Mixing
Geometric multiplicity of the largest eigenvalue
$begingroup$
Let
$$
A= begin{bmatrix}
a & 2f & 0 \
2f & b & 3f \
0 & 3f & c
end{bmatrix},
$$
where $a$, $b$, $c$, $f$ are real numbers and $fneq 0$. Find the geometric multiplicity of the largest eigenvalue of $A$.
I don't think I have to use the characteristic equation. Or do I?
linear-algebra eigenvalues-eigenvectors
edited Jan 25 at 10:21
amWhy
1
asked Jan 17 at 8:01
A.CA.C
213
$endgroup$
add a comment |
$begingroup$
Let
$$
A= begin{bmatrix}
a & 2f & 0 \
2f & b & 3f \
0 & 3f & c
end{bmatrix},
$$
where $a$, $b$, $c$, $f$ are real numbers and $fneq 0$. Find the geometric multiplicity of the largest eigenvalue of $A$.
I don't think I have to use the characteristic equation. Or do I?
linear-algebra eigenvalues-eigenvectors
edited Jan 25 at 10:21
amWhy
1
asked Jan 17 at 8:01
A.CA.C
213
$endgroup$
$begingroup$
Welcome to MSE. You do have a math editor. It is Latex. See math.meta.stackexchange.com/questions/5020/….
$endgroup$
– mathcounterexamples.net
Jan 17 at 9:13
$begingroup$
This same question was asked a couple of weeks ago (math.stackexchange.com/q/3063378/265466) and got no response.
$endgroup$
– amd
Jan 17 at 22:50
add a comment |
$begingroup$
Let
$$
A= begin{bmatrix}
a & 2f & 0 \
2f & b & 3f \
0 & 3f & c
end{bmatrix},
$$
where $a$, $b$, $c$, $f$ are real numbers and $fneq 0$. Find the geometric multiplicity of the largest eigenvalue of $A$.
I don't think I have to use the characteristic equation. Or do I?
linear-algebra eigenvalues-eigenvectors
edited Jan 25 at 10:21
amWhy
1
asked Jan 17 at 8:01
A.CA.C
213
$endgroup$
Let
$$
A= begin{bmatrix}
a & 2f & 0 \
2f & b & 3f \
0 & 3f & c
end{bmatrix},
$$
where $a$, $b$, $c$, $f$ are real numbers and $fneq 0$. Find the geometric multiplicity of the largest eigenvalue of $A$.
I don't think I have to use the characteristic equation. Or do I?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Jan 25 at 10:21
amWhy
1
asked Jan 17 at 8:01
A.CA.C
213
edited Jan 25 at 10:21
amWhy
1
asked Jan 17 at 8:01
A.CA.C
213
edited Jan 25 at 10:21
amWhy
1
edited Jan 25 at 10:21
amWhy
1
edited Jan 25 at 10:21
amWhy
1
1
asked Jan 17 at 8:01
A.CA.C
213
asked Jan 17 at 8:01
A.CA.C
213
asked Jan 17 at 8:01
A.CA.C
213
213
$begingroup$
Welcome to MSE. You do have a math editor. It is Latex. See math.meta.stackexchange.com/questions/5020/….
$endgroup$
– mathcounterexamples.net
Jan 17 at 9:13
$begingroup$
This same question was asked a couple of weeks ago (math.stackexchange.com/q/3063378/265466) and got no response.
$endgroup$
– amd
Jan 17 at 22:50
add a comment |
$begingroup$
Welcome to MSE. You do have a math editor. It is Latex. See math.meta.stackexchange.com/questions/5020/….
$endgroup$
– mathcounterexamples.net
Jan 17 at 9:13
$begingroup$
This same question was asked a couple of weeks ago (math.stackexchange.com/q/3063378/265466) and got no response.
$endgroup$
– amd
Jan 17 at 22:50
$begingroup$
Welcome to MSE. You do have a math editor. It is Latex. See math.meta.stackexchange.com/questions/5020/….
$endgroup$
– mathcounterexamples.net
Jan 17 at 9:13
$begingroup$
Welcome to MSE. You do have a math editor. It is Latex. See math.meta.stackexchange.com/questions/5020/….
$endgroup$
– mathcounterexamples.net
Jan 17 at 9:13
$begingroup$
This same question was asked a couple of weeks ago (math.stackexchange.com/q/3063378/265466) and got no response.
$endgroup$
– amd
Jan 17 at 22:50
$begingroup$
This same question was asked a couple of weeks ago (math.stackexchange.com/q/3063378/265466) and got no response.
$endgroup$
– amd
Jan 17 at 22:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
let $lambda $ be an eigen value then
$$A - lambda I =
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda \
end{bmatrix}
$$
then $ Rank(A-lambda I )$ = 2 because
Minor wrt $a_{31}$ = $begin{vmatrix} 2f & 0 \
b-lambda & 3f \
end{vmatrix} = 6f^{2} ne 0$ as $ f ne 0$
So geometric multiplicity of $ lambda = 3 - Rank(A-lambda I ) = 1$ for all Eigen Values of this matrix irrespective of $ lambda$.
