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本帖最后由 hbghlyj 于 2022-2-10 22:55 编辑 (如果不限制"对称"的话,立方根不唯一,例如,单位矩阵的立方根,可以是旋转角为$±\frac{2π}3$的旋转矩阵)
math.stackexchange.com/questions/2326596/uniqueness-of-symmetric ... eal-symmetric-matrix
(user1511的回答)
这不仅适用于实对称三次根，而且适用于所有实对称奇数$n$次根时也是如此。通过改变正交基，我们可以假设$A$是一个实对角矩阵。由于 $n$ 是奇数，因此实数具有唯一的实$n$次根。因此，$A$的每个实对称$n$次根都可以正交对角化为$QA^{1/n}Q^T$，其中$Q$是某个实正交矩阵，$A^{1/n}$是对角矩阵$A$的$n$次根。根据定义，这个$n$次根的$n$次幂必为$A$。所以，我们有 $A=(QA^{1/n}Q^T)^n=QAQ^T$。 令 $\lambda_1,\ldots,\lambda_k$ 为 $A$ 的不同特征值，$f$ 为拉格朗日插值多项式，使得 $f(\lambda_i)=\lambda_i^{1/n}$ 对于每个 $i$。那么$f(A)=A^{1/n}$（因为$A$是对角矩阵）和$QA^{1/n}Q^T=Qf(A)Q^T=f(QAQ^T) =f(A)$。因此，$A$ 的实对称$n$次根是唯一的，因为它必须等于 $f(A)$。
(hardmath的回答)
我在[Luce and Perry], "A Method of Matrix Analysis of Group Structure," (心理测量学,卷14, No. 1, 1949年3月)中发现了这个陈述和其证明:
Theorem $7$: If $n$ is a positive odd integer and $S$ a real symmetric matrix, then there is one and only one real symmetric $n^{th}$ root of $S$.
结果是他们的论文事后才想到的，它只需要立方根情况（正如这里的问题所要求的那样）。 但只要给出细节，我们不妨把所有正奇数$n$的情况都处理一下。
(这里附件限制是1M 只好把它分成两份:1-10页
1-10.pdf
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,11-22页
11-22.pdf
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)
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 $S$. |
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业余的业余
发表于 2022-2-10 22:44
