Mixing
Why the following set of eigenvalues will not work for any allowed values of a and b? [closed]
$begingroup$
Following markov chain has transition matrix:
$
begin{pmatrix} 0.7&a&0.3-a\b&0.5-b&0.5\1&0&0end{pmatrix}
$
Explain why the following set of eigenvalues is impossible for any allowed values of a and b: {1, 1, 0.5}
Does it relate to the fact that two eigenvalues have the same value.
eigenvalues-eigenvectors markov-chains markov-process
asked Feb 1 at 23:32
luffyluffy
92
$endgroup$
closed as off-topic by Did, Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, ancientmathematician Feb 5 at 11:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Following markov chain has transition matrix:
$
begin{pmatrix} 0.7&a&0.3-a\b&0.5-b&0.5\1&0&0end{pmatrix}
$
Explain why the following set of eigenvalues is impossible for any allowed values of a and b: {1, 1, 0.5}
Does it relate to the fact that two eigenvalues have the same value.
eigenvalues-eigenvectors markov-chains markov-process
asked Feb 1 at 23:32
luffyluffy
92
$endgroup$
closed as off-topic by Did, Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, ancientmathematician Feb 5 at 11:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Following markov chain has transition matrix:
$
begin{pmatrix} 0.7&a&0.3-a\b&0.5-b&0.5\1&0&0end{pmatrix}
$
Explain why the following set of eigenvalues is impossible for any allowed values of a and b: {1, 1, 0.5}
Does it relate to the fact that two eigenvalues have the same value.
eigenvalues-eigenvectors markov-chains markov-process
asked Feb 1 at 23:32
luffyluffy
92
$endgroup$
Following markov chain has transition matrix:
$
begin{pmatrix} 0.7&a&0.3-a\b&0.5-b&0.5\1&0&0end{pmatrix}
$
Explain why the following set of eigenvalues is impossible for any allowed values of a and b: {1, 1, 0.5}
Does it relate to the fact that two eigenvalues have the same value.
eigenvalues-eigenvectors markov-chains markov-process
eigenvalues-eigenvectors markov-chains markov-process
asked Feb 1 at 23:32
luffyluffy
92
asked Feb 1 at 23:32
luffyluffy
92
asked Feb 1 at 23:32
luffyluffy
92
asked Feb 1 at 23:32
luffyluffy
92
asked Feb 1 at 23:32
luffyluffy
92
92
closed as off-topic by Did, Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, ancientmathematician Feb 5 at 11:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, ancientmathematician Feb 5 at 11:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cesareo, José Carlos Santos, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, it's not about two eigenvalues being the same. That's perfectly possible in a Markov chain. (From context, it appears this version multiplies a row by this matrix on the right - that's what row sum $1$ leads to.)
Instead, consider the trace. It's equal to the sum of the eigenvalues, so it would be $2.5$ - but it's also equal to the sum of the diagonal elements. With only one that's allowed to vary, the center element of the matrix would be $1.8$. Nope.
answered Feb 1 at 23:39
jmerryjmerry
17k11633
$endgroup$
$begingroup$
So, if writing an explanation, how would you explain it in simple terms.
$endgroup$
– luffy
Feb 2 at 0:26
1
$begingroup$
Why don't you try writing it yourself, and put it up in either a comment, an edit to your post, or an answer? That way, we can critique your writing and see if there are any gaps in your understanding that still need to be addressed.
$endgroup$
– jmerry
Feb 2 at 0:32
$begingroup$
well, if two eigenvalues are the same, then in the markov chain, they form a closed set and the other value is absorbing.
$endgroup$
– luffy
Feb 2 at 0:37
1
$begingroup$
Uh - what does that have to do with anything?
$endgroup$
– jmerry
Feb 2 at 0:46
$begingroup$
going by what you say. the trace of the matrix doesn't equal our set of eigenvalues
$endgroup$
– luffy
Feb 2 at 1:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it's not about two eigenvalues being the same. That's perfectly possible in a Markov chain. (From context, it appears this version multiplies a row by this matrix on the right - that's what row sum $1$ leads to.)
Instead, consider the trace. It's equal to the sum of the eigenvalues, so it would be $2.5$ - but it's also equal to the sum of the diagonal elements. With only one that's allowed to vary, the center element of the matrix would be $1.8$. Nope.
answered Feb 1 at 23:39
jmerryjmerry
17k11633
$endgroup$
$begingroup$
So, if writing an explanation, how would you explain it in simple terms.
$endgroup$
– luffy
Feb 2 at 0:26
1
$begingroup$
Why don't you try writing it yourself, and put it up in either a comment, an edit to your post, or an answer? That way, we can critique your writing and see if there are any gaps in your understanding that still need to be addressed.
$endgroup$
– jmerry
Feb 2 at 0:32
$begingroup$
well, if two eigenvalues are the same, then in the markov chain, they form a closed set and the other value is absorbing.
$endgroup$
– luffy
Feb 2 at 0:37
1
$begingroup$
Uh - what does that have to do with anything?
$endgroup$
– jmerry
Feb 2 at 0:46
$begingroup$
going by what you say. the trace of the matrix doesn't equal our set of eigenvalues
$endgroup$
– luffy
Feb 2 at 1:53
add a comment |
$begingroup$
No, it's not about two eigenvalues being the same. That's perfectly possible in a Markov chain. (From context, it appears this version multiplies a row by this matrix on the right - that's what row sum $1$ leads to.)
