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A linear transformation $T$ such that $(v_1, dots, v_k, Tv_1, dots, Tv_k)$ is a basis for $mathbb F^2k$
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Let $T: mathbb R^2k to mathbb R^2k$ be a bijective linear transformation. We define a set
beginalign*
S = (v_1, dots, v_k) subset mathbb R^2k: (v_1, dots, v_k, Tv_1, dots, Tv_k) text is a basis for $mathbb R^2k$ .
endalign*
Is it possible to characterize the set $S$? In particular, the topological property, such as connectedness?
linear-algebra general-topology matrices
edited Jul 25 at 22:57
Omnomnomnom
121k784170
asked Jul 25 at 22:44
user9527
923524
 |Â
show 1 more comment
up vote
2
down vote
favorite
Let $T: mathbb R^2k to mathbb R^2k$ be a bijective linear transformation. We define a set
beginalign*
S = (v_1, dots, v_k) subset mathbb R^2k: (v_1, dots, v_k, Tv_1, dots, Tv_k) text is a basis for $mathbb R^2k$ .
endalign*
Is it possible to characterize the set $S$? In particular, the topological property, such as connectedness?
linear-algebra general-topology matrices
edited Jul 25 at 22:57
Omnomnomnom
121k784170
asked Jul 25 at 22:44
user9527
923524
What kind of topology would you impose on this space? How would you define the distance between two bases?
– Omnomnomnom
Jul 25 at 22:48
Presumably, you could interpret $S$ as some kind of subspace of the Grassmannian $Gr(k,Bbb R^2k)$, but you should say that this is what you have in mind
– Omnomnomnom
Jul 25 at 22:51
1
@Prism $(v_1,dots,v_k,Tv_1,dots,Tv_k)$ makes no sense if each $v_k$ is a scalar
– Omnomnomnom
Jul 25 at 22:52
Since it is finite dimension vector space, I was in mind some norm topology on $R^2k$.
– user9527
Jul 25 at 22:54
@Omnomnomnom: By your comments below, I can see the set is closed as intersections of closed sets. But how many connected components can it have?
– user9527
Jul 26 at 3:01
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $T: mathbb R^2k to mathbb R^2k$ be a bijective linear transformation. We define a set
beginalign*
S = (v_1, dots, v_k) subset mathbb R^2k: (v_1, dots, v_k, Tv_1, dots, Tv_k) text is a basis for $mathbb R^2k$ .
endalign*
Is it possible to characterize the set $S$? In particular, the topological property, such as connectedness?
linear-algebra general-topology matrices
edited Jul 25 at 22:57
Omnomnomnom
121k784170
asked Jul 25 at 22:44
user9527
923524
Let $T: mathbb R^2k to mathbb R^2k$ be a bijective linear transformation. We define a set
beginalign*
S = (v_1, dots, v_k) subset mathbb R^2k: (v_1, dots, v_k, Tv_1, dots, Tv_k) text is a basis for $mathbb R^2k$ .
endalign*
Is it possible to characterize the set $S$? In particular, the topological property, such as connectedness?
linear-algebra general-topology matrices
edited Jul 25 at 22:57
Omnomnomnom
121k784170
asked Jul 25 at 22:44
user9527
923524
edited Jul 25 at 22:57
Omnomnomnom
121k784170
edited Jul 25 at 22:57
Omnomnomnom
121k784170
edited Jul 25 at 22:57
Omnomnomnom
121k784170
121k784170
asked Jul 25 at 22:44
user9527
923524
asked Jul 25 at 22:44
user9527
923524
asked Jul 25 at 22:44
user9527
923524
923524
What kind of topology would you impose on this space? How would you define the distance between two bases?
– Omnomnomnom
Jul 25 at 22:48
Presumably, you could interpret $S$ as some kind of subspace of the Grassmannian $Gr(k,Bbb R^2k)$, but you should say that this is what you have in mind
– Omnomnomnom
Jul 25 at 22:51
1
@Prism $(v_1,dots,v_k,Tv_1,dots,Tv_k)$ makes no sense if each $v_k$ is a scalar
– Omnomnomnom
Jul 25 at 22:52
Since it is finite dimension vector space, I was in mind some norm topology on $R^2k$.