Hence the geometric multiplicity of the largest eigenvalue of A also equals 1.
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
$endgroup$
$begingroup$
You have written $N=R$ . I think $N+R=3$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. Principal minor is wrt diagonal entries, I suppose.
$endgroup$
– Widawensen
Jan 28 at 16:35
$begingroup$
@Widawensen I am new to this forum. Please correct me If I am missing something here 1. $N(A-lambda I) = Rank(A-lambda I )$ because Null space here will be equal to rank of the matrix span by $A-lambda$ right ? thanks for pointing out mistake for minor. I corrected the same
$endgroup$
– Pradeep Bihani
Jan 28 at 16:36
1
$begingroup$
From wikipedia - Nullity, the dimension of the kernel of a mathematical operator or null space of a matrix. Nullity is just =1 in this case, rank =2.
$endgroup$
– Widawensen
Jan 28 at 16:39
$begingroup$
Yes got your point. Thanks for sharing.
$endgroup$
– Pradeep Bihani
Jan 28 at 16:40
$begingroup$
O.k. Now answer is accurate.
$endgroup$
– Widawensen
Jan 28 at 16:42
add a comment |
$begingroup$
We can check the matrix for some values (matrix is symmetric so it has only real eigenvalues)
Substitute for example $a=b=c=f=1$.
For this you have $λ_1 = 1 + sqrt{13}$, $λ_2 = 1 - sqrt{13}$ , $λ_3=1$.
More generally for $a=b=c=n$ and $f $
(matrix of the form $nI+fB$ where $B= begin{bmatrix}
0 & 2 & 0 \
2 & 0 & 3 \
0 & 3 & 0
end{bmatrix}$)
you have $λ_1 = n+fsqrt{13}$, $λ_2 = n -fsqrt{13}$ , $λ_3=n$.
Evidently at least for some pattern of values multiplicity is just $1$.
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
$endgroup$
$begingroup$
See also for eigendecomposition of sym. matrices en.wikipedia.org/wiki/…
$endgroup$
– Widawensen
Jan 25 at 16:42
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
votes
$begingroup$
let $lambda $ be an eigen value then
$$A - lambda I =
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda \
end{bmatrix}
$$
then $ Rank(A-lambda I )$ = 2 because
Minor wrt $a_{31}$ = $begin{vmatrix} 2f & 0 \
b-lambda & 3f \
end{vmatrix} = 6f^{2} ne 0$ as $ f ne 0$
So geometric multiplicity of $ lambda = 3 - Rank(A-lambda I ) = 1$ for all Eigen Values of this matrix irrespective of $ lambda$.
Hence the geometric multiplicity of the largest eigenvalue of A also equals 1.
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
$endgroup$
$begingroup$
You have written $N=R$ . I think $N+R=3$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. Principal minor is wrt diagonal entries, I suppose.
$endgroup$
– Widawensen
Jan 28 at 16:35
$begingroup$
@Widawensen I am new to this forum. Please correct me If I am missing something here 1. $N(A-lambda I) = Rank(A-lambda I )$ because Null space here will be equal to rank of the matrix span by $A-lambda$ right ? thanks for pointing out mistake for minor. I corrected the same
$endgroup$
– Pradeep Bihani
Jan 28 at 16:36
1
$begingroup$
From wikipedia - Nullity, the dimension of the kernel of a mathematical operator or null space of a matrix. Nullity is just =1 in this case, rank =2.
$endgroup$
– Widawensen
Jan 28 at 16:39
$begingroup$
Yes got your point. Thanks for sharing.
$endgroup$
– Pradeep Bihani
Jan 28 at 16:40
$begingroup$
O.k. Now answer is accurate.
$endgroup$
– Widawensen
Jan 28 at 16:42
add a comment |
$begingroup$
let $lambda $ be an eigen value then
$$A - lambda I =
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda \
end{bmatrix}
$$
then $ Rank(A-lambda I )$ = 2 because
Minor wrt $a_{31}$ = $begin{vmatrix} 2f & 0 \
b-lambda & 3f \
end{vmatrix} = 6f^{2} ne 0$ as $ f ne 0$
So geometric multiplicity of $ lambda = 3 - Rank(A-lambda I ) = 1$ for all Eigen Values of this matrix irrespective of $ lambda$.
Hence the geometric multiplicity of the largest eigenvalue of A also equals 1.
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
$endgroup$
$begingroup$
You have written $N=R$ . I think $N+R=3$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. Principal minor is wrt diagonal entries, I suppose.
$endgroup$
– Widawensen
Jan 28 at 16:35
$begingroup$
@Widawensen I am new to this forum. Please correct me If I am missing something here 1. $N(A-lambda I) = Rank(A-lambda I )$ because Null space here will be equal to rank of the matrix span by $A-lambda$ right ? thanks for pointing out mistake for minor. I corrected the same
$endgroup$
– Pradeep Bihani
Jan 28 at 16:36
1
$begingroup$
From wikipedia - Nullity, the dimension of the kernel of a mathematical operator or null space of a matrix. Nullity is just =1 in this case, rank =2.
