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Constructing a continuous path between two matrices
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Let $A_1, A_2 in GL_n(mathbb R)$ be two fixed matrices with eigenvalues lying on the open left half plane of $mathbb C$, i.e., with negative real parts and $A_1 neq A_2$. We may assume $A_1, A_2$ are diagonalizable if necessary. Let $(A_1)_alpha$ and $(A_2)_beta$ be the corresponding matrix that scales the eigenvalues by $alpha in mathbb R$ and $beta in mathbb R$ respectively. That is if $(lambda_1, dots, lambda_n)$ are eigenvalues of $A_1$, then $(alpha lambda_1, dots, alpha lambda_n)$ are eigenvalues of $(A_1)_alpha$.
I am wondering whether it is possible to construct a continuous path $gamma: [0,1] to GL_n(mathbb R)$ with eigenvalues all lying on the left half plane along the path: by choosing a sequence of nonzero real numbers, $(alpha_n)$ and $(beta_n)$, $gamma$ is piecewise linear. For example, suppose there exist some finite sequences $(alpha_1, dots, alpha_k)$ and $(beta_1, dots, beta_l)$, then the path would be
beginalign*
A_1 xrightarrow (A_1)_alpha_1 xrightarrow dots xrightarrow (A_1)_alpha_k xrightarrow (A_2)_beta_1 xrightarrow dots xrightarrow (A_2)_beta_l xrightarrow A_2.
endalign*
In between, each arrow stands for a convex path. Assume we can construct a continuous path between $A$ and $(A)_alpha$.
If we take a convex combination of $A_1, A_2$, i.e., $A(t) = (1-t)A_1 + tA_2$. Then $det(A(t))$ is a nonconstant polynomial in $t$ and has at most $n$ zeros. Intuitively I am thinking of taking the path until we meet a singular point of $t$ (by continuity, all eigenvalues should stay in the left half plane) and change path to some $(A_1)_alpha$ and doing this repeatedly. Not sure whether this will work.
p.s. The reason for my consideration was because I need to guarantee certain matrix structure along the path. I have worked out scaling eigenvalues can be done within the structure of interest.
linear-algebra general-topology path-connected matrix-analysis
asked Aug 1 at 20:40
user9527
923524
 |Â
show 4 more comments
up vote
5
down vote
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Let $A_1, A_2 in GL_n(mathbb R)$ be two fixed matrices with eigenvalues lying on the open left half plane of $mathbb C$, i.e., with negative real parts and $A_1 neq A_2$. We may assume $A_1, A_2$ are diagonalizable if necessary. Let $(A_1)_alpha$ and $(A_2)_beta$ be the corresponding matrix that scales the eigenvalues by $alpha in mathbb R$ and $beta in mathbb R$ respectively. That is if $(lambda_1, dots, lambda_n)$ are eigenvalues of $A_1$, then $(alpha lambda_1, dots, alpha lambda_n)$ are eigenvalues of $(A_1)_alpha$.
I am wondering whether it is possible to construct a continuous path $gamma: [0,1] to GL_n(mathbb R)$ with eigenvalues all lying on the left half plane along the path: by choosing a sequence of nonzero real numbers, $(alpha_n)$ and $(beta_n)$, $gamma$ is piecewise linear. For example, suppose there exist some finite sequences $(alpha_1, dots, alpha_k)$ and $(beta_1, dots, beta_l)$, then the path would be
beginalign*
A_1 xrightarrow (A_1)_alpha_1 xrightarrow dots xrightarrow (A_1)_alpha_k xrightarrow (A_2)_beta_1 xrightarrow dots xrightarrow (A_2)_beta_l xrightarrow A_2.
endalign*
In between, each arrow stands for a convex path. Assume we can construct a continuous path between $A$ and $(A)_alpha$.
If we take a convex combination of $A_1, A_2$, i.e., $A(t) = (1-t)A_1 + tA_2$. Then $det(A(t))$ is a nonconstant polynomial in $t$ and has at most $n$ zeros. Intuitively I am thinking of taking the path until we meet a singular point of $t$ (by continuity, all eigenvalues should stay in the left half plane) and change path to some $(A_1)_alpha$ and doing this repeatedly. Not sure whether this will work.
p.s. The reason for my consideration was because I need to guarantee certain matrix structure along the path. I have worked out scaling eigenvalues can be done within the structure of interest.
linear-algebra general-topology path-connected matrix-analysis
asked Aug 1 at 20:40
user9527
923524
I don't get your $A_alpha$. How is it defined exactly? Is is simply $alpha A$?
– user2520938
Aug 1 at 22:34
@user2520938: $(A)_alpha$ has eigenvalues which are a factor of eigenvalues of $A$. $(A)_alpha neq alpha A$ in general.
– user9527
Aug 1 at 22:39
But that is not uniquely defined.... You need to be more specific
– user2520938
Aug 1 at 22:39
Yes. As I commented at the end, $A_1, A_2$ has some special structure and I would like to construct a path within this structure. I figured out we can scale eigenvalues within the structure.
