Union of Jordan chains
$begingroup$
Let $mathcal{V}_i=mathcal{R}(A^i)capmathcal{N}(A)$, where $A$ is a nilpotent matrix of order $k$ in $C^{ntimes n}$ ($mathcal{R}(cdot),mathcal{N}(cdot)$ are the image and null spaces of a matrix, respectively). Define disjoint sets $mathcal{S}_0,mathcal{S}_1,cdots,mathcal{S}_{k-1}$ such that for $i=0,cdots,k-1$, $mathcal{S}_icupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_i$. In particular, $mathcal{S}=mathcal{S}_0cupmathcal{S}_1cupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_0=mathcal{N}(A)$.
Then for $i =1,cdots,k-1$ and any $sinmathcal{S}_i$, if $v$ is such that $A^iv=s$, we call $mathcal{J}_s={A^iv,cdots,Av,v}$ a Jordan chain for $A$.
Enough with definitions: my question is what is, really, $bigcup_{sinmathcal{S}}mathcal{J}_s$? Let ${s_1,s_2,cdots,s_r}$ be a basis for $mathcal{S}$. Are we able to write
$$bigcup_{sinmathcal{S}}mathcal{J}_s={A^rv_1,A^{r-1}v_1,cdots,Av_1,v_1}cup{A^rv_2,cdots,Av_2,v_2}cupcdotscup{A^rv_r,cdots,Av_r,v_r}$$
where $s_i=A^rv_i$ for $i=1,cdots,r$.
It seems like that would be the sensible thing. I'm essentially having a hard time figuring out what the union of the Jordan chains is.
linear-algebra matrices
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
$endgroup$
add a comment |
$begingroup$
Let $mathcal{V}_i=mathcal{R}(A^i)capmathcal{N}(A)$, where $A$ is a nilpotent matrix of order $k$ in $C^{ntimes n}$ ($mathcal{R}(cdot),mathcal{N}(cdot)$ are the image and null spaces of a matrix, respectively). Define disjoint sets $mathcal{S}_0,mathcal{S}_1,cdots,mathcal{S}_{k-1}$ such that for $i=0,cdots,k-1$, $mathcal{S}_icupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_i$. In particular, $mathcal{S}=mathcal{S}_0cupmathcal{S}_1cupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_0=mathcal{N}(A)$.
Then for $i =1,cdots,k-1$ and any $sinmathcal{S}_i$, if $v$ is such that $A^iv=s$, we call $mathcal{J}_s={A^iv,cdots,Av,v}$ a Jordan chain for $A$.
Enough with definitions: my question is what is, really, $bigcup_{sinmathcal{S}}mathcal{J}_s$? Let ${s_1,s_2,cdots,s_r}$ be a basis for $mathcal{S}$. Are we able to write
$$bigcup_{sinmathcal{S}}mathcal{J}_s={A^rv_1,A^{r-1}v_1,cdots,Av_1,v_1}cup{A^rv_2,cdots,Av_2,v_2}cupcdotscup{A^rv_r,cdots,Av_r,v_r}$$
where $s_i=A^rv_i$ for $i=1,cdots,r$.
It seems like that would be the sensible thing. I'm essentially having a hard time figuring out what the union of the Jordan chains is.
linear-algebra matrices
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
$endgroup$
add a comment |
$begingroup$
Let $mathcal{V}_i=mathcal{R}(A^i)capmathcal{N}(A)$, where $A$ is a nilpotent matrix of order $k$ in $C^{ntimes n}$ ($mathcal{R}(cdot),mathcal{N}(cdot)$ are the image and null spaces of a matrix, respectively). Define disjoint sets $mathcal{S}_0,mathcal{S}_1,cdots,mathcal{S}_{k-1}$ such that for $i=0,cdots,k-1$, $mathcal{S}_icupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_i$. In particular, $mathcal{S}=mathcal{S}_0cupmathcal{S}_1cupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_0=mathcal{N}(A)$.
Then for $i =1,cdots,k-1$ and any $sinmathcal{S}_i$, if $v$ is such that $A^iv=s$, we call $mathcal{J}_s={A^iv,cdots,Av,v}$ a Jordan chain for $A$.
Enough with definitions: my question is what is, really, $bigcup_{sinmathcal{S}}mathcal{J}_s$? Let ${s_1,s_2,cdots,s_r}$ be a basis for $mathcal{S}$. Are we able to write
$$bigcup_{sinmathcal{S}}mathcal{J}_s={A^rv_1,A^{r-1}v_1,cdots,Av_1,v_1}cup{A^rv_2,cdots,Av_2,v_2}cupcdotscup{A^rv_r,cdots,Av_r,v_r}$$
where $s_i=A^rv_i$ for $i=1,cdots,r$.
It seems like that would be the sensible thing. I'm essentially having a hard time figuring out what the union of the Jordan chains is.
linear-algebra matrices
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
$endgroup$
Let $mathcal{V}_i=mathcal{R}(A^i)capmathcal{N}(A)$, where $A$ is a nilpotent matrix of order $k$ in $C^{ntimes n}$ ($mathcal{R}(cdot),mathcal{N}(cdot)$ are the image and null spaces of a matrix, respectively). Define disjoint sets $mathcal{S}_0,mathcal{S}_1,cdots,mathcal{S}_{k-1}$ such that for $i=0,cdots,k-1$, $mathcal{S}_icupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_i$. In particular, $mathcal{S}=mathcal{S}_0cupmathcal{S}_1cupcdotscupmathcal{S}_{k-1}$ is a basis of $mathcal{V}_0=mathcal{N}(A)$.
Then for $i =1,cdots,k-1$ and any $sinmathcal{S}_i$, if $v$ is such that $A^iv=s$, we call $mathcal{J}_s={A^iv,cdots,Av,v}$ a Jordan chain for $A$.
Enough with definitions: my question is what is, really, $bigcup_{sinmathcal{S}}mathcal{J}_s$? Let ${s_1,s_2,cdots,s_r}$ be a basis for $mathcal{S}$. Are we able to write
$$bigcup_{sinmathcal{S}}mathcal{J}_s={A^rv_1,A^{r-1}v_1,cdots,Av_1,v_1}cup{A^rv_2,cdots,Av_2,v_2}cupcdotscup{A^rv_r,cdots,Av_r,v_r}$$
where $s_i=A^rv_i$ for $i=1,cdots,r$.
It seems like that would be the sensible thing. I'm essentially having a hard time figuring out what the union of the Jordan chains is.
linear-algebra matrices
linear-algebra matrices
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
edited Jan 7 at 11:00
AstlyDichrar
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
asked Jan 7 at 10:53
AstlyDichrarAstlyDichrar
41738
41738
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