Prove that there exists an $m \times m$ matrix $X$ such that $AX-XA=C.$
Let $A$ be a real symmetric $m \times m$ matrix with $m$ distinct eigenvalues and $v_1,v_2, \cdots , v_m$ be the corresponding eigenvectors. Let $C$ be an $m \times m$ matrix such that $\left \langle Cv_j,v_j \right \rangle = 0$ for all $1 \leq j \leq m.$ Prove that there exists an $m \times m$ matrix $X$ such that $AX-XA=C.$
How do I prove that? Any help will be highly appreciated.
Thank you very much.
$\endgroup$ 63 Answers
$\begingroup$Take the equation$$ AX-XA=C $$Pick $v_i, v_j$ and calculate the following:$$ \langle v_i,(AX-XA)v_j\rangle=\langle v_i,AXv_j\rangle-\langle v_i,XAv_j\rangle $$Since $A$ is symmetric, we can move it to $v_i$ in the first term. So we get:$$ \langle v_i,(AX-XA)v_j\rangle=(\lambda_i-\lambda_j)\langle v_i,Xv_j\rangle $$Motivated by this, we define $X$ by$$ \langle v_i,Xv_j\rangle=\frac{1}{\lambda_i-\lambda_j}\langle v_i,Cv_j\rangle $$if $i\neq j$ and $0$ otherwise. This makes $X$ well-defined since $\langle v_i,Xv_j\rangle$ are its components in an orthogonal basis. Now we have:$$ \langle v_i,(AX-XA)v_j\rangle=(\lambda_i-\lambda_j)\langle v_i,Xv_j\rangle=\langle v_i,Cv_j\rangle $$Since $\langle v_i,(AX-XA)v_j\rangle$ are just the components of $AX-XA$ in some orthogonal basis and they are equal to the components of $C$ in this same basis, $AX-XA = C$.
$\endgroup$ $\begingroup$In the basis $v_k$ we see that $A$ is diagonal with diagonal elements $\lambda_k$. In this basis, we see that $[C]_{kk} = 0$.
Consider the operator $L(X) = AX-XA$. It is not hard to see (in the basis $v_i v_j^T$) that$[L(X)]_{ij} = (\lambda_i - \lambda_j)[X]_{ij}$ and hence${\cal R} L = \{ B | B \text{ has zero diagonal} \}$.
Returning to the original question, we see that $C \in {\cal R}L$ iff $C$ has a zero diagonal in the basis $v_k$.
$\endgroup$ 2 $\begingroup$Hint: We may assume $v_i$ has unit length so $v_1,\dots,v_m$ is an orthonormal basis. With respect to this orthonormal basis we need to solve $DY-YD=B$ for $Y$ where $B$ has all zeros on the diagonal and $D$ is diagonal with distinct entries. Now write out the entries of $DY-YD$ to see you can solve this for $Y$.
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