$endgroup$
– Widawensen
Jan 28 at 16:39
$begingroup$
Yes got your point. Thanks for sharing.
$endgroup$
– Pradeep Bihani
Jan 28 at 16:40
$begingroup$
O.k. Now answer is accurate.
$endgroup$
– Widawensen
Jan 28 at 16:42
add a comment |
$begingroup$
let $lambda $ be an eigen value then
$$A - lambda I =
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda \
end{bmatrix}
$$
then $ Rank(A-lambda I )$ = 2 because
Minor wrt $a_{31}$ = $begin{vmatrix} 2f & 0 \
b-lambda & 3f \
end{vmatrix} = 6f^{2} ne 0$ as $ f ne 0$
So geometric multiplicity of $ lambda = 3 - Rank(A-lambda I ) = 1$ for all Eigen Values of this matrix irrespective of $ lambda$.
Hence the geometric multiplicity of the largest eigenvalue of A also equals 1.
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
$endgroup$
let $lambda $ be an eigen value then
$$A - lambda I =
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda \
end{bmatrix}
$$
then $ Rank(A-lambda I )$ = 2 because
Minor wrt $a_{31}$ = $begin{vmatrix} 2f & 0 \
b-lambda & 3f \
end{vmatrix} = 6f^{2} ne 0$ as $ f ne 0$
So geometric multiplicity of $ lambda = 3 - Rank(A-lambda I ) = 1$ for all Eigen Values of this matrix irrespective of $ lambda$.
Hence the geometric multiplicity of the largest eigenvalue of A also equals 1.
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
edited Jan 28 at 16:40
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
answered Jan 26 at 13:19
Pradeep BihaniPradeep Bihani
499
499
$begingroup$
You have written $N=R$ . I think $N+R=3$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. Principal minor is wrt diagonal entries, I suppose.
$endgroup$
– Widawensen
Jan 28 at 16:35
$begingroup$
@Widawensen I am new to this forum. Please correct me If I am missing something here 1. $N(A-lambda I) = Rank(A-lambda I )$ because Null space here will be equal to rank of the matrix span by $A-lambda$ right ? thanks for pointing out mistake for minor. I corrected the same
$endgroup$
– Pradeep Bihani
Jan 28 at 16:36
1
$begingroup$
From wikipedia - Nullity, the dimension of the kernel of a mathematical operator or null space of a matrix. Nullity is just =1 in this case, rank =2.
$endgroup$
– Widawensen
Jan 28 at 16:39
$begingroup$
Yes got your point. Thanks for sharing.
$endgroup$
– Pradeep Bihani
Jan 28 at 16:40
$begingroup$
O.k. Now answer is accurate.
$endgroup$
– Widawensen
Jan 28 at 16:42
add a comment |
$begingroup$
You have written $N=R$ . I think $N+R=3$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. Principal minor is wrt diagonal entries, I suppose.
$endgroup$
– Widawensen
Jan 28 at 16:35
$begingroup$
@Widawensen I am new to this forum. Please correct me If I am missing something here 1. $N(A-lambda I) = Rank(A-lambda I )$ because Null space here will be equal to rank of the matrix span by $A-lambda$ right ? thanks for pointing out mistake for minor. I corrected the same
$endgroup$
– Pradeep Bihani
Jan 28 at 16:36
1
$begingroup$
From wikipedia - Nullity, the dimension of the kernel of a mathematical operator or null space of a matrix. Nullity is just =1 in this case, rank =2.
$endgroup$
– Widawensen
Jan 28 at 16:39
$begingroup$
Yes got your point. Thanks for sharing.
$endgroup$
– Pradeep Bihani
Jan 28 at 16:40
$begingroup$
O.k. Now answer is accurate.
$endgroup$
– Widawensen
Jan 28 at 16:42
$begingroup$
You have written $N=R$ . I think $N+R=3$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. Principal minor is wrt diagonal entries, I suppose.
$endgroup$
– Widawensen
Jan 28 at 16:35
$begingroup$
You have written $N=R$ . I think $N+R=3$ See en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem. Principal minor is wrt diagonal entries, I suppose.
$endgroup$
– Widawensen
Jan 28 at 16:35
$begingroup$
@Widawensen I am new to this forum. Please correct me If I am missing something here 1. $N(A-lambda I) = Rank(A-lambda I )$ because Null space here will be equal to rank of the matrix span by $A-lambda$ right ? thanks for pointing out mistake for minor. I corrected the same
$endgroup$
– Pradeep Bihani
Jan 28 at 16:36
$begingroup$
@Widawensen I am new to this forum. Please correct me If I am missing something here 1. $N(A-lambda I) = Rank(A-lambda I )$ because Null space here will be equal to rank of the matrix span by $A-lambda$ right ? thanks for pointing out mistake for minor. I corrected the same
$endgroup$
– Pradeep Bihani
Jan 28 at 16:36
1
1
$begingroup$
From wikipedia - Nullity, the dimension of the kernel of a mathematical operator or null space of a matrix. Nullity is just =1 in this case, rank =2.
$endgroup$
– Widawensen
Jan 28 at