– user9527
Jul 25 at 22:54
@Omnomnomnom: By your comments below, I can see the set is closed as intersections of closed sets. But how many connected components can it have?
– user9527
Jul 26 at 3:01
 |Â
show 1 more comment
What kind of topology would you impose on this space? How would you define the distance between two bases?
– Omnomnomnom
Jul 25 at 22:48
Presumably, you could interpret $S$ as some kind of subspace of the Grassmannian $Gr(k,Bbb R^2k)$, but you should say that this is what you have in mind
– Omnomnomnom
Jul 25 at 22:51
1
@Prism $(v_1,dots,v_k,Tv_1,dots,Tv_k)$ makes no sense if each $v_k$ is a scalar
– Omnomnomnom
Jul 25 at 22:52
Since it is finite dimension vector space, I was in mind some norm topology on $R^2k$.
– user9527
Jul 25 at 22:54
@Omnomnomnom: By your comments below, I can see the set is closed as intersections of closed sets. But how many connected components can it have?
– user9527
Jul 26 at 3:01
What kind of topology would you impose on this space? How would you define the distance between two bases?
– Omnomnomnom
Jul 25 at 22:48
What kind of topology would you impose on this space? How would you define the distance between two bases?
– Omnomnomnom
Jul 25 at 22:48
Presumably, you could interpret $S$ as some kind of subspace of the Grassmannian $Gr(k,Bbb R^2k)$, but you should say that this is what you have in mind
– Omnomnomnom
Jul 25 at 22:51
Presumably, you could interpret $S$ as some kind of subspace of the Grassmannian $Gr(k,Bbb R^2k)$, but you should say that this is what you have in mind
– Omnomnomnom
Jul 25 at 22:51
1
1
@Prism $(v_1,dots,v_k,Tv_1,dots,Tv_k)$ makes no sense if each $v_k$ is a scalar
– Omnomnomnom
Jul 25 at 22:52
@Prism $(v_1,dots,v_k,Tv_1,dots,Tv_k)$ makes no sense if each $v_k$ is a scalar
– Omnomnomnom
Jul 25 at 22:52
Since it is finite dimension vector space, I was in mind some norm topology on $R^2k$.
– user9527
Jul 25 at 22:54
Since it is finite dimension vector space, I was in mind some norm topology on $R^2k$.
– user9527
Jul 25 at 22:54
@Omnomnomnom: By your comments below, I can see the set is closed as intersections of closed sets. But how many connected components can it have?
– user9527
Jul 26 at 3:01
@Omnomnomnom: By your comments below, I can see the set is closed as intersections of closed sets. But how many connected components can it have?
– user9527
Jul 26 at 3:01
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
A bit too long for a comment:
Per this construction, we can identify the $k$ dimensional subspaces of $Bbb R^k$ with the corresponding orthogonal projections, i.e. the symmetric $2ktimes 2k$ matrices satisfying $X^2 = X$ and $operatornametr(X) = k$. A set $B = (v_1,dots,v_k)$ will be contained in $S$ if and only if the subspace $U$ spanned by $B$ satisfies
$$
U cap T(U) = 0
$$
This is in turn equivalent to the statement that the projection $X$ onto $U$ satisfies
$$
XTX = 0
$$
So, if you'd like to characterize the space of $k$-dimensional subspaces of $V$ satisfying $V = U oplus T(U)$, then you could identify this set with the set of projection matrices $X$ given by
$$
X in Bbb R^2k times 2k : X = X^T, X = X^2, operatornametr(X) = k, XTX = 0
$$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A bit too long for a comment:
Per this construction, we can identify the $k$ dimensional subspaces of $Bbb R^k$ with the corresponding orthogonal projections, i.e. the symmetric $2ktimes 2k$ matrices satisfying $X^2 = X$ and $operatornametr(X) = k$. A set $B = (v_1,dots,v_k)$ will be contained in $S$ if and only if the subspace $U$ spanned by $B$ satisfies
