0

Let $V$ be the set of all symmetric $n\times n$ matrices over $\Bbb C$ $(A'=A)$,and $U$ be the set of all antisymmetric $n\times n$ matrices over $\Bbb C$ $(A'=-A)$. For $A \in V$, define $$ \varphi : V \rightarrow U, X \mapsto {AX} - {XA}. $$ Prove that $\varphi$ is surjective if and only if the characteristic function of $A$ has no multiple root in $\mathbb{C}$.

$\Leftarrow$: It is easy, since $A$ is diagonalizable, invertible $P$ exists such that $P^{-1}AP=diag(a_1,\cdots,a_n)$. Let $Y=P^{-1}XP, C=P^{-1}BP$ for any $B\in U$, $y_{ii}=0, y_{ij}=\frac{c_{ij}}{a_i-a_j}$ then $X=PYP^{-1}$ satisfies $AX-XA=B$.

$\Rightarrow$: Can we use the $\Leftarrow$ part? We do not know the diagonalizability of $A$. The symmetry of $A$, how to use?

jjagmath
  • 22,582
xldd
  • 3,785
  • Title: No, it is not surjective (take $A=0$). What is the dot in "$XA.$ is surjective"? – Dietrich Burde Jan 24 '25 at 11:37
  • Do you mean hermitian or symmetric over $\mathbb C$? – Christophe Boilley Jan 24 '25 at 12:18
  • more generally this problem statement is falsified by any defective $A$ with a degree $n$ minimal polynomial. In such a case https://math.stackexchange.com/questions/57308/commutation-when-minimal-and-characteristic-polynomial-agree/ tells you $\dim \ker \varphi =n$ so $\varphi$ is surjective by rank-nullity – user8675309 Jan 24 '25 at 21:42

2 Answers2

1

[EDIT:Assuming we replace $\mathbb C$ by $\mathbb R$, as the question uses the word "symmetric" and not "hermitian".]

For all $X\in V$, the matrix $P^{-1}\phi(X)P$ has null coefficients over each block corresponding to eigenspaces. So if such an eigenspace has multiplicity greater than 1, $\mathrm{rk}(\mathrm{Im}(\phi))<\frac{n(n-1)}2$.

Christophe Boilley
  • 8,996
1

The problem statement is False.

Consider nilpotent $A:= \left[\begin{matrix}1 & i\\i & -1\end{matrix}\right]$ and $A^*\mapsto AA^*-A^*A= \left[\begin{matrix}0 & - 2 i\\2 i & 0\end{matrix}\right]$ and $\dim U=1$ so the mapping $T(X)=AX-XA$ is surjective.

A possible fix would be to change any mention of $\mathbb C$ to $\mathbb R$; then the problem reduces (via rank-nullity) to checking that $\varphi$ cannot be injective when $A$ has repeated eigenvalues. Let $\big\{\mathbf u_i\big\}$ be an orthonormal set of eigenvectors for $A$, then $T(\mathbf u_i\mathbf u_i^T\big) = \mathbf 0$ so $\dim \ker T\geq n$; if $A\mathbf u_1 = \lambda \mathbf u_1$ and $A\mathbf u_2 = \lambda \mathbf u_2$ then $\big(\mathbf u_1\mathbf u_2^T+\mathbf u_2\mathbf u_1^T\big) \in \ker T$ and
$\mathbf 0=\alpha\big(\mathbf u_1\mathbf u_2^T+\mathbf u_2\mathbf u_1^T\big)+\sum_{i=1}^n \alpha_i\mathbf u_i\mathbf u_i^T$
$\implies$ all coefficients are zero so $\ker T \gt n$

user8675309
  • 12,193
  • Note this problem cannot be fixed by changing mention from complex (anti) symmetric to complex (anti) Hermitian matrices as vector spaces (with real scalars) since $T(I) = 0 = T(0)=T(A)$ so $T$ is not injective hence not surjective (by rank-nullity) when the domain and co-domain have the same dimension. – user8675309 Jan 24 '25 at 18:29