How to find a matrix with characteristic equation? [duplicate]
$begingroup$
This question already has an answer here:
Determining a matrix from its characteristic polynomial
3 answers
I'll write it again but the users below don't seem to understand ... THIS IS NOT A DUPLICATE . PLEASE READ MY QUESTION THOROUGLY.
I need to find a matrix with characteristic equation given by
λ² − λ − 1 = 0
I know that Since the polynomial p(λ) = λ² − λ − 1 has degree 2,so we look for a 2 × 2 matrix.
I let A be any 2 × 2 matrix, where a, b, c, d ∈ R.
A=$$
begin{pmatrix}
a & b \
c & d \
end{pmatrix}
$$
The characteristic equation of A is:
p(λ) = det(A − λId) = λ² − (a + d)λ + ad − bc
And here is what I don't get . On the answer sheet is says that this polynomial coincides with λ² − λ − 1 if and only if:
−(a + d) = −1 and ad − bc = −1.
I have dyscalculia and even the things that might seem simple are hard for me to understand. Can somebody please explain where they are getting
−(a + d) = −1 and ad − bc = −1
and why , for example, they are not setting λ²=-1? What does that mean?
Please be as simple and clear as possible! I really appreciate the help ! Thanks guys
Edit: please do not mark this as a duplicate. This is a UNIQUE question in which i am trying to understand ONE of the stages to determine the characteristic equation .
matrix-equations matrix-calculus
asked Dec 10 '18 at 17:07
BM97BM97
758
$endgroup$
marked as duplicate by José Carlos Santos, John Douma, Dietrich Burde, John Hughes, amWhy Dec 10 '18 at 20:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 1 more comment
$begingroup$
This question already has an answer here:
Determining a matrix from its characteristic polynomial
3 answers
I'll write it again but the users below don't seem to understand ... THIS IS NOT A DUPLICATE . PLEASE READ MY QUESTION THOROUGLY.
I need to find a matrix with characteristic equation given by
λ² − λ − 1 = 0
I know that Since the polynomial p(λ) = λ² − λ − 1 has degree 2,so we look for a 2 × 2 matrix.
I let A be any 2 × 2 matrix, where a, b, c, d ∈ R.
A=$$
begin{pmatrix}
a & b \
c & d \
end{pmatrix}
$$
The characteristic equation of A is:
p(λ) = det(A − λId) = λ² − (a + d)λ + ad − bc
And here is what I don't get . On the answer sheet is says that this polynomial coincides with λ² − λ − 1 if and only if:
−(a + d) = −1 and ad − bc = −1.
I have dyscalculia and even the things that might seem simple are hard for me to understand. Can somebody please explain where they are getting
−(a + d) = −1 and ad − bc = −1
and why , for example, they are not setting λ²=-1? What does that mean?
Please be as simple and clear as possible! I really appreciate the help ! Thanks guys
Edit: please do not mark this as a duplicate. This is a UNIQUE question in which i am trying to understand ONE of the stages to determine the characteristic equation .
matrix-equations matrix-calculus
asked Dec 10 '18 at 17:07
BM97BM97
758
$endgroup$
marked as duplicate by José Carlos Santos, John Douma, Dietrich Burde, John Hughes, amWhy Dec 10 '18 at 20:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Look at en.wikibooks.org/wiki/Control_Systems/… for example.
$endgroup$
– copper.hat
Dec 10 '18 at 17:11
$begingroup$
No ! It is not a duplicate ! I am asking a specific question reguarding my problem ! please check thoroughly both questions before writing down random things.
$endgroup$
– BM97
Dec 10 '18 at 17:20
$begingroup$
Yes i have, and no one appears to answer my question . All the solutions i've seen on this site take a different approach on answering the question . I have one type of method on answering the question which i really haven't seen here and i would really really like it if one of you guys could help me out ? I mean you the time to mark my question as a duplicate but not answer it?
$endgroup$
– BM97
Dec 10 '18 at 17:24
$begingroup$
I mean is is that hard to answer?
