Eigenvalues and eigenvector of a matrix containing $sin theta$ and $cos theta$

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$$
A
= beginbmatrix
0 & 1 & -sin theta \
-1 & 0 & cos theta \
-sin theta & cos theta & 0
endbmatrix
$$
Given this matrix I want to calculate its eigenvalues and eigenvectors. What I don't understand is how this can be done because I find a characteristic polynomial of $lambda^3 = 0$ which means that its eigenvalue is $0$ with algebraic multiplicity of $3$. When I perform the row operations to find the eigenvectors I can't because the top left element is $0$ so I can't eliminate the elements below it.







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  • 1




    "which means that its eigenvalue is $0$ with geometric multiplicity of $3$" Nope, it is the algebraic multiplicity that's $3$. The geometric multiplicity is $1$.
    – amsmath
    Aug 26 at 11:54











  • Exchanging rows is an elementary row operation.
    – amd
    Aug 26 at 16:27














up vote
1
down vote

favorite












$$
A
= beginbmatrix
0 & 1 & -sin theta \
-1 & 0 & cos theta \
-sin theta & cos theta & 0
endbmatrix
$$
Given this matrix I want to calculate its eigenvalues and eigenvectors. What I don't understand is how this can be done because I find a characteristic polynomial of $lambda^3 = 0$ which means that its eigenvalue is $0$ with algebraic multiplicity of $3$. When I perform the row operations to find the eigenvectors I can't because the top left element is $0$ so I can't eliminate the elements below it.







share|cite|improve this question


















  • 1




    "which means that its eigenvalue is $0$ with geometric multiplicity of $3$" Nope, it is the algebraic multiplicity that's $3$. The geometric multiplicity is $1$.
    – amsmath
    Aug 26 at 11:54











  • Exchanging rows is an elementary row operation.
    – amd
    Aug 26 at 16:27












up vote
1
down vote

favorite









up vote
1
down vote

favorite











$$
A
= beginbmatrix
0 & 1 & -sin theta \
-1 & 0 & cos theta \
-sin theta & cos theta & 0
endbmatrix
$$
Given this matrix I want to calculate its eigenvalues and eigenvectors. What I don't understand is how this can be done because I find a characteristic polynomial of $lambda^3 = 0$ which means that its eigenvalue is $0$ with algebraic multiplicity of $3$. When I perform the row operations to find the eigenvectors I can't because the top left element is $0$ so I can't eliminate the elements below it.







share|cite|improve this question














$$
A
= beginbmatrix
0 & 1 & -sin theta \
-1 & 0 & cos theta \
-sin theta & cos theta & 0
endbmatrix
$$
Given this matrix I want to calculate its eigenvalues and eigenvectors. What I don't understand is how this can be done because I find a characteristic polynomial of $lambda^3 = 0$ which means that its eigenvalue is $0$ with algebraic multiplicity of $3$. When I perform the row operations to find the eigenvectors I can't because the top left element is $0$ so I can't eliminate the elements below it.









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share|cite|improve this question




share|cite|improve this question








edited Aug 26 at 11:59

























asked Aug 26 at 11:34









Harry Touloupas

104




104







  • 1




    "which means that its eigenvalue is $0$ with geometric multiplicity of $3$" Nope, it is the algebraic multiplicity that's $3$. The geometric multiplicity is $1$.
    – amsmath
    Aug 26 at 11:54











  • Exchanging rows is an elementary row operation.
    – amd
    Aug 26 at 16:27












  • 1




    "which means that its eigenvalue is $0$ with geometric multiplicity of $3$" Nope, it is the algebraic multiplicity that's $3$. The geometric multiplicity is $1$.
    – amsmath
    Aug 26 at 11:54











  • Exchanging rows is an elementary row operation.
    – amd
    Aug 26 at 16:27







1




1




"which means that its eigenvalue is $0$ with geometric multiplicity of $3$" Nope, it is the algebraic multiplicity that's $3$. The geometric multiplicity is $1$.
– amsmath
Aug 26 at 11:54





