Let $A, B \in M_{n}(\mathbb{C})$. Suppose that all of the eigenvalues of $A$ and $B$ are positive real numbers. If $A^{4}=B^{4}$, prove that $A=B$.

Question

I'm stumped at the last part of this problem. Could anyone advise, please? Thank you.

My attempt: Note that $A^{4}$ and $B^{4}$ share same eigenvalues and eigenvectors, let $A^{4}v_{i}=B^{4}v_{i}=\lambda_{i}^{4}v_{i}$ for $i=1,2,...,n$. Since $\lambda_{i}$ are positive and real, we have $Av_{i}=Bv_{i}=\lambda_{i}v_{i}$, so $(A-B)v_{i}=0$. However, it doesn't say the matrices are symmetric, so $\{v_{1},...,v_{n}\}$ cannot form a basis of $\mathbb{C}^{n}$, i.e., it may not imply $A-B$ are zero matrix.

Answer 1

Using the hint in the comments, it suffices to consider the case when $A^2=B^2$.

Consider the linear map $T:M_n\big(\mathbb C\big)\longrightarrow M_n\big(\mathbb C\big)$ given by
$T\big(v\big) = A^{-1}vB$ and in particular consider $w:=A^{-1}B -I_n$
$\implies T(w) = A^{-1}\big(A^{-1}B -I_n\big)B=A^{-2}B^2-A^{-1}B=I_n-A^{-1}B=-w$
$\implies A =B$, because $w$ must be zero since $T$ has all positive eigenvalues (this can be verified analytically but algebraically we can use the isomorphism $\text{vec}: M_n\big(\mathbb C\big)\longrightarrow \mathbb C^{n^2}$ and evaluate the spectrum of $T':=\text{vec}\circ T\circ\text{vec}^{-1}$ also given by $T'\Big(\text{vec}\big(v\big)\Big)= \big(B^T\otimes A^{-1}\big)\text{vec}\big(v\big)$ where $\otimes$ denotes the Kronecker product)

generalizing
in general when $A^k = B^k$ for some $k\in \mathbb N$ and they both have only positive eigenvalues, then $w:=I_n$ $\underbrace{T\circ T\circ \ldots \circ T}_{T \text{ occurs k-times}}(w)=T^k(w)=w \implies T w=w$
Thus $A^{-1}B= T(w)=T(I_n)=I_n\implies B=A$

For avoidance of doubt, Kronecker products notation can help to make the above explicit, i.e. it says
$\big(B^T\otimes A^{-1}\big)^k\text{vec}(w)=\text{vec}(w)\implies \prod_{r=1}^k\big(B^T\otimes A^{-1} -\omega^r I_{n^2}\big)\text{vec}(w)=\mathbf 0$
where $\omega$ is a generating $k$th root of unity
$\implies \big(B^T\otimes A^{-1} -I_{n^2}\big)\text{vec}(w)=\mathbf 0$ since $\big(B^T\otimes A^{-1} -\omega^r I_{n^2}\big)$ has non-zero determinant for $r\lt k$. That is, $T'\big(\text{vec}(w)\big) =\text{vec}(w)\iff T(w)=w$

Answer 2

Let $M=A^4=B^4$. Then $M$ has a positive spectrum and $f:z\mapsto z^{1/4}$ is analytic in some domain containing the spectrum of $M$. Therefore $f$ is “defined on the spectrum of $M$” (in the sense of definition 1.1 in Higham, Functions of Matrices) and $M$ has a matrix quartic root $C$ with a positive spectrum such that $C=p(M)$ for some Hermite interpolation polynomial $p$ (see definition 1.4 in Higham).

As $A^4=C^4$, the multi-set of the fourth powers of the Jordan blocks in the Jordan form of $A$ is equal to the multi-set of the fourth powers of the fourth powers of the Jordan blocks in the Jordan form of $C$. Clearly, the fourth powers of two Jordan blocks for positive eigenvalues are equal only if the two Jordan blocks are equal. Therefore the Jordan forms of $A$ and $C$ must be equal, i.e., $A$ and $C$ are similar. Let $A=V^{-1}CV$. Then $$ V^{-1}MV=V^{-1}C^4V=(V^{-1}CV)^4=A^4=M. $$ Hence $$ A=V^{-1}CV=V^{-1}p(M)V=p(V^{-1}MV)=p(M)=C. $$ By a similar argument, we also have $B=C$. Therefore $A=B$.

Answer 3

Too long for a comment, I just want to elaborate the last step of @user8675309's excellent answer (+1), namely how to argue $w = 0$ from the equation $(T + I_{(n)})w = A^{-1}wB + w = 0$, as this involves several interesting properties of the Kronecker product and the vectorization operator.

Applying the vectorization operator $\operatorname{vec}$ on both sides of $A^{-1}wB + w = 0$ yields (where we used the property $\operatorname{vec}(ABC) = (C^\top \otimes A)\operatorname{vec}(B)$): \begin{align} (B^\top \otimes A^{-1} + I_{(n^2)})\operatorname{vec}(w) = 0. \end{align} By condition, let $\lambda_1, \ldots, \lambda_n$ be all the positive eigenvalues of $A$ and $\mu_1, \ldots, \mu_n$ be all the positive eigenvalues of $B$, then $\{\lambda_i^{-1}\mu_j, 1 \leq i , j \leq n\}$ are eigenvalues of the order $n^2$ matrix $B^\top \otimes A^{-1}$, which are obviously all positive. Consequently, the eigenvalues of $B^\top \otimes A^{-1} + I_{(n^2)}$ are $1 + \lambda_i^{-1}\mu_j > 0$, $1 \leq i, j \leq n$, implying that the matrix $B^\top \otimes A^{-1} + I_{(n^2)}$ is invertible, whence $\operatorname{vec}(w) = 0$, i.e., $w = 0$.

Answer 4

Let $n\in\mathbb{N}$ and $A$, $B\in M_n(\mathbf{C})$ such that all eigenvalues of $A$ and $B$ are positive. Then for any $m\in\mathbb{N}$, $A^m=B^m$ implies $A=B$.

Proof. Since $A^m$ and $B^m$ have the same Jordan form, and all eigenvalues of $A$ and $B$ are positive, we can conclude that $A$ and $B$ have the same Jordan form (in fact, since all eigenvalues of $A$ are nonzero, the Jordan form of $A$ and $A^m$ have the same sizes of Jordan blocks; see here). Hence there is an invertible $P\in M_n(\mathbf{C})$ such that $A=P^{-1}BP$.

Denote $C(B)$ and $C(B^m)$ the centrailizers of $B$ and $B^m$, respectively. Claerly $C(B)\subseteq C(B^m)$. Since $B$ and $B^m$ have the same sizes of jordan blocks, $C(B)$ and $C(B^m)$ have the same dimensions (in fact, the dimension of the centralizer of a matrix depends only on the sizes of Jordan blocks of its Jordan form; see here). Hence $C(B)=C(B^m)$. Since $B^m=A^m=P^{-1}B^mP$, we see that $P\in C(B^m)$ and thus $P\in C(B)$. Therefore $A=P^{-1}BP=B$, as desired.