$endgroup$
– BM97
Dec 10 '18 at 17:32
$begingroup$
Is it possible...just possible ... that the highly-experienced users here are more adept at recognizing a duplicate than you are? Also: I think you may have forgotten exactly who cares about this question. One answer: not the people you insult as having offered you insufficient or inadequate free help. You ask "Is it that hard to answer?" The answer (in my case) is "No, it's not that hard. But I'm choosing not to."
$endgroup$
– John Hughes
Dec 10 '18 at 18:01
|
show 1 more comment
$begingroup$
This question already has an answer here:
Determining a matrix from its characteristic polynomial
3 answers
I'll write it again but the users below don't seem to understand ... THIS IS NOT A DUPLICATE . PLEASE READ MY QUESTION THOROUGLY.
I need to find a matrix with characteristic equation given by
λ² − λ − 1 = 0
I know that Since the polynomial p(λ) = λ² − λ − 1 has degree 2,so we look for a 2 × 2 matrix.
I let A be any 2 × 2 matrix, where a, b, c, d ∈ R.
A=$$
begin{pmatrix}
a & b \
c & d \
end{pmatrix}
$$
The characteristic equation of A is:
p(λ) = det(A − λId) = λ² − (a + d)λ + ad − bc
And here is what I don't get . On the answer sheet is says that this polynomial coincides with λ² − λ − 1 if and only if:
−(a + d) = −1 and ad − bc = −1.
I have dyscalculia and even the things that might seem simple are hard for me to understand. Can somebody please explain where they are getting
−(a + d) = −1 and ad − bc = −1
and why , for example, they are not setting λ²=-1? What does that mean?
Please be as simple and clear as possible! I really appreciate the help ! Thanks guys
Edit: please do not mark this as a duplicate. This is a UNIQUE question in which i am trying to understand ONE of the stages to determine the characteristic equation .
matrix-equations matrix-calculus
asked Dec 10 '18 at 17:07
BM97BM97
758
$endgroup$
This question already has an answer here:
Determining a matrix from its characteristic polynomial
3 answers
I'll write it again but the users below don't seem to understand ... THIS IS NOT A DUPLICATE . PLEASE READ MY QUESTION THOROUGLY.
I need to find a matrix with characteristic equation given by
λ² − λ − 1 = 0
I know that Since the polynomial p(λ) = λ² − λ − 1 has degree 2,so we look for a 2 × 2 matrix.
I let A be any 2 × 2 matrix, where a, b, c, d ∈ R.
A=$$
begin{pmatrix}
a & b \
c & d \
end{pmatrix}
$$
The characteristic equation of A is:
p(λ) = det(A − λId) = λ² − (a + d)λ + ad − bc
And here is what I don't get . On the answer sheet is says that this polynomial coincides with λ² − λ − 1 if and only if:
−(a + d) = −1 and ad − bc = −1.
I have dyscalculia and even the things that might seem simple are hard for me to understand. Can somebody please explain where they are getting
−(a + d) = −1 and ad − bc = −1
and why , for example, they are not setting λ²=-1? What does that mean?
Please be as simple and clear as possible! I really appreciate the help ! Thanks guys
Edit: please do not mark this as a duplicate. This is a UNIQUE question in which i am trying to understand ONE of the stages to determine the characteristic equation .
This question already has an answer here:
Determining a matrix from its characteristic polynomial
3 answers
matrix-equations matrix-calculus
matrix-equations matrix-calculus
asked Dec 10 '18 at 17:07
BM97BM97
758
asked Dec 10 '18 at 17:07
BM97BM97
758
edited Dec 10 '18 at 17:37
BM97
asked Dec 10 '18 at 17:07
BM97BM97
758
asked Dec 10 '18 at 17:07
BM97BM97
758
asked Dec 10 '18 at 17:07
BM97BM97
758
758
marked as duplicate by José Carlos Santos, John Douma, Dietrich Burde, John Hughes, amWhy Dec 10 '18 at 20:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, John Douma, Dietrich Burde, John Hughes, amWhy Dec 10 '18 at 20:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Look at en.wikibooks.org/wiki/Control_Systems/… for example.
$endgroup$
– copper.hat
Dec 10 '18 at 17:11
$begingroup$
No ! It is not a duplicate ! I am asking a specific question reguarding my problem ! please check thoroughly both questions before writing down random things.