– user9527
Aug 1 at 22:43
1
So you have a set of matrices $M$. You know that if $Ain M$, then for any $ainmathbb R$ there exists a matrix $A_ain M$ whose eigenvalues are $a$ times those of $A$, and which can be connected to $A$ via a continuous path in $M$. You want to know if this is enough to prove that $A_1$ and $A_2$ (which are in $M$) can be path-connected?
– Jack M
Aug 2 at 8:04
 |Â
show 4 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $A_1, A_2 in GL_n(mathbb R)$ be two fixed matrices with eigenvalues lying on the open left half plane of $mathbb C$, i.e., with negative real parts and $A_1 neq A_2$. We may assume $A_1, A_2$ are diagonalizable if necessary. Let $(A_1)_alpha$ and $(A_2)_beta$ be the corresponding matrix that scales the eigenvalues by $alpha in mathbb R$ and $beta in mathbb R$ respectively. That is if $(lambda_1, dots, lambda_n)$ are eigenvalues of $A_1$, then $(alpha lambda_1, dots, alpha lambda_n)$ are eigenvalues of $(A_1)_alpha$.
I am wondering whether it is possible to construct a continuous path $gamma: [0,1] to GL_n(mathbb R)$ with eigenvalues all lying on the left half plane along the path: by choosing a sequence of nonzero real numbers, $(alpha_n)$ and $(beta_n)$, $gamma$ is piecewise linear. For example, suppose there exist some finite sequences $(alpha_1, dots, alpha_k)$ and $(beta_1, dots, beta_l)$, then the path would be
beginalign*
A_1 xrightarrow (A_1)_alpha_1 xrightarrow dots xrightarrow (A_1)_alpha_k xrightarrow (A_2)_beta_1 xrightarrow dots xrightarrow (A_2)_beta_l xrightarrow A_2.
endalign*
In between, each arrow stands for a convex path. Assume we can construct a continuous path between $A$ and $(A)_alpha$.
If we take a convex combination of $A_1, A_2$, i.e., $A(t) = (1-t)A_1 + tA_2$. Then $det(A(t))$ is a nonconstant polynomial in $t$ and has at most $n$ zeros. Intuitively I am thinking of taking the path until we meet a singular point of $t$ (by continuity, all eigenvalues should stay in the left half plane) and change path to some $(A_1)_alpha$ and doing this repeatedly. Not sure whether this will work.
p.s. The reason for my consideration was because I need to guarantee certain matrix structure along the path. I have worked out scaling eigenvalues can be done within the structure of interest.
linear-algebra general-topology path-connected matrix-analysis
asked Aug 1 at 20:40
user9527
923524
Let $A_1, A_2 in GL_n(mathbb R)$ be two fixed matrices with eigenvalues lying on the open left half plane of $mathbb C$, i.e., with negative real parts and $A_1 neq A_2$. We may assume $A_1, A_2$ are diagonalizable if necessary. Let $(A_1)_alpha$ and $(A_2)_beta$ be the corresponding matrix that scales the eigenvalues by $alpha in mathbb R$ and $beta in mathbb R$ respectively. That is if $(lambda_1, dots, lambda_n)$ are eigenvalues of $A_1$, then $(alpha lambda_1, dots, alpha lambda_n)$ are eigenvalues of $(A_1)_alpha$.
I am wondering whether it is possible to construct a continuous path $gamma: [0,1] to GL_n(mathbb R)$ with eigenvalues all lying on the left half plane along the path: by choosing a sequence of nonzero real numbers, $(alpha_n)$ and $(beta_n)$, $gamma$ is piecewise linear. For example, suppose there exist some finite sequences $(alpha_1, dots, alpha_k)$ and $(beta_1, dots, beta_l)$, then the path would be
beginalign*
A_1 xrightarrow (A_1)_alpha_1 xrightarrow dots xrightarrow (A_1)_alpha_k xrightarrow (A_2)_beta_1 xrightarrow dots xrightarrow (A_2)_beta_l xrightarrow A_2.
endalign*
In between, each arrow stands for a convex path. Assume we can construct a continuous path between $A$ and $(A)_alpha$.
If we take a convex combination of $A_1, A_2$, i.e., $A(t) = (1-t)A_1 + tA_2$. Then $det(A(t))$ is a nonconstant polynomial in $t$ and has at most $n$ zeros. Intuitively I am thinking of taking the path until we meet a singular point of $t$ (by continuity, all eigenvalues should stay in the left half plane) and change path to some $(A_1)_alpha$ and doing this repeatedly. Not sure whether this will work.
p.s. The reason for my consideration was because I need to guarantee certain matrix structure along the path. I have worked out scaling eigenvalues can be done within the structure of interest.
linear-algebra general-topology path-connected matrix-analysis
asked Aug 1 at 20:40
user9527
923524
edited Aug 1 at 23:07
asked Aug 1 at 20:40
user9527
923524
asked Aug 1 at 20:40
user9527
923524
asked Aug 1 at 20:40
user9527
923524
923524
I don't get your $A_alpha$. How is it defined exactly? Is is simply $alpha A$?
