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Let $X$ be a vector spaces, $X_1,...,X_n$ -its linear subspaces such that $X=X_1\bigoplus...\bigoplus X_n$; let $a_1,...,a_n$ be scalars. We put

$T(x_1+...+x_n)=a_1x_1+...+a_nx_n$ for $x_1\in X_1, ...,x_n\in X_n$.

What is general form of an invariant subspace of operator $T$? Is it $V:=V_1 +...+V_n$, where $V_1$ is a subspace of $X_1$,..,$V_n$ is a subspace of $X_n$?

Thanks

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Your guess is right (assuming that $a_1,\ldots,a_n$ are pairwise distinct). Indeed, let $U\subset X$ be an invariant subspace and assume $u=x_1+\ldots+x_n$ belongs to $U$, for some $x_i\in X_i$. Then, $T(u)$, $T(T(u))$, $...$ also belong to $U$, so that $a_1^kx_1+\ldots+a_n^kx_n\in U$, for all $k$. Since the matrix $(v_{pq})=(a_p^{q-1})$, where $p,q\in\{1,\ldots,n\}$, is invertible, we conclude that $x_i\in U$.

Therefore, denoting by $U_1$ the set of all $x_1\in X_1$ such that $x_1+y_2\ldots+y_n\in U$, for some $y_j\in X_j$, we get $U=U_1+\ldots+U_n$.

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  • $\begingroup$ Why for example $x_1\in U$? $\endgroup$ – Richard Oct 11 '14 at 15:51
  • $\begingroup$ I added the explanation to the answer. $\endgroup$ – user2097 Oct 11 '14 at 17:07
  • $\begingroup$ Thanks for the explanation. $\endgroup$ – Richard Oct 11 '14 at 18:14

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