$$
U cap T(U) = 0
$$
This is in turn equivalent to the statement that the projection $X$ onto $U$ satisfies
$$
XTX = 0
$$
So, if you'd like to characterize the space of $k$-dimensional subspaces of $V$ satisfying $V = U oplus T(U)$, then you could identify this set with the set of projection matrices $X$ given by
$$
X in Bbb R^2k times 2k : X = X^T, X = X^2, operatornametr(X) = k, XTX = 0
$$
add a comment |Â
up vote
2
down vote
accepted
A bit too long for a comment:
Per this construction, we can identify the $k$ dimensional subspaces of $Bbb R^k$ with the corresponding orthogonal projections, i.e. the symmetric $2ktimes 2k$ matrices satisfying $X^2 = X$ and $operatornametr(X) = k$. A set $B = (v_1,dots,v_k)$ will be contained in $S$ if and only if the subspace $U$ spanned by $B$ satisfies
$$
U cap T(U) = 0
$$
This is in turn equivalent to the statement that the projection $X$ onto $U$ satisfies
$$
XTX = 0
$$
So, if you'd like to characterize the space of $k$-dimensional subspaces of $V$ satisfying $V = U oplus T(U)$, then you could identify this set with the set of projection matrices $X$ given by
$$
X in Bbb R^2k times 2k : X = X^T, X = X^2, operatornametr(X) = k, XTX = 0
$$
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A bit too long for a comment:
Per this construction, we can identify the $k$ dimensional subspaces of $Bbb R^k$ with the corresponding orthogonal projections, i.e. the symmetric $2ktimes 2k$ matrices satisfying $X^2 = X$ and $operatornametr(X) = k$. A set $B = (v_1,dots,v_k)$ will be contained in $S$ if and only if the subspace $U$ spanned by $B$ satisfies
$$
U cap T(U) = 0
$$
This is in turn equivalent to the statement that the projection $X$ onto $U$ satisfies
$$
XTX = 0
$$
So, if you'd like to characterize the space of $k$-dimensional subspaces of $V$ satisfying $V = U oplus T(U)$, then you could identify this set with the set of projection matrices $X$ given by
$$
X in Bbb R^2k times 2k : X = X^T, X = X^2, operatornametr(X) = k, XTX = 0
$$
A bit too long for a comment:
Per this construction, we can identify the $k$ dimensional subspaces of $Bbb R^k$ with the corresponding orthogonal projections, i.e. the symmetric $2ktimes 2k$ matrices satisfying $X^2 = X$ and $operatornametr(X) = k$. A set $B = (v_1,dots,v_k)$ will be contained in $S$ if and only if the subspace $U$ spanned by $B$ satisfies
$$
U cap T(U) = 0
$$
This is in turn equivalent to the statement that the projection $X$ onto $U$ satisfies
$$
XTX = 0
$$
So, if you'd like to characterize the space of $k$-dimensional subspaces of $V$ satisfying $V = U oplus T(U)$, then you could identify this set with the set of projection matrices $X$ given by
$$
X in Bbb R^2k times 2k : X = X^T, X = X^2, operatornametr(X) = k, XTX = 0
$$
answered Jul 25 at 23:11
community wiki
Omnomnomnom
add a comment |Â
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What kind of topology would you impose on this space? How would you define the distance between two bases?
– Omnomnomnom
Jul 25 at 22:48
Presumably, you could interpret $S$ as some kind of subspace of the Grassmannian $Gr(k,Bbb R^2k)$, but you should say that this is what you have in mind
– Omnomnomnom
Jul 25 at 22:51
1
@Prism $(v_1,dots,v_k,Tv_1,dots,Tv_k)$ makes no sense if each $v_k$ is a scalar
– Omnomnomnom
Jul 25 at 22:52
Since it is finite dimension vector space, I was in mind some norm topology on $R^2k$.
– user9527
Jul 25 at 22:54
@Omnomnomnom: By your comments below, I can see the set is closed as intersections of closed sets. But how many connected components can it have?
– user9527
Jul 26 at 3:01