Instead, consider the trace. It's equal to the sum of the eigenvalues, so it would be $2.5$ - but it's also equal to the sum of the diagonal elements. With only one that's allowed to vary, the center element of the matrix would be $1.8$. Nope.
answered Feb 1 at 23:39
jmerryjmerry
17k11633
$endgroup$
$begingroup$
So, if writing an explanation, how would you explain it in simple terms.
$endgroup$
– luffy
Feb 2 at 0:26
1
$begingroup$
Why don't you try writing it yourself, and put it up in either a comment, an edit to your post, or an answer? That way, we can critique your writing and see if there are any gaps in your understanding that still need to be addressed.
$endgroup$
– jmerry
Feb 2 at 0:32
$begingroup$
well, if two eigenvalues are the same, then in the markov chain, they form a closed set and the other value is absorbing.
$endgroup$
– luffy
Feb 2 at 0:37
1
$begingroup$
Uh - what does that have to do with anything?
$endgroup$
– jmerry
Feb 2 at 0:46
$begingroup$
going by what you say. the trace of the matrix doesn't equal our set of eigenvalues
$endgroup$
– luffy
Feb 2 at 1:53
add a comment |
$begingroup$
No, it's not about two eigenvalues being the same. That's perfectly possible in a Markov chain. (From context, it appears this version multiplies a row by this matrix on the right - that's what row sum $1$ leads to.)
Instead, consider the trace. It's equal to the sum of the eigenvalues, so it would be $2.5$ - but it's also equal to the sum of the diagonal elements. With only one that's allowed to vary, the center element of the matrix would be $1.8$. Nope.
answered Feb 1 at 23:39
jmerryjmerry
17k11633
$endgroup$
No, it's not about two eigenvalues being the same. That's perfectly possible in a Markov chain. (From context, it appears this version multiplies a row by this matrix on the right - that's what row sum $1$ leads to.)
Instead, consider the trace. It's equal to the sum of the eigenvalues, so it would be $2.5$ - but it's also equal to the sum of the diagonal elements. With only one that's allowed to vary, the center element of the matrix would be $1.8$. Nope.
answered Feb 1 at 23:39
jmerryjmerry
17k11633
answered Feb 1 at 23:39
jmerryjmerry
17k11633
answered Feb 1 at 23:39
jmerryjmerry
17k11633
answered Feb 1 at 23:39
jmerryjmerry
17k11633
17k11633
$begingroup$
So, if writing an explanation, how would you explain it in simple terms.
$endgroup$
– luffy
Feb 2 at 0:26
1
$begingroup$
Why don't you try writing it yourself, and put it up in either a comment, an edit to your post, or an answer? That way, we can critique your writing and see if there are any gaps in your understanding that still need to be addressed.
$endgroup$
– jmerry
Feb 2 at 0:32
$begingroup$
well, if two eigenvalues are the same, then in the markov chain, they form a closed set and the other value is absorbing.
$endgroup$
– luffy
Feb 2 at 0:37
1
$begingroup$
Uh - what does that have to do with anything?
$endgroup$
– jmerry
Feb 2 at 0:46
$begingroup$
going by what you say. the trace of the matrix doesn't equal our set of eigenvalues
$endgroup$
– luffy
Feb 2 at 1:53
add a comment |
$begingroup$
So, if writing an explanation, how would you explain it in simple terms.
$endgroup$
– luffy
Feb 2 at 0:26
1
$begingroup$
Why don't you try writing it yourself, and put it up in either a comment, an edit to your post, or an answer? That way, we can critique your writing and see if there are any gaps in your understanding that still need to be addressed.
$endgroup$
– jmerry
Feb 2 at 0:32
$begingroup$
well, if two eigenvalues are the same, then in the markov chain, they form a closed set and the other value is absorbing.
$endgroup$
– luffy
Feb 2 at 0:37
1
$begingroup$
Uh - what does that have to do with anything?
$endgroup$
– jmerry
Feb 2 at 0:46
$begingroup$
going by what you say. the trace of the matrix doesn't equal our set of eigenvalues
$endgroup$
– luffy
Feb 2 at 1:53
$begingroup$
So, if writing an explanation, how would you explain it in simple terms.
$endgroup$
– luffy
Feb 2 at 0:26
$begingroup$
So, if writing an explanation, how would you explain it in simple terms.
$endgroup$
– luffy
Feb 2 at 0:26
1
1
$begingroup$
Why don't you try writing it yourself, and put it up in either a comment, an edit to your post, or an answer? That way, we can critique your writing and see if there are any gaps in your understanding that still need to be addressed.
$endgroup$
– jmerry
Feb 2 at 0:32
$begingroup$
Why don't you try writing it yourself, and put it up in either a comment, an edit to your post, or an answer? That way, we can critique your writing and see if there are any gaps in your understanding that still need to be addressed.
$endgroup$
– jmerry
Feb 2 at 0:32
$begingroup$
well, if two eigenvalues are the same, then in the markov chain, they form a closed set and the other value is absorbing.
$endgroup$
– luffy
Feb 2 at 0:37
$begingroup$
well, if two eigenvalues are the same, then in the markov chain, they form a closed set and the other value is absorbing.
$endgroup$
– luffy
Feb 2 at 0:37
1
1
$begingroup$
Uh - what does that have to do with anything?
$endgroup$
– jmerry
Feb 2 at 0:46
$begingroup$
Uh - what does that have to do with anything?
$endgroup$
– jmerry
Feb 2 at 0:46
$begingroup$
going by what you say. the trace of the matrix doesn't equal our set of eigenvalues
$endgroup$
– luffy
Feb 2 at 1:53
$begingroup$
going by what you say. the trace of the matrix doesn't equal our set of eigenvalues
$endgroup$
– luffy
Feb 2 at 1:53
add a comment |
