Find $c$ such that you get real eigenvalues
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let
A=$beginbmatrix
1& c\
1& -1
endbmatrix$
find all eigenvalues, but do not use characteristic polynomial. Find c for which eigenvalue will be real.
I use $detA=-1-c$ and $tr(A)=0$, since $detA=lambda_1lambda_2$ and $tr(A)=lambda_1+lambda_2$ I get that $lambda_1=-lambda_2$ so $lambda_2=pmsqrt1+c$ and $lambda_1=mpsqrt1+c$, $c$ must be $geq-1$ if I want real eigenvalue, I think that this is only way to not use characteristic polynomial , what you think?
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:55
Marko Škorić
4008
add a comment |Â
up vote
1
down vote
favorite
Let
A=$beginbmatrix
1& c\
1& -1
endbmatrix$
find all eigenvalues, but do not use characteristic polynomial. Find c for which eigenvalue will be real.
I use $detA=-1-c$ and $tr(A)=0$, since $detA=lambda_1lambda_2$ and $tr(A)=lambda_1+lambda_2$ I get that $lambda_1=-lambda_2$ so $lambda_2=pmsqrt1+c$ and $lambda_1=mpsqrt1+c$, $c$ must be $geq-1$ if I want real eigenvalue, I think that this is only way to not use characteristic polynomial , what you think?
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:55
Marko Škorić
4008
2
You're right about using the trace and det of $A$ to deduce the range of $c$. But I cannot see how it differs from "use the characteristic polynomial"...
– Vim
Sep 8 at 9:06
Well, how do you know these relations without using the characteristic polynomial? Iirc these come from applying Vieta's formulae.
– Guido A.
Sep 8 at 9:06
I do not know is there some other way, It must be somethin in linear algebra that I can use to prove it that but I do know
– Marko Å korić
Sep 8 at 9:10
It seems a bit of a circular argument for you are just not using the characteristic polynomial explicitly ...
– Bruce
Sep 8 at 9:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let
A=$beginbmatrix
1& c\
1& -1
endbmatrix$
find all eigenvalues, but do not use characteristic polynomial. Find c for which eigenvalue will be real.
I use $detA=-1-c$ and $tr(A)=0$, since $detA=lambda_1lambda_2$ and $tr(A)=lambda_1+lambda_2$ I get that $lambda_1=-lambda_2$ so $lambda_2=pmsqrt1+c$ and $lambda_1=mpsqrt1+c$, $c$ must be $geq-1$ if I want real eigenvalue, I think that this is only way to not use characteristic polynomial , what you think?
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:55
Marko Škorić
4008
Let
A=$beginbmatrix
1& c\
1& -1
endbmatrix$
find all eigenvalues, but do not use characteristic polynomial. Find c for which eigenvalue will be real.
I use $detA=-1-c$ and $tr(A)=0$, since $detA=lambda_1lambda_2$ and $tr(A)=lambda_1+lambda_2$ I get that $lambda_1=-lambda_2$ so $lambda_2=pmsqrt1+c$ and $lambda_1=mpsqrt1+c$, $c$ must be $geq-1$ if I want real eigenvalue, I think that this is only way to not use characteristic polynomial , what you think?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:55
Marko Škorić
4008
asked Sep 8 at 8:55
Marko Škorić
4008
edited Sep 8 at 9:01
asked Sep 8 at 8:55
Marko Škorić
4008
asked Sep 8 at 8:55
Marko Škorić
4008
asked Sep 8 at 8:55
Marko Škorić
4008
4008
2
You're right about using the trace and det of $A$ to deduce the range of $c$. But I cannot see how it differs from "use the characteristic polynomial"...
– Vim
Sep 8 at 9:06
Well, how do you know these relations without using the characteristic polynomial? Iirc these come from applying Vieta's formulae.
– Guido A.
Sep 8 at 9:06
I do not know is there some other way, It must be somethin in linear algebra that I can use to prove it that but I do know
– Marko Å korić
Sep 8 at 9:10
It seems a bit of a circular argument for you are just not using the characteristic polynomial explicitly ...
– Bruce
Sep 8 at 9:10
add a comment |Â
2
You're right about using the trace and det of $A$ to deduce the range of $c$. But I cannot see how it differs from "use the characteristic polynomial"...
– Vim
Sep 8 at 9:06
Well, how do you know these relations without using the characteristic polynomial? Iirc these come from applying Vieta's formulae.
– Guido A.
