{-1}, \end{equation*} where $R = PQP^{-1}$. Thus, if $A$ has (a)pal characteristic polynomial, it is similar to its inverse.

The converse follows immediately by observing that similar matrices have the same characteristic polynomial. $\qquad \square$

Note: The permuting matrix $Q$ for transforming $J$ to $J^{-1}$ can be obtained by writing the identity matrix $I$ of the same order as $J$ in the block diagonal form $I = \operatorname{diag}(I_1, \ldots, I_p)$, where $I_k$ is the identity matrix of the same order as $J_k$, and then applying the same permutation on these respective blocks as applied on the blocks of $J$ to obtain $J^{-1}$. That is, if $$J^{-1} = \operatorname{diag}(J_{\sigma(1)}, \ldots, J_{\sigma(p)}),$$ where $\sigma$ denotes the permutation applied to the blocks, then $$Q = \operatorname{diag}(I_{\sigma(1)}, \ldots, I_{\sigma(p)}).$$


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