Uniqueness of symmetric cube root of real symmetric matrix

Question

I managed to prove that if $A$ is a real symmetric matrix, then there exists a real symmetric matrix $B$ such that $B^3=A$. But I am not sure wether such $B$ is unique.

Summary of my proof:Let $A \in M_n(\Bbb R)$ such that $A^t = A.$ Using the spectral theorem, there is an orthogonal matrix $P$ such that $P^{-1}AP = D$ (diagonal matrix). Then $(D')^3$ = $D$, and $B = PD'P^{-1}$ satisfies the condition. I could take a different orthogonal matrix $Q$ such that $Q^{-1}AQ=D_1$, then $(D'_1)^3=D_1$, and $B_1=QD_1Q^{-1}$. It seems like $B=B_1$, but how can i justify this?

Answer 1

As modified, the claim is true. I found the statement and its proof in Luce and Perry, "A Method of Matrix Analysis of Group Structure," (Psychometrika, v. 14, No. 1, March 1949):

Theorem $7$: If $n$ is a positive odd integer and $S$ a real symmetric matnix, then there is one and only one real symmetric $n^{th}$ root of $S$.

The result is something of an afterthought to their paper, which needs only the cube root case (as called for in the Question here). But as long as the details are to be given, we may as well treat all odd integer $n\gt 0$ cases.

The existence half of their proof is predictable, based on diagonalizing $m\times m$ real symmetric $S$ and taking real $n^{th}$ roots of the resulting diagonal entries (as the Asker is presumed to have done here for the cube root). That is, if $S = PDP^T$ for diagonal $D$ and orthogonal $P$, then obtain $R$ from $D$ by extracting the $n^{th}$ root of its diagonal entries. Now $B = PRP^T$ is real symmetric and satisfies:

$$ B^n = (PRP^T)^n = PR^nP^T = PDP^T = S $$

For specificity we can impose that the $m$ diagonal entries of $D$ are ordered ascendingly:

$$ d_{11} \le d_{22} \le \ldots \le d_{mm} $$

By monotonicity of the $n^{th}$ root function the diagonal entries of $R$ are in corresponding order.

Suppose there potentially exists $C \neq B$ which is also an $n^{th}$ root of $S$, $C^n = S$. Then there exists an orthogonal matrix $Q$ such that $Q^T C Q = T$ is diagonal, with the same ascending order of diagonal entries as before. Now:

$$ T^n = Q^T C^n Q = (Q^T P) R^n (P^T Q) = (Q^T P) D (P^T Q) $$

Since $T^n$ and $D$ are (orthogonally) similar, they share the same characteristic roots, and because the diagonal entries of $T$ are ascending, so too are the entries of the diagonal for $T^n$. Thus $T^n = D$.

But also by construction $R^n = D$, and because the $n^{th}$ root is strictly monotone ($1-1$), from $R^n = T^n$ we infer $R=T$.

Let $U = P^T Q$, and rewrite the orthogonal similarity above for $T^n = D$:

$$ D = U^T D U $$

In other words the diagonal matrix $D$ commutes with the orthogonal matrix $U$, $UD = DU$. From the definition of matrix multiplication this amounts to:

$$ u_{ik} d_{kk} = d_{ii} u_{ik} $$

for all $1\le i,k \le m$, since all the off-diagonal entries of $D$ are zeros. In other words:

$$ u_{ik}[d_{kk} - d_{ii}] = 0 $$

It follows (once again using that taking $n^{th}$ roots is injective) that:

$$ u_{ik}[d_{kk}^{1/n} - d_{ii}^{1/n}] = 0 $$

That is, $UR = RU$. Recalling $T=R$ and $U = P^T Q$, we have:

$$ P^T QT = R P^T Q $$

$$ Q T Q^T = P R P^T $$

$$ C = B $$

This contradicts the choice of $C \neq B$ and shows that $B$ is uniquely the real symmetric $n^{th}$ root of $S$.

Answer 2

This is true not only for real symmetric cubic roots, but for all real symmetric $n$-th roots as well when $n\ge1$ is odd.

By a change of orthonormal basis, we may assume that $A$ is a real diagonal matrix. Since $n$ is odd, real numbers have unique real $n$-th roots. Hence every real symmetric $n$-th root of $A$ can be orthogonally diagonalised as $QA^{1/n}Q^T$, where $Q$ is some real orthogonal matrix and $A^{1/n}$ is the entrywise $n$-th root of the diagonal matrix $A$. By definition, the $n$-th power of this $n$-th root must give us back $A$. So, we have $A=(QA^{1/n}Q^T)^n=QAQ^T$.

Let $\lambda_1,\ldots,\lambda_k$ be the distinct eigenvalues of $A$ and $f$ be the Lagrange interpolating polynomial such that $f(\lambda_i)=\lambda_i^{1/n}$ for each $i$. Then $f(A)=A^{1/n}$ (because $A$ is diagonal) and $QA^{1/n}Q^T=Qf(A)Q^T=f(QAQ^T)=f(A)$. Thus, the real symmetric $n$-th root of $A$ is unique because it must be equal to $f(A)$.