Eigenvalues and eigenvector of a matrix containing $sin theta$ and $cos theta$
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1
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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.
linear-algebra matrices eigenvalues-eigenvectors
add a comment |Â
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.
linear-algebra matrices eigenvalues-eigenvectors
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 $1$.
â amsmath
Aug 26 at 11:54
Exchanging rows is an elementary row operation.
â amd
Aug 26 at 16:27
add a comment |Â
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.
linear-algebra matrices eigenvalues-eigenvectors
$$
A
= beginbmatrix
0 & 1 & -sin theta \
-1 & 0 & cos theta \
-sin theta & cos theta & 0
endbmatrix
$$
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.
linear-algebra matrices eigenvalues-eigenvectors
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 $1$.
â amsmath
Aug 26 at 11:54
Exchanging rows is an elementary row operation.
â amd
Aug 26 at 16:27
add a comment |Â
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 $1$.
â amsmath
Aug 26 at 11:54
Exchanging rows is an elementary row operation.
â 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 $1$.
â 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 $1$.
â amsmath
Aug 26 at 11:54
Exchanging rows is an elementary row operation.
â amd
Aug 26 at 16:27
Exchanging rows is an elementary row operation.
â amd
Aug 26 at 16:27
add a comment |Â
1 Answer
1
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 $lambda(costheta,sintheta,1)$.
My question is can this be done when I am trying to find the eigenvectors of any matrix?
â Harry Touloupas
Aug 26 at 11:54
@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
â 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 $A$.
â 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 $k>=3$?
â Harry Touloupas
Aug 26 at 12:15
@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
â José Carlos Santos
Aug 26 at 12:31
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
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 $lambda(costheta,sintheta,1)$.
My question is can this be done when I am trying to find the eigenvectors of any matrix?
â Harry Touloupas
Aug 26 at 11:54
@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
â 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 $A$.
â 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 $k>=3$?
â Harry Touloupas
Aug 26 at 12:15
@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
â José Carlos Santos
Aug 26 at 12:31
 |Â
show 2 more comments
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 $lambda(costheta,sintheta,1)$.
My question is can this be done when I am trying to find the eigenvectors of any matrix?
â Harry Touloupas
Aug 26 at 11:54
@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
â 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 $A$.
â 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 $k>=3$?
â Harry Touloupas
Aug 26 at 12:15
@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
â José Carlos Santos
Aug 26 at 12:31
 |Â
show 2 more comments
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 $lambda(costheta,sintheta,1)$.
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 $lambda(costheta,sintheta,1)$.
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?
â Harry Touloupas
Aug 26 at 11:54
@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
â 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 $A$.
â 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 $k>=3$?
â Harry Touloupas
Aug 26 at 12:15
@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
â José Carlos Santos
Aug 26 at 12:31
 |Â
show 2 more comments
My question is can this be done when I am trying to find the eigenvectors of any matrix?
â Harry Touloupas
Aug 26 at 11:54
@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
â 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 $A$.
â 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 $k>=3$?
â Harry Touloupas
Aug 26 at 12:15
@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
â 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?
â 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?
â Harry Touloupas
Aug 26 at 11:54
@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
â José Carlos Santos
Aug 26 at 11:55
@HarryTouloupas No. But it can be done when the eigenvalue is $0$.
â 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 $A$.
â 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 $A$.
â 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 $k>=3$?
â 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 $k>=3$?
â Harry Touloupas
Aug 26 at 12:15
@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
â José Carlos Santos
Aug 26 at 12:31
@HarryTouloupas Because$$A^2=beginbmatrix-cos ^2(theta ) & -sin (theta ) cos (theta ) & cos (theta ) \-sin (theta ) cos (theta ) & -sin ^2(theta ) & sin (theta ) \ -cos (theta ) & -sin (theta ) & 1endbmatrixneq0.$$
â José Carlos Santos
Aug 26 at 12:31
 |Â
show 2 more comments
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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 $1$.
â amsmath
Aug 26 at 11:54
Exchanging rows is an elementary row operation.
â amd
Aug 26 at 16:27