$endgroup$
– BM97
Dec 10 '18 at 17:20
$begingroup$
Yes i have, and no one appears to answer my question . All the solutions i've seen on this site take a different approach on answering the question . I have one type of method on answering the question which i really haven't seen here and i would really really like it if one of you guys could help me out ? I mean you the time to mark my question as a duplicate but not answer it?
$endgroup$
– BM97
Dec 10 '18 at 17:24
$begingroup$
I mean is is that hard to answer?
$endgroup$
– BM97
Dec 10 '18 at 17:32
$begingroup$
Is it possible...just possible ... that the highly-experienced users here are more adept at recognizing a duplicate than you are? Also: I think you may have forgotten exactly who cares about this question. One answer: not the people you insult as having offered you insufficient or inadequate free help. You ask "Is it that hard to answer?" The answer (in my case) is "No, it's not that hard. But I'm choosing not to."
$endgroup$
– John Hughes
Dec 10 '18 at 18:01
|
show 1 more comment
$begingroup$
Look at en.wikibooks.org/wiki/Control_Systems/… for example.
$endgroup$
– copper.hat
Dec 10 '18 at 17:11
$begingroup$
No ! It is not a duplicate ! I am asking a specific question reguarding my problem ! please check thoroughly both questions before writing down random things.
$endgroup$
– BM97
Dec 10 '18 at 17:20
$begingroup$
Yes i have, and no one appears to answer my question . All the solutions i've seen on this site take a different approach on answering the question . I have one type of method on answering the question which i really haven't seen here and i would really really like it if one of you guys could help me out ? I mean you the time to mark my question as a duplicate but not answer it?
$endgroup$
– BM97
Dec 10 '18 at 17:24
$begingroup$
I mean is is that hard to answer?
$endgroup$
– BM97
Dec 10 '18 at 17:32
$begingroup$
Is it possible...just possible ... that the highly-experienced users here are more adept at recognizing a duplicate than you are? Also: I think you may have forgotten exactly who cares about this question. One answer: not the people you insult as having offered you insufficient or inadequate free help. You ask "Is it that hard to answer?" The answer (in my case) is "No, it's not that hard. But I'm choosing not to."
$endgroup$
– John Hughes
Dec 10 '18 at 18:01
$begingroup$
Look at en.wikibooks.org/wiki/Control_Systems/… for example.
$endgroup$
– copper.hat
Dec 10 '18 at 17:11
$begingroup$
Look at en.wikibooks.org/wiki/Control_Systems/… for example.
$endgroup$
– copper.hat
Dec 10 '18 at 17:11
$begingroup$
No ! It is not a duplicate ! I am asking a specific question reguarding my problem ! please check thoroughly both questions before writing down random things.
$endgroup$
– BM97
Dec 10 '18 at 17:20
$begingroup$
No ! It is not a duplicate ! I am asking a specific question reguarding my problem ! please check thoroughly both questions before writing down random things.
$endgroup$
– BM97
Dec 10 '18 at 17:20
$begingroup$
Yes i have, and no one appears to answer my question . All the solutions i've seen on this site take a different approach on answering the question . I have one type of method on answering the question which i really haven't seen here and i would really really like it if one of you guys could help me out ? I mean you the time to mark my question as a duplicate but not answer it?
$endgroup$
– BM97
Dec 10 '18 at 17:24
$begingroup$
Yes i have, and no one appears to answer my question . All the solutions i've seen on this site take a different approach on answering the question . I have one type of method on answering the question which i really haven't seen here and i would really really like it if one of you guys could help me out ? I mean you the time to mark my question as a duplicate but not answer it?
$endgroup$
– BM97
Dec 10 '18 at 17:24
$begingroup$
I mean is is that hard to answer?
$endgroup$
– BM97
Dec 10 '18 at 17:32
$begingroup$
I mean is is that hard to answer?
$endgroup$
– BM97
Dec 10 '18 at 17:32
$begingroup$
Is it possible...just possible ... that the highly-experienced users here are more adept at recognizing a duplicate than you are? Also: I think you may have forgotten exactly who cares about this question. One answer: not the people you insult as having offered you insufficient or inadequate free help. You ask "Is it that hard to answer?" The answer (in my case) is "No, it's not that hard. But I'm choosing not to."
