Characteristic polynomials of $P(x)=A+ x B$ and $Q(x)=A+ x C$ are equal. If $P(x)$ is block diagonal show $Q(x)$ is block diagonal.

Question

Begin with a diagonal $N\times N$ matrix, $A$, with distinct eigenvalues $a_i$, and a real symmetric block diagonal matrix $\mathbf{B}$: $$ \mathbf{B}=\left(\begin{matrix} \mathbf{B_{11}} & \mathbf{0}\\ \mathbf{0} & \mathbf{B}_{22}\end{matrix} \right) $$ where $\mathbf{B}_{11}$ is a real symmetric $M\times M$ matrix ($M<N$) and $\mathbf{B}_{22}$ is a real symmetric $(N-M)\times (N-M)$ matrix. I am interested in the polynomial matrix $\mathbf{P}(x)=A+ x \mathbf{B}$. Suppose we have another real symmetric matrix, $\mathbf{C}$, and we know that $\mathbf{Q}(x)= A + x \mathbf{C}$ has the same characteristic polynomial as $\mathbf{P}(x)$ (i.e. $\det(\mathbf{Q}(x)-y \mathbf{I})=\det(\mathbf{P}(x)-y \mathbf{I}$). Can we show that $\mathbf{C}$ must also be block diagonal?

Things I know:

1. The characteristic polynomial for $\mathbf{Q}(x)$ is reducible. This point is what makes me most strongly suspect the conjecture is true, but doesn't really point to any way of showing it to be so.
2. The eigenvalue functions, $E_P(x)$, of $\mathbf{P}(x)$ are the same as the eigenvalue functions, $E_Q(x)$, of $\mathbf{Q}(x)$.
3. $\mathbf{P}(x)$ is similar to $\mathbf{Q}(x)$. If that similarity transformation were constant, we could prove the conjecture. But in principle the similarity transform could depend on $x$.
4. $\mathbf{C}$ must be similar to $\mathbf{B}$ (from the $x\rightarrow \infty$ limit of #3), which means that $\mathbf{B}$ and $\mathbf{C}$ share the same eigenvalues.
5. The trace of any function of $\mathbf{P}(x)$ is equal to the trace of the same function of $\mathbf{Q}(x)$: $Tr f(\mathbf{P}(x)) = Tr f(\mathbf{Q}(x))$.
6. As a corollary of #4: All of the symmetrized traces of powers of $A$ with powers of $\mathbf{B}$ are the same as the corresponding symmetrized power traces of $A$ with $\mathbf{C}$. That is: $Tr~(A^i \mathbf{B}^j)_S= Tr (A^i \mathbf{C}^j)_S $ where $(...)_S$ indicates a symmetrized matrix product (e.g. $(A \mathbf{B}^2)_S= A\mathbf{BB}+\mathbf{B}A\mathbf{B}+\mathbf{BB}A$).
7. $\mathbf{P}(x)$ and $\mathbf{Q}(x)$ are members of the ring of polynomial matrices.

None of the above seem to provide a route to proving the conjecture. I have also looked for potential contradictions. I believe I can show it is true for $N=3$ and the algebra for $N=4$ is ... intimidating. Any suggestions for other ways to proceed to either prove or falsify this conjecture?

[Note that we could also restrict $\mathbf{B}$ to have distinct eigenvalues if that would help.]

Answer 1

Edit: Based on some Sage computations for $N=4$, I believe that a much stronger result is true:

Theorem. Let $A, B, C$ be real matrices of size $N\times N$. Assume that $A$ is diagonal with distinct eigenvalues, and that $B$, $C$ are symmetric. Then the following are equivalent:

1. $\det(A + xB - y) = \det(A + xC - y)$ in $\mathbb R[x,y]$;
2. there exists an invertible diagonal matrix $R$ such that $B = R^{-1}CR$.

While it is easy to see that $(2)$ implies $(1)$, I don't know how to prove the reverse. A possible method would be to show that there exists an invertible matrix $R$ such that $RB = CR$ and $RA = AR$. Indeed, since $A$ has distinct eigenvalues, any matrix that commutes with it must be diagonal.

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Previous answer:
Here is a collection of observations.

Let $b_i$, $c_i$ be the diagonal entries of $B$, $C$ respectively, and let \begin{align} p(x,y) &:= \det(A - y + xB) = \det(A - y + xC) \end{align} be the characteristic polynomial of $P$ and $Q$. We will consider the coefficient of $x$ in $p(x,y)$, denoted $[x]p(x,y) \in \mathbb R[y]$. It is not hard to see that $$ [x]p(x,y) = \sum_{i = 1}^n b_i\prod_{j\ne i}(a_j-y) = \sum_{i = 1}^n c_i\prod_{j\ne i}(a_j-y). $$ By substituting $y = a_i$, since the $a_i$ are pairwise distinct, all summands but one vanish, and we obtain $$\boxed{\forall i = 1,\dots,n,\, b_i = c_i.}\tag{$1$}$$

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Since $B\sim C$, we also have $$ \lVert B \rVert_2^2 = \operatorname{tr}(B^T B)=\operatorname{tr}(B^2) =\operatorname{tr}(C^2)=\operatorname{tr}(C^T C)=\lVert C \rVert_2^2, \tag{$2$} $$ where $\lVert \cdot \rVert_2$ is the Frobenius (Euclidean) norm. Hence, by $(1)$, if B is diagonal then $C$ must also be diagonal, proving the conjecture. I do not know if this argument generalizes easily. The following is an attempt at constraining the off-diagonal entries.