"which means that its eigenvalue is $0$ with geometric multiplicity of $3$" Nope, it is the algebraic multiplicity that's $3$. The geometric multiplicity is $1$.
– amsmath
Aug 26 at 11:54













Exchanging rows is an elementary row operation.
– amd
Aug 26 at 16:27




Exchanging rows is an elementary row operation.
– amd
Aug 26 at 16:27










1 Answer
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0
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Since you have a $0$ there, exchange the first and the second rows, thereby getting$$beginbmatrix-1&0&costheta\0&1&sintheta\-sintheta&costheta&0endbmatrix.$$Now, proceed in the usual way. In the end, you will get that the eigenvectors with eigenvalue $a$ are those vectors of the form $lambda(costheta,sintheta,1)$.






share|cite|improve this answer




















  • My question is can this be done when I am trying to find the eigenvectors of any matrix?
    – Harry Touloupas
    Aug 26 at 11:54










  • @HarryTouloupas No. But it can be done when the eigenvalue is $0$.
    – José Carlos Santos
    Aug 26 at 11:55










  • @HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $lambda$ you can perform row operations (also exchange of rows) on $A-lambda I$ in order to find an eigenvector. Here, $lambda$ is zero, so you do the stuff with $A$.
    – amsmath
    Aug 26 at 11:58











  • Also in this example why is $A^k = vec0$ for every $k>=3$ ? I can see why $A^k = vec0$ but why do we need the $k>=3$?
    – Harry Touloupas
    Aug 26 at 12:15











  • @HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
    – José Carlos Santos
    Aug 26 at 12:31











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Since you have a $0$ there, exchange the first and the second rows, thereby getting$$beginbmatrix-1&0&costheta\0&1&sintheta\-sintheta&costheta&0endbmatrix.$$Now, proceed in the usual way. In the end, you will get that the eigenvectors with eigenvalue $a$ are those vectors of the form $lambda(costheta,sintheta,1)$.






share|cite|improve this answer




















  • My question is can this be done when I am trying to find the eigenvectors of any matrix?
    – Harry Touloupas
    Aug 26 at 11:54










  • @HarryTouloupas No. But it can be done when the eigenvalue is $0$.
    – José Carlos Santos
    Aug 26 at 11:55










  • @HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $lambda$ you can perform row operations (also exchange of rows) on $A-lambda I$ in order to find an eigenvector. Here, $lambda$ is zero, so you do the stuff with $A$.
    – amsmath
    Aug 26 at 11:58











  • Also in this example why is $A^k = vec0$ for every $k>=3$ ? I can see why $A^k = vec0$ but why do we need the $k>=3$?
    – Harry Touloupas
    Aug 26 at 12:15











  • @HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
    – José Carlos Santos
    Aug 26 at 12:31















up vote
0
down vote



accepted










Since you have a $0$ there, exchange the first and the second rows, thereby getting$$beginbmatrix-1&0&costheta\0&1&sintheta\-sintheta&costheta&0endbmatrix.$$Now, proceed in the usual way. In the end, you will get that the eigenvectors with eigenvalue $a$ are those vectors of the form $lambda(costheta,sintheta,1)$.






share|cite|improve this answer




















  • My question is can this be done when I am trying to find the eigenvectors of any matrix?
    – Harry Touloupas
    Aug 26 at 11:54










  • @HarryTouloupas No. But it can be done when the eigenvalue is $0$.
    – José Carlos Santos
    Aug 26 at 11:55










  • @HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $lambda$ you can perform row operations (also exchange of rows) on $A-lambda I$ in order to find an eigenvector. Here, $lambda$ is zero, so you do the stuff with $A$.
    – amsmath
    Aug 26 at 11:58











  • Also in this example why is $A^k = vec0$ for every $k>=3$ ? I can see why $A^k = vec0$ but why do we need the $k>=3$?
    – Harry Touloupas
    Aug 26 at 12:15