$endgroup$
– John Hughes
Dec 10 '18 at 18:01
$begingroup$
Is it possible...just possible ... that the highly-experienced users here are more adept at recognizing a duplicate than you are? Also: I think you may have forgotten exactly who cares about this question. One answer: not the people you insult as having offered you insufficient or inadequate free help. You ask "Is it that hard to answer?" The answer (in my case) is "No, it's not that hard. But I'm choosing not to."
$endgroup$
– John Hughes
Dec 10 '18 at 18:01
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
You have
$$lambda^2-lambda-1=lambda^2color{red}{-1}cdotlambda+color{blue}{(-1)},$$
you agree? Now the number which is multiplied by $lambda$ is $color{red}{-1}$, the number which is just added is $color{blue}{-1}$.
In
$$lambda^2color{red}{-(a+d)}cdotlambda+
color{blue}{ad-bc}$$
the number which is multiplied by $lambda$ is $color{red}{-(a+d)}$ and the number which is just added is $color{blue}{ad-bc}$.
From here
$$color{red}{-1}=color{red}{-(a+d)}qquadtext{and}quadcolor{blue}{-1}=color{blue}{ad-bc}.
$$
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
$begingroup$
Thank you !! I really appreciate your help
$endgroup$
– BM97
Dec 10 '18 at 18:05
$begingroup$
You're welcome.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:06
$begingroup$
One question.... is it fair to say that we are then only taking into consideration the parts of the equation that do not involve λ?
$endgroup$
– BM97
Dec 10 '18 at 18:17
$begingroup$
Well, sort of, yes.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:58
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have
$$lambda^2-lambda-1=lambda^2color{red}{-1}cdotlambda+color{blue}{(-1)},$$
you agree? Now the number which is multiplied by $lambda$ is $color{red}{-1}$, the number which is just added is $color{blue}{-1}$.
In
$$lambda^2color{red}{-(a+d)}cdotlambda+
color{blue}{ad-bc}$$
the number which is multiplied by $lambda$ is $color{red}{-(a+d)}$ and the number which is just added is $color{blue}{ad-bc}$.
From here
$$color{red}{-1}=color{red}{-(a+d)}qquadtext{and}quadcolor{blue}{-1}=color{blue}{ad-bc}.
$$
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
$begingroup$
Thank you !! I really appreciate your help
$endgroup$
– BM97
Dec 10 '18 at 18:05
$begingroup$
You're welcome.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:06
$begingroup$
One question.... is it fair to say that we are then only taking into consideration the parts of the equation that do not involve λ?
$endgroup$
– BM97
Dec 10 '18 at 18:17
$begingroup$
Well, sort of, yes.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:58
add a comment |
$begingroup$
You have
$$lambda^2-lambda-1=lambda^2color{red}{-1}cdotlambda+color{blue}{(-1)},$$
you agree? Now the number which is multiplied by $lambda$ is $color{red}{-1}$, the number which is just added is $color{blue}{-1}$.
In
$$lambda^2color{red}{-(a+d)}cdotlambda+
color{blue}{ad-bc}$$
the number which is multiplied by $lambda$ is $color{red}{-(a+d)}$ and the number which is just added is $color{blue}{ad-bc}$.
From here
$$color{red}{-1}=color{red}{-(a+d)}qquadtext{and}quadcolor{blue}{-1}=color{blue}{ad-bc}.
$$
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
$begingroup$
Thank you !! I really appreciate your help
$endgroup$
– BM97
Dec 10 '18 at 18:05
$begingroup$
You're welcome.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:06
$begingroup$
One question.... is it fair to say that we are then only taking into consideration the parts of the equation that do not involve λ?
$endgroup$
– BM97
Dec 10 '18 at 18:17
$begingroup$
Well, sort of, yes.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:58
add a comment |
$begingroup$
You have
$$lambda^2-lambda-1=lambda^2color{red}{-1}cdotlambda+color{blue}{(-1)},$$
you agree? Now the number which is multiplied by $lambda$ is $color{red}{-1}$, the number which is just added is $color{blue}{-1}$.
