Are they similar matrix












0












$begingroup$


Do $begin{bmatrix}
0&i&0\0&0&1\0&0&0
end{bmatrix} $ and $begin{bmatrix}
0&0&0\-i&0&0\0&1&0
end{bmatrix} $ are similar.Is this True/false



Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me






linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why negative vote
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:25










  • $begingroup$
    Welcome to MSE! It is more likely that you'll receive an answer if you showed us that you've made an effort
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:38










  • $begingroup$
    Sorry i dont know how to start
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:39










  • $begingroup$
    Oh don't mind the downvoters, they have a habit of downvoting posts that doesn't show efforts (and without explaining why)
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39












  • $begingroup$
    I've made up for it by upvoting. Just make sure you show that you've tried next time
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39


















0












$begingroup$


Do $begin{bmatrix}
0&i&0\0&0&1\0&0&0
end{bmatrix} $ and $begin{bmatrix}
0&0&0\-i&0&0\0&1&0
end{bmatrix} $ are similar.Is this True/false



Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me






linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why negative vote
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:25










  • $begingroup$
    Welcome to MSE! It is more likely that you'll receive an answer if you showed us that you've made an effort
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:38










  • $begingroup$
    Sorry i dont know how to start
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:39










  • $begingroup$
    Oh don't mind the downvoters, they have a habit of downvoting posts that doesn't show efforts (and without explaining why)
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39












  • $begingroup$
    I've made up for it by upvoting. Just make sure you show that you've tried next time
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39
















0












0








0





$begingroup$


Do $begin{bmatrix}
0&i&0\0&0&1\0&0&0
end{bmatrix} $ and $begin{bmatrix}
0&0&0\-i&0&0\0&1&0
end{bmatrix} $ are similar.Is this True/false



Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me






linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector







share|cite|improve this question











$endgroup$




Do $begin{bmatrix}
0&i&0\0&0&1\0&0&0
end{bmatrix} $ and $begin{bmatrix}
0&0&0\-i&0&0\0&1&0
end{bmatrix} $ are similar.Is this True/false



Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me






linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector




linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 6:38







Vasanth Kris

















asked Dec 10 '18 at 6:19









Vasanth KrisVasanth Kris

6




6












  • $begingroup$
    Why negative vote
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:25










  • $begingroup$
    Welcome to MSE! It is more likely that you'll receive an answer if you showed us that you've made an effort
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:38










  • $begingroup$
    Sorry i dont know how to start
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:39










  • $begingroup$
    Oh don't mind the downvoters, they have a habit of downvoting posts that doesn't show efforts (and without explaining why)
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39












  • $begingroup$
    I've made up for it by upvoting. Just make sure you show that you've tried next time
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39




















  • $begingroup$
    Why negative vote
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:25










  • $begingroup$
    Welcome to MSE! It is more likely that you'll receive an answer if you showed us that you've made an effort
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:38










  • $begingroup$
    Sorry i dont know how to start
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:39










  • $begingroup$
    Oh don't mind the downvoters, they have a habit of downvoting posts that doesn't show efforts (and without explaining why)
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39












  • $begingroup$
    I've made up for it by upvoting. Just make sure you show that you've tried next time
    $endgroup$
    – Vee Hua Zhi
    Dec 10 '18 at 6:39


















$begingroup$
Why negative vote
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:25




$begingroup$
Why negative vote
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:25












$begingroup$
Welcome to MSE! It is more likely that you'll receive an answer if you showed us that you've made an effort
$endgroup$
– Vee Hua Zhi
Dec 10 '18 at 6:38




$begingroup$
Welcome to MSE! It is more likely that you'll receive an answer if you showed us that you've made an effort
$endgroup$
– Vee Hua Zhi
Dec 10 '18 at 6:38












$begingroup$
Sorry i dont know how to start
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:39




$begingroup$
Sorry i dont know how to start
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:39












$begingroup$
Oh don't mind the downvoters, they have a habit of downvoting posts that doesn't show efforts (and without explaining why)
$endgroup$
– Vee Hua Zhi
Dec 10 '18 at 6:39






$begingroup$
Oh don't mind the downvoters, they have a habit of downvoting posts that doesn't show efforts (and without explaining why)
$endgroup$
– Vee Hua Zhi
Dec 10 '18 at 6:39














$begingroup$
I've made up for it by upvoting. Just make sure you show that you've tried next time
$endgroup$
– Vee Hua Zhi
Dec 10 '18 at 6:39






$begingroup$
I've made up for it by upvoting. Just make sure you show that you've tried next time
$endgroup$
– Vee Hua Zhi
Dec 10 '18 at 6:39












1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint: both matrices have the same characteristic polynomial $p(x)=x^3$, and for both that is also the minimal polynomial. What can you conclude about their Jordan canonical forms?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I dont know jordan form any other way?
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:41










  • $begingroup$
    Well, that's one of the reasons why people on this forum ask to show what you tried-because that way it will also be clear how much do you know about the problem. I'll try to think of another solution.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:44