  • @HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
    – José Carlos Santos
    Aug 26 at 12:31













up vote
0
down vote



accepted







up vote
0
down vote



accepted






Since you have a $0$ there, exchange the first and the second rows, thereby getting$$beginbmatrix-1&0&costheta\0&1&sintheta\-sintheta&costheta&0endbmatrix.$$Now, proceed in the usual way. In the end, you will get that the eigenvectors with eigenvalue $a$ are those vectors of the form $lambda(costheta,sintheta,1)$.






share|cite|improve this answer












Since you have a $0$ there, exchange the first and the second rows, thereby getting$$beginbmatrix-1&0&costheta\0&1&sintheta\-sintheta&costheta&0endbmatrix.$$Now, proceed in the usual way. In the end, you will get that the eigenvectors with eigenvalue $a$ are those vectors of the form $lambda(costheta,sintheta,1)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 26 at 11:50









José Carlos Santos

119k16101182




119k16101182











  • My question is can this be done when I am trying to find the eigenvectors of any matrix?
    – Harry Touloupas
    Aug 26 at 11:54










  • @HarryTouloupas No. But it can be done when the eigenvalue is $0$.
    – José Carlos Santos
    Aug 26 at 11:55










  • @HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $lambda$ you can perform row operations (also exchange of rows) on $A-lambda I$ in order to find an eigenvector. Here, $lambda$ is zero, so you do the stuff with $A$.
    – amsmath
    Aug 26 at 11:58











  • Also in this example why is $A^k = vec0$ for every $k>=3$ ? I can see why $A^k = vec0$ but why do we need the $k>=3$?
    – Harry Touloupas
    Aug 26 at 12:15











  • @HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
    – José Carlos Santos
    Aug 26 at 12:31

















  • My question is can this be done when I am trying to find the eigenvectors of any matrix?
    – Harry Touloupas
    Aug 26 at 11:54










  • @HarryTouloupas No. But it can be done when the eigenvalue is $0$.
    – José Carlos Santos
    Aug 26 at 11:55










  • @HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $lambda$ you can perform row operations (also exchange of rows) on $A-lambda I$ in order to find an eigenvector. Here, $lambda$ is zero, so you do the stuff with $A$.
    – amsmath
    Aug 26 at 11:58











  • Also in this example why is $A^k = vec0$ for every $k>=3$ ? I can see why $A^k = vec0$ but why do we need the $k>=3$?
    – Harry Touloupas
    Aug 26 at 12:15











  • @HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
    – José Carlos Santos
    Aug 26 at 12:31
















My question is can this be done when I am trying to find the eigenvectors of any matrix?
– Harry Touloupas
Aug 26 at 11:54




My question is can this be done when I am trying to find the eigenvectors of any matrix?
– Harry Touloupas
Aug 26 at 11:54












@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
– José Carlos Santos
Aug 26 at 11:55




@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
– José Carlos Santos
Aug 26 at 11:55












@HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $lambda$ you can perform row operations (also exchange of rows) on $A-lambda I$ in order to find an eigenvector. Here, $lambda$ is zero, so you do the stuff with $A$.
– amsmath
Aug 26 at 11:58





@HarryTouloupas For an arbitrary matrix $A$ with eigenvalue $lambda$ you can perform row operations (also exchange of rows) on $A-lambda I$ in order to find an eigenvector. Here, $lambda$ is zero, so you do the stuff with $A$.
– amsmath
Aug 26 at 11:58













Also in this example why is $A^k = vec0$ for every $k>=3$ ? I can see why $A^k = vec0$ but why do we need the $k>=3$?
– Harry Touloupas
Aug 26 at 12:15





Also in this example why is $A^k = vec0$ for every $k>=3$ ? I can see why $A^k = vec0$ but why do we need the $k>=3$?
– Harry Touloupas
Aug 26 at 12:15













@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
– José Carlos Santos
Aug 26 at 12:31





@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
– José Carlos Santos
Aug 26 at 12:31


















 

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