In
$$lambda^2color{red}{-(a+d)}cdotlambda+
color{blue}{ad-bc}$$
the number which is multiplied by $lambda$ is $color{red}{-(a+d)}$ and the number which is just added is $color{blue}{ad-bc}$.
From here
$$color{red}{-1}=color{red}{-(a+d)}qquadtext{and}quadcolor{blue}{-1}=color{blue}{ad-bc}.
$$
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
You have
$$lambda^2-lambda-1=lambda^2color{red}{-1}cdotlambda+color{blue}{(-1)},$$
you agree? Now the number which is multiplied by $lambda$ is $color{red}{-1}$, the number which is just added is $color{blue}{-1}$.
In
$$lambda^2color{red}{-(a+d)}cdotlambda+
color{blue}{ad-bc}$$
the number which is multiplied by $lambda$ is $color{red}{-(a+d)}$ and the number which is just added is $color{blue}{ad-bc}$.
From here
$$color{red}{-1}=color{red}{-(a+d)}qquadtext{and}quadcolor{blue}{-1}=color{blue}{ad-bc}.
$$
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
edited Dec 10 '18 at 18:05
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
answered Dec 10 '18 at 17:47
Michael HoppeMichael Hoppe
10.8k31834
10.8k31834
$begingroup$
Thank you !! I really appreciate your help
$endgroup$
– BM97
Dec 10 '18 at 18:05
$begingroup$
You're welcome.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:06
$begingroup$
One question.... is it fair to say that we are then only taking into consideration the parts of the equation that do not involve λ?
$endgroup$
– BM97
Dec 10 '18 at 18:17
$begingroup$
Well, sort of, yes.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:58
add a comment |
$begingroup$
Thank you !! I really appreciate your help
$endgroup$
– BM97
Dec 10 '18 at 18:05
$begingroup$
You're welcome.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:06
$begingroup$
One question.... is it fair to say that we are then only taking into consideration the parts of the equation that do not involve λ?
$endgroup$
– BM97
Dec 10 '18 at 18:17
$begingroup$
Well, sort of, yes.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:58
$begingroup$
Thank you !! I really appreciate your help
$endgroup$
– BM97
Dec 10 '18 at 18:05
$begingroup$
Thank you !! I really appreciate your help
$endgroup$
– BM97
Dec 10 '18 at 18:05
$begingroup$
You're welcome.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:06
$begingroup$
You're welcome.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:06
$begingroup$
One question.... is it fair to say that we are then only taking into consideration the parts of the equation that do not involve λ?
$endgroup$
– BM97
Dec 10 '18 at 18:17
$begingroup$
One question.... is it fair to say that we are then only taking into consideration the parts of the equation that do not involve λ?
$endgroup$
– BM97
Dec 10 '18 at 18:17
$begingroup$
Well, sort of, yes.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:58
$begingroup$
Well, sort of, yes.
$endgroup$
– Michael Hoppe
Dec 10 '18 at 18:58
add a comment |
$begingroup$
Look at en.wikibooks.org/wiki/Control_Systems/… for example.
$endgroup$
– copper.hat
Dec 10 '18 at 17:11
$begingroup$
No ! It is not a duplicate ! I am asking a specific question reguarding my problem ! please check thoroughly both questions before writing down random things.
$endgroup$
– BM97
Dec 10 '18 at 17:20
$begingroup$
Yes i have, and no one appears to answer my question . All the solutions i've seen on this site take a different approach on answering the question . I have one type of method on answering the question which i really haven't seen here and i would really really like it if one of you guys could help me out ? I mean you the time to mark my question as a duplicate but not answer it?
$endgroup$
– BM97
Dec 10 '18 at 17:24
$begingroup$
I mean is is that hard to answer?
$endgroup$
– BM97
Dec 10 '18 at 17:32
$begingroup$
Is it possible...just possible ... that the highly-experienced users here are more adept at recognizing a duplicate than you are? Also: I think you may have forgotten exactly who cares about this question. One answer: not the people you insult as having offered you insufficient or inadequate free help. You ask "Is it that hard to answer?" The answer (in my case) is "No, it's not that hard. But I'm choosing not to."
$endgroup$
– John Hughes
Dec 10 '18 at 18:01