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Consider the coefficient $[x^2]p(x,y)$. We have $$ [x^2]p(x,y) = \sum_{i<j} (b_ib_j - b_{ij}^2)\prod_{k\not \in \{i,j\}}(a_k-y) = \sum_{i<j} (c_ic_j - c_{ij}^2)\prod_{k\not \in \{i,j\}}(a_k-y). $$ Moreover, we can simplify the $b_ib_j$ and $c_ic_j$ terms, obtaining $$ \sum_{i<j} b_{ij}^2 \prod_{k\not \in \{i,j\}}(a_k-y) = \sum_{i<j} c_{ij}^2 \prod_{k\not \in \{i,j\}}(a_k-y). $$ Dividing by $\prod_{k=1}^n (a_k-y)$, we get the identity $$ \sum_{i<j} \frac{b_{ij}^2}{(a_i-y)(a_j-y)} = \sum_{i<j} \frac{c_{ij}^2}{(a_i-y)(a_j-y)}. \tag{$3$} $$ Unfortunately, as $y$ varies, this only appears to give a constraint of codimension $n-1$ on the values of $b_{ij}^2-c_{ij}^2$.

Answer 2

This is by no means a formal proof but here are some ideas for proving the problem:

My opinion is that the crucial condition is that $\mathbf{A}$ is a diagonal matrix with distinct eigenvalues, or that it has a 1D eigenspaces.

We can try proof by contradiction by expressing $\mathbf{Q}(x)$ (where $\mathbf C_{12}\neq\mathbf0$):

$$\mathbf Q(x) = \begin{pmatrix} \mathbf A_1 + x \mathbf C_{11} & x\mathbf C_{12}\\ x\mathbf C_{12}^⊤ & \mathbf A_2 + x \mathbf C_{22} \end{pmatrix}$$

As we can see the $x\mathbf C_{12}$ terms can cause coupling between the eigenspaces of $\mathbf A_1$ and $\mathbf A_2$, causing eigenvectors of $\mathbf Q(x)$ to be a linear combination of that of $\mathbf A_1$ and $\mathbf A_2$. When compared to $\mathbf P(x)$: $$\mathbf P(x) = \begin{pmatrix} \mathbf A_1 + x \mathbf B_{11} & \mathbf 0\\ \mathbf 0 & \mathbf A_2 + x \mathbf B_{22} \end{pmatrix}$$

$$\det(\mathbf P(x)-y\mathbf I) = \det(\mathbf A_1 + x \mathbf B_{11}-y\mathbf I_m)\cdot \det(\mathbf A_2 + x \mathbf B_{22}-y\mathbf I_{n-m})$$

, the eigenvectors of $\mathbf A_1$ and $\mathbf A_2$ are not mixed together. This indicate that the characteristic polynomial of $\mathbf Q(x)$ cannot be equal to $\mathbf P(x)$ unless $\mathbf C_{12} = \mathbf 0$ which will complete the proof.

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A second approach is to try diagonalising $\mathbf C$:

$$\mathbf C = \mathbf U^⊤\mathbf\Lambda\mathbf U$$

, where $\mathbf \Lambda$ is some diagonal matrix.

We can then transform $\mathbf Q(x)$:

$$\mathbf U^⊤\mathbf Q(x)\mathbf U=\mathbf U^⊤\mathbf A\mathbf U +x\mathbf\Lambda$$

Since we know that $\mathbf{A}$ is a diagonal matrix with distinct eigenvalues, $\mathbf U^⊤\mathbf A\mathbf U$ cannot be block diagonal unless $\mathbf U$ is block diagonal.

This again causes coupling between eigenspaces of $\mathbf A_1$ and $\mathbf A_2$ so it would require a similar proof as above.

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Another attempt could be using the Schur Complement:

$$\det(\mathbf Q(x) - y\mathbf I) = \det(\mathbf D_1)\cdot\det(\mathbf D_2 - x^2\mathbf C_{12}^⊤\mathbf D_1^{-1}\mathbf C_{12})$$

, where:

$$\begin{equation}\begin{cases} \mathbf D_1 = \mathbf A_1 + x\mathbf C_11-y\mathbf I_m\\ \mathbf D_2 = \mathbf A_2 + x\mathbf C_22-y\mathbf I_{n-m} \end{cases}\end{equation}$$

As there is no $x^2$ dependent terms within the characteristic polynomial of $\mathbf P(x)$, we would expect $\mathbf C_{12}^⊤\mathbf D_1^{-1}\mathbf C_{12} = \mathbf 0$.

Then we can perhaps use some quadratic form $\mathcal Q (\mathbf v)$, with some $\mathbf v \in \mathbb R^{n-m}$:

$$\mathcal Q (\mathbf v) = (\mathbf C_{12} \mathbf v)^⊤ \mathbf D_1^{-1}(\mathbf C_{12} \mathbf v)$$

If we can prove that above always equal to $0$ for all $\mathbf v$, then we can prove that $\mathbf C_{12} = \mathbf 0$ and hence proving the original conjecture.

Answer 3

No, $C$ is not necessarily block-diagonal. A counterexample can be obtained by slightly modifying A. A. Albert’s example that you have mentioned in your self-answered question. Let $$ A=\pmatrix{1\\ &-1\\ &&0},\quad B=\pmatrix{0&1\\ 1&0\\ &&0},\quad C=\pmatrix{0&0&\frac{1}{\sqrt{2}}\\ 0&0&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0}. $$ Then $B$ is block-diagonal, $C$ is not (as it is irreducible) and $$ \det(A+xB-tI)=\det(A+xC-tI) =-t(t^2-x^2-1). $$ (This also serves as a counterexample to your aforementioned self-answered question.)