Sep 8 at 9:06
I do not know is there some other way, It must be somethin in linear algebra that I can use to prove it that but I do know
– Marko Å korić
Sep 8 at 9:10
It seems a bit of a circular argument for you are just not using the characteristic polynomial explicitly ...
– Bruce
Sep 8 at 9:10
2
2
You're right about using the trace and det of $A$ to deduce the range of $c$. But I cannot see how it differs from "use the characteristic polynomial"...
– Vim
Sep 8 at 9:06
You're right about using the trace and det of $A$ to deduce the range of $c$. But I cannot see how it differs from "use the characteristic polynomial"...
– Vim
Sep 8 at 9:06
Well, how do you know these relations without using the characteristic polynomial? Iirc these come from applying Vieta's formulae.
– Guido A.
Sep 8 at 9:06
Well, how do you know these relations without using the characteristic polynomial? Iirc these come from applying Vieta's formulae.
– Guido A.
Sep 8 at 9:06
I do not know is there some other way, It must be somethin in linear algebra that I can use to prove it that but I do know
– Marko Å korić
Sep 8 at 9:10
I do not know is there some other way, It must be somethin in linear algebra that I can use to prove it that but I do know
– Marko Å korić
Sep 8 at 9:10
It seems a bit of a circular argument for you are just not using the characteristic polynomial explicitly ...
– Bruce
Sep 8 at 9:10
It seems a bit of a circular argument for you are just not using the characteristic polynomial explicitly ...
– Bruce
Sep 8 at 9:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
We have
$$beginbmatrix
lambda x\
lambda yendbmatrix = lambdabeginbmatrix
x\
yendbmatrix = beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
yendbmatrix = beginbmatrix
x+cy\
x-yendbmatrix$$
Therefore $lambda y = x-y$ so $(lambda + 1)y = x$.
Also $lambda x = x+cy$ so $(lambda - 1)x = cy$ or $$(lambda^2-1 - c)y = 0$$ If $y = 0$, then also $x = 0$ so $lambda$ is not an eigenvalue. Therefore $lambda^2 - 1 - c = 0$ so $lambda = pm sqrt1+c$.
The eigenvalues are real if and only if $c ge -1$.
answered Sep 8 at 9:49
mechanodroid
24.6k62245
That’s precisely the solution by the definition, that is solve the system $Ax=lambda x$ with $xneq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear.
– gimusi
Sep 8 at 10:43
Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway?
– amd
Sep 8 at 21:07
@amd Yes, but it's more elementary as it avoids determinants.
– mechanodroid
Sep 8 at 21:17
add a comment |Â
up vote
2
down vote
By direct method we need to solve
$$beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
y
endbmatrix=lambdabeginbmatrix
x\
y
endbmatrix iff beginbmatrix
1-lambda& c\
1& -1-lambda
endbmatrixbeginbmatrix
x\
y
endbmatrix=beginbmatrix
0\
0
endbmatrix $$
then use elimination method in order to exclude the trivial solution.
answered Sep 8 at 9:09
gimusi
74.1k73889
Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way
– Marko Å korić
Sep 8 at 9:11
@MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials.
– gimusi
Sep 8 at 9:13
Yes but that is my teacher he give that on exam
– Marko Å korić
Sep 8 at 9:16
3
@MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr).
– gimusi
Sep 8 at 9:19
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We have
$$beginbmatrix
lambda x\
lambda yendbmatrix = lambdabeginbmatrix
x\
yendbmatrix = beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
yendbmatrix = beginbmatrix
x+cy\
x-yendbmatrix$$
Therefore $lambda y = x-y$ so $(lambda + 1)y = x$.
Also $lambda x = x+cy$ so $(lambda - 1)x = cy$ or $$(lambda^2-1 - c)y = 0$$ If $y = 0$, then also $x = 0$ so $lambda$ is not an eigenvalue. Therefore $lambda^2 - 1 - c = 0$ so $lambda = pm sqrt1+c$.
The eigenvalues are real if and only if $c ge -1$.
answered Sep 8 at 9:49
mechanodroid
24.6k62245
That’s precisely the solution by the definition, that is solve the system $Ax=lambda x$ with $xneq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear.
– gimusi
Sep 8 at 10:43
Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway?
– amd
Sep 8 at 21:07
@amd Yes, but it's more elementary as it avoids determinants.
– mechanodroid
Sep 8 at 21:17
add a comment |Â
up vote
2
down vote
accepted
We have
$$beginbmatrix
lambda x\
lambda yendbmatrix = lambdabeginbmatrix
x\
yendbmatrix = beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
yendbmatrix = beginbmatrix
x+cy\
x-yendbmatrix$$
Therefore $lambda y = x-y$ so $(lambda + 1)y = x$.