  • $begingroup$
    Btw what is answer true or false.... Next time if i ask i will add details as much as i xan
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:46










  • $begingroup$
    They are similar. Matrices of order at most $3times 3$ (not true for higher dimension matrices) are similar if and only if their characteristic and minimal polynomials are the same. I never thought about proving this lemma without Jordan form though.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:48













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint: both matrices have the same characteristic polynomial $p(x)=x^3$, and for both that is also the minimal polynomial. What can you conclude about their Jordan canonical forms?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I dont know jordan form any other way?
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:41










  • $begingroup$
    Well, that's one of the reasons why people on this forum ask to show what you tried-because that way it will also be clear how much do you know about the problem. I'll try to think of another solution.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:44












  • $begingroup$
    Btw what is answer true or false.... Next time if i ask i will add details as much as i xan
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:46










  • $begingroup$
    They are similar. Matrices of order at most $3times 3$ (not true for higher dimension matrices) are similar if and only if their characteristic and minimal polynomials are the same. I never thought about proving this lemma without Jordan form though.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:48


















0












$begingroup$

Hint: both matrices have the same characteristic polynomial $p(x)=x^3$, and for both that is also the minimal polynomial. What can you conclude about their Jordan canonical forms?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I dont know jordan form any other way?
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:41










  • $begingroup$
    Well, that's one of the reasons why people on this forum ask to show what you tried-because that way it will also be clear how much do you know about the problem. I'll try to think of another solution.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:44












  • $begingroup$
    Btw what is answer true or false.... Next time if i ask i will add details as much as i xan
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:46










  • $begingroup$
    They are similar. Matrices of order at most $3times 3$ (not true for higher dimension matrices) are similar if and only if their characteristic and minimal polynomials are the same. I never thought about proving this lemma without Jordan form though.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:48
















0












0








0





$begingroup$

Hint: both matrices have the same characteristic polynomial $p(x)=x^3$, and for both that is also the minimal polynomial. What can you conclude about their Jordan canonical forms?






share|cite|improve this answer









$endgroup$



Hint: both matrices have the same characteristic polynomial $p(x)=x^3$, and for both that is also the minimal polynomial. What can you conclude about their Jordan canonical forms?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 6:38









MarkMark

10.4k1622




10.4k1622












  • $begingroup$
    I dont know jordan form any other way?
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:41










  • $begingroup$
    Well, that's one of the reasons why people on this forum ask to show what you tried-because that way it will also be clear how much do you know about the problem. I'll try to think of another solution.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:44












  • $begingroup$
    Btw what is answer true or false.... Next time if i ask i will add details as much as i xan
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:46










  • $begingroup$
    They are similar. Matrices of order at most $3times 3$ (not true for higher dimension matrices) are similar if and only if their characteristic and minimal polynomials are the same. I never thought about proving this lemma without Jordan form though.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:48




















  • $begingroup$
    I dont know jordan form any other way?
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:41










  • $begingroup$
    Well, that's one of the reasons why people on this forum ask to show what you tried-because that way it will also be clear how much do you know about the problem. I'll try to think of another solution.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:44












  • $begingroup$
    Btw what is answer true or false.... Next time if i ask i will add details as much as i xan
    $endgroup$
    – Vasanth Kris
    Dec 10 '18 at 6:46










  • $begingroup$
    They are similar. Matrices of order at most $3times 3$ (not true for higher dimension matrices) are similar if and only if their characteristic and minimal polynomials are the same. I never thought about proving this lemma without Jordan form though.
    $endgroup$
    – Mark
    Dec 10 '18 at 6:48


















$begingroup$
I dont know jordan form any other way?
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:41




$begingroup$
I dont know jordan form any other way?
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:41












$begingroup$
Well, that's one of the reasons why people on this forum ask to show what you tried-because that way it will also be clear how much do you know about the problem. I'll try to think of another solution.
$endgroup$
– Mark
Dec 10 '18 at 6:44






$begingroup$
Well, that's one of the reasons why people on this forum ask to show what you tried-because that way it will also be clear how much do you know about the problem. I'll try to think of another solution.
$endgroup$
– Mark
Dec 10 '18 at 6:44














$begingroup$
Btw what is answer true or false.... Next time if i ask i will add details as much as i xan
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:46




$begingroup$
Btw what is answer true or false.... Next time if i ask i will add details as much as i xan
$endgroup$
– Vasanth Kris
Dec 10 '18 at 6:46












$begingroup$
They are similar. Matrices of order at most $3times 3$ (not true for higher dimension matrices) are similar if and only if their characteristic and minimal polynomials are the same. I never thought about proving this lemma without Jordan form though.
$endgroup$
– Mark
Dec 10 '18 at 6:48






$begingroup$
They are similar. Matrices of order at most $3times 3$ (not true for higher dimension matrices) are similar if and only if their characteristic and minimal polynomials are the same. I never thought about proving this lemma without Jordan form though.
$endgroup$
– Mark
Dec 10 '18 at 6:48




















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