– user2520938
Aug 1 at 22:34
@user2520938: $(A)_alpha$ has eigenvalues which are a factor of eigenvalues of $A$. $(A)_alpha neq alpha A$ in general.
– user9527
Aug 1 at 22:39
But that is not uniquely defined.... You need to be more specific
– user2520938
Aug 1 at 22:39
Yes. As I commented at the end, $A_1, A_2$ has some special structure and I would like to construct a path within this structure. I figured out we can scale eigenvalues within the structure.
– user9527
Aug 1 at 22:43
1
So you have a set of matrices $M$. You know that if $Ain M$, then for any $ainmathbb R$ there exists a matrix $A_ain M$ whose eigenvalues are $a$ times those of $A$, and which can be connected to $A$ via a continuous path in $M$. You want to know if this is enough to prove that $A_1$ and $A_2$ (which are in $M$) can be path-connected?
– Jack M
Aug 2 at 8:04
 |Â
show 4 more comments
I don't get your $A_alpha$. How is it defined exactly? Is is simply $alpha A$?
– user2520938
Aug 1 at 22:34
@user2520938: $(A)_alpha$ has eigenvalues which are a factor of eigenvalues of $A$. $(A)_alpha neq alpha A$ in general.
– user9527
Aug 1 at 22:39
But that is not uniquely defined.... You need to be more specific
– user2520938
Aug 1 at 22:39
Yes. As I commented at the end, $A_1, A_2$ has some special structure and I would like to construct a path within this structure. I figured out we can scale eigenvalues within the structure.
– user9527
Aug 1 at 22:43
1
So you have a set of matrices $M$. You know that if $Ain M$, then for any $ainmathbb R$ there exists a matrix $A_ain M$ whose eigenvalues are $a$ times those of $A$, and which can be connected to $A$ via a continuous path in $M$. You want to know if this is enough to prove that $A_1$ and $A_2$ (which are in $M$) can be path-connected?
– Jack M
Aug 2 at 8:04
I don't get your $A_alpha$. How is it defined exactly? Is is simply $alpha A$?
– user2520938
Aug 1 at 22:34
I don't get your $A_alpha$. How is it defined exactly? Is is simply $alpha A$?
– user2520938
Aug 1 at 22:34
@user2520938: $(A)_alpha$ has eigenvalues which are a factor of eigenvalues of $A$. $(A)_alpha neq alpha A$ in general.
– user9527
Aug 1 at 22:39
@user2520938: $(A)_alpha$ has eigenvalues which are a factor of eigenvalues of $A$. $(A)_alpha neq alpha A$ in general.
– user9527
Aug 1 at 22:39
But that is not uniquely defined.... You need to be more specific
– user2520938
Aug 1 at 22:39
But that is not uniquely defined.... You need to be more specific
– user2520938
Aug 1 at 22:39
Yes. As I commented at the end, $A_1, A_2$ has some special structure and I would like to construct a path within this structure. I figured out we can scale eigenvalues within the structure.
– user9527
Aug 1 at 22:43
Yes. As I commented at the end, $A_1, A_2$ has some special structure and I would like to construct a path within this structure. I figured out we can scale eigenvalues within the structure.
– user9527
Aug 1 at 22:43
1
1
So you have a set of matrices $M$. You know that if $Ain M$, then for any $ainmathbb R$ there exists a matrix $A_ain M$ whose eigenvalues are $a$ times those of $A$, and which can be connected to $A$ via a continuous path in $M$. You want to know if this is enough to prove that $A_1$ and $A_2$ (which are in $M$) can be path-connected?
– Jack M
Aug 2 at 8:04
So you have a set of matrices $M$. You know that if $Ain M$, then for any $ainmathbb R$ there exists a matrix $A_ain M$ whose eigenvalues are $a$ times those of $A$, and which can be connected to $A$ via a continuous path in $M$. You want to know if this is enough to prove that $A_1$ and $A_2$ (which are in $M$) can be path-connected?
– Jack M
Aug 2 at 8:04
 |Â
show 4 more comments
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I don't get your $A_alpha$. How is it defined exactly? Is is simply $alpha A$?
– user2520938
Aug 1 at 22:34
@user2520938: $(A)_alpha$ has eigenvalues which are a factor of eigenvalues of $A$. $(A)_alpha neq alpha A$ in general.
– user9527
Aug 1 at 22:39
But that is not uniquely defined.... You need to be more specific
– user2520938
Aug 1 at 22:39
Yes. As I commented at the end, $A_1, A_2$ has some special structure and I would like to construct a path within this structure. I figured out we can scale eigenvalues within the structure.
– user9527
Aug 1 at 22:43
1
So you have a set of matrices $M$. You know that if $Ain M$, then for any $ainmathbb R$ there exists a matrix $A_ain M$ whose eigenvalues are $a$ times those of $A$, and which can be connected to $A$ via a continuous path in $M$. You want to know if this is enough to prove that $A_1$ and $A_2$ (which are in $M$) can be path-connected?
– Jack M
Aug 2 at 8:04