Also $lambda x = x+cy$ so $(lambda - 1)x = cy$ or $$(lambda^2-1 - c)y = 0$$ If $y = 0$, then also $x = 0$ so $lambda$ is not an eigenvalue. Therefore $lambda^2 - 1 - c = 0$ so $lambda = pm sqrt1+c$.
The eigenvalues are real if and only if $c ge -1$.
answered Sep 8 at 9:49
mechanodroid
24.6k62245
That’s precisely the solution by the definition, that is solve the system $Ax=lambda x$ with $xneq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear.
– gimusi
Sep 8 at 10:43
Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway?
– amd
Sep 8 at 21:07
@amd Yes, but it's more elementary as it avoids determinants.
– mechanodroid
Sep 8 at 21:17
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We have
$$beginbmatrix
lambda x\
lambda yendbmatrix = lambdabeginbmatrix
x\
yendbmatrix = beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
yendbmatrix = beginbmatrix
x+cy\
x-yendbmatrix$$
Therefore $lambda y = x-y$ so $(lambda + 1)y = x$.
Also $lambda x = x+cy$ so $(lambda - 1)x = cy$ or $$(lambda^2-1 - c)y = 0$$ If $y = 0$, then also $x = 0$ so $lambda$ is not an eigenvalue. Therefore $lambda^2 - 1 - c = 0$ so $lambda = pm sqrt1+c$.
The eigenvalues are real if and only if $c ge -1$.
answered Sep 8 at 9:49
mechanodroid
24.6k62245
We have
$$beginbmatrix
lambda x\
lambda yendbmatrix = lambdabeginbmatrix
x\
yendbmatrix = beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
yendbmatrix = beginbmatrix
x+cy\
x-yendbmatrix$$
Therefore $lambda y = x-y$ so $(lambda + 1)y = x$.
Also $lambda x = x+cy$ so $(lambda - 1)x = cy$ or $$(lambda^2-1 - c)y = 0$$ If $y = 0$, then also $x = 0$ so $lambda$ is not an eigenvalue. Therefore $lambda^2 - 1 - c = 0$ so $lambda = pm sqrt1+c$.
The eigenvalues are real if and only if $c ge -1$.
answered Sep 8 at 9:49
mechanodroid
24.6k62245
answered Sep 8 at 9:49
mechanodroid
24.6k62245
answered Sep 8 at 9:49
mechanodroid
24.6k62245
answered Sep 8 at 9:49
mechanodroid
24.6k62245
24.6k62245
That’s precisely the solution by the definition, that is solve the system $Ax=lambda x$ with $xneq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear.
– gimusi
Sep 8 at 10:43
Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway?
– amd
Sep 8 at 21:07
@amd Yes, but it's more elementary as it avoids determinants.
– mechanodroid
Sep 8 at 21:17
add a comment |Â
That’s precisely the solution by the definition, that is solve the system $Ax=lambda x$ with $xneq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear.
– gimusi
Sep 8 at 10:43
Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway?
– amd
Sep 8 at 21:07
@amd Yes, but it's more elementary as it avoids determinants.
– mechanodroid
Sep 8 at 21:17
That’s precisely the solution by the definition, that is solve the system $Ax=lambda x$ with $xneq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear.
– gimusi
Sep 8 at 10:43
That’s precisely the solution by the definition, that is solve the system $Ax=lambda x$ with $xneq 0$. I’ve also suggested that way but maybe I wasn’t sufficiently clear.
– gimusi
Sep 8 at 10:43
Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway?
– amd
Sep 8 at 21:07
Hmm... isn’t this a slightly roundabout way to end up at the characteristic polynomial, anyway?
– amd
Sep 8 at 21:07
@amd Yes, but it's more elementary as it avoids determinants.
– mechanodroid
Sep 8 at 21:17
@amd Yes, but it's more elementary as it avoids determinants.
– mechanodroid
Sep 8 at 21:17
add a comment |Â
up vote
2
down vote
By direct method we need to solve
$$beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
y
endbmatrix=lambdabeginbmatrix
x\
y
endbmatrix iff beginbmatrix
1-lambda& c\
1& -1-lambda
endbmatrixbeginbmatrix
x\
y
endbmatrix=beginbmatrix
0\
0
endbmatrix $$
then use elimination method in order to exclude the trivial solution.
answered Sep 8 at 9:09
gimusi
74.1k73889
Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way
– Marko Å korić
Sep 8 at 9:11
@MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials.
– gimusi
Sep 8 at 9:13
Yes but that is my teacher he give that on exam
– Marko Å korić
Sep 8 at 9:16
3
@MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr).
– gimusi
Sep 8 at 9:19
add a comment |Â
up vote
2
down vote
By direct method we need to solve
$$beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
y
endbmatrix=lambdabeginbmatrix
x\
y
endbmatrix iff beginbmatrix
1-lambda& c\
1& -1-lambda
endbmatrixbeginbmatrix
x\
y
endbmatrix=beginbmatrix
0\
0
endbmatrix $$
then use elimination method in order to exclude the trivial solution.
answered Sep 8 at 9:09
gimusi
74.1k73889
Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way
– Marko Å korić
Sep 8 at 9:11
@MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials.
– gimusi
Sep 8 at 9:13
Yes but that is my teacher he give that on exam
– Marko Å korić
Sep 8 at 9:16
3
@MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr).
– gimusi
Sep 8 at 9:19
add a comment |Â
up vote
2
down vote
up vote
2
down vote
By direct method we need to solve
$$beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
y
endbmatrix=lambdabeginbmatrix
x\
y
endbmatrix iff beginbmatrix
1-lambda& c\
1& -1-lambda
endbmatrixbeginbmatrix
x\
y
endbmatrix=beginbmatrix
0\
0
endbmatrix $$
then use elimination method in order to exclude the trivial solution.
answered Sep 8 at 9:09
gimusi
74.1k73889
By direct method we need to solve
$$beginbmatrix
1& c\
1& -1
endbmatrixbeginbmatrix
x\
y
endbmatrix=lambdabeginbmatrix
x\
y
endbmatrix iff beginbmatrix
1-lambda& c\
1& -1-lambda
endbmatrixbeginbmatrix
x\
y
endbmatrix=beginbmatrix
0\
0
endbmatrix $$
then use elimination method in order to exclude the trivial solution.
answered Sep 8 at 9:09
gimusi
74.1k73889
answered Sep 8 at 9:09
gimusi
74.1k73889
answered Sep 8 at 9:09
gimusi
74.1k73889
answered Sep 8 at 9:09
gimusi
74.1k73889
74.1k73889
Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way
– Marko Å korić
Sep 8 at 9:11
@MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials.
– gimusi
Sep 8 at 9:13
Yes but that is my teacher he give that on exam
– Marko Å korić
Sep 8 at 9:16
3
@MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr).
– gimusi
Sep 8 at 9:19
add a comment |Â
Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way
– Marko Å korić
Sep 8 at 9:11
@MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials.
– gimusi
Sep 8 at 9:13
Yes but that is my teacher he give that on exam
– Marko Å korić
Sep 8 at 9:16
3
@MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr).
– gimusi
Sep 8 at 9:19
Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way
– Marko Å korić
Sep 8 at 9:11
Yes it is good but, still it is look that I use characteristic polynomials, but I do not know other way
– Marko Å korić
Sep 8 at 9:11
@MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials.
– gimusi
Sep 8 at 9:13
@MarkoŠkorić Yes at the end we always find something equivalent to that but here we can proceed by basic tools without introducing the concept of characteristic polynomials.
– gimusi
Sep 8 at 9:13
Yes but that is my teacher he give that on exam
– Marko Å korić
Sep 8 at 9:16
Yes but that is my teacher he give that on exam
– Marko Å korić
Sep 8 at 9:16
3
3
@MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr).
– gimusi
Sep 8 at 9:19
@MarkoŠkorić That's the basic way coming directly by the definition of eigenvalues, I can't really see something simpler that that. I guess that the teacher was aimed to apply directly the definition without using pedentatly the algorithmic way to derive the result by the roots of the characteristic polynomials (or det and Tr).
– gimusi
Sep 8 at 9:19
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2909435%2ffind-c-such-that-you-get-real-eigenvalues%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
You're right about using the trace and det of $A$ to deduce the range of $c$. But I cannot see how it differs from "use the characteristic polynomial"...
– Vim
Sep 8 at 9:06
Well, how do you know these relations without using the characteristic polynomial? Iirc these come from applying Vieta's formulae.
– Guido A.
Sep 8 at 9:06
I do not know is there some other way, It must be somethin in linear algebra that I can use to prove it that but I do know
– Marko Å korić
Sep 8 at 9:10
It seems a bit of a circular argument for you are just not using the characteristic polynomial explicitly ...
– Bruce
Sep 8 at 9:10