Decomposition of invariant subspaces
$begingroup$
Consider a set of $d$ linearly independent generalized eigenvectors of some matrix $A in mathbb{C}^{d times d}$. Suppose this set is decomposed into $M$ distinct Jordan chains and the $m$-th Jordan chain has $N(m)$ vectors in it.
Consider $A$-invariant sets of the form
$$
U_n^m=operatorname{span}(x_1^m, dots , x^m_n),
$$
where $m in {1, dots, M}$, $n in {1, dots, N(m)}$ and $x^m_n$ is the $n$-th vector in the $m$-th Jordan chain.
Question. Is it true that any nontrivial $A$-invariant subspace $U$ of $mathbb{C}^d$ can be decomposed into a direct sum of such sets?
$$
AU subset U quad overset{?}{Leftrightarrow} quad U= bigoplus_{ overset{Large m, n}{rm small (some)} } U^m_n
$$
linear-algebra matrices invariant-subspace generalizedeigenvector
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Consider a set of $d$ linearly independent generalized eigenvectors of some matrix $A in mathbb{C}^{d times d}$. Suppose this set is decomposed into $M$ distinct Jordan chains and the $m$-th Jordan chain has $N(m)$ vectors in it.
Consider $A$-invariant sets of the form
$$
U_n^m=operatorname{span}(x_1^m, dots , x^m_n),
$$
where $m in {1, dots, M}$, $n in {1, dots, N(m)}$ and $x^m_n$ is the $n$-th vector in the $m$-th Jordan chain.
Question. Is it true that any nontrivial $A$-invariant subspace $U$ of $mathbb{C}^d$ can be decomposed into a direct sum of such sets?
$$
AU subset U quad overset{?}{Leftrightarrow} quad U= bigoplus_{ overset{Large m, n}{rm small (some)} } U^m_n
$$
linear-algebra matrices invariant-subspace generalizedeigenvector
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
$endgroup$
add a comment |
$begingroup$
Consider a set of $d$ linearly independent generalized eigenvectors of some matrix $A in mathbb{C}^{d times d}$. Suppose this set is decomposed into $M$ distinct Jordan chains and the $m$-th Jordan chain has $N(m)$ vectors in it.
Consider $A$-invariant sets of the form
$$
U_n^m=operatorname{span}(x_1^m, dots , x^m_n),
$$
where $m in {1, dots, M}$, $n in {1, dots, N(m)}$ and $x^m_n$ is the $n$-th vector in the $m$-th Jordan chain.
Question. Is it true that any nontrivial $A$-invariant subspace $U$ of $mathbb{C}^d$ can be decomposed into a direct sum of such sets?
$$
AU subset U quad overset{?}{Leftrightarrow} quad U= bigoplus_{ overset{Large m, n}{rm small (some)} } U^m_n
$$
linear-algebra matrices invariant-subspace generalizedeigenvector
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
$endgroup$
Consider a set of $d$ linearly independent generalized eigenvectors of some matrix $A in mathbb{C}^{d times d}$. Suppose this set is decomposed into $M$ distinct Jordan chains and the $m$-th Jordan chain has $N(m)$ vectors in it.
Consider $A$-invariant sets of the form
$$
U_n^m=operatorname{span}(x_1^m, dots , x^m_n),
$$
where $m in {1, dots, M}$, $n in {1, dots, N(m)}$ and $x^m_n$ is the $n$-th vector in the $m$-th Jordan chain.
Question. Is it true that any nontrivial $A$-invariant subspace $U$ of $mathbb{C}^d$ can be decomposed into a direct sum of such sets?
$$
AU subset U quad overset{?}{Leftrightarrow} quad U= bigoplus_{ overset{Large m, n}{rm small (some)} } U^m_n
$$
linear-algebra matrices invariant-subspace generalizedeigenvector
linear-algebra matrices invariant-subspace generalizedeigenvector
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
asked Dec 24 '18 at 21:15
ZeeklessZeekless
577111
577111
add a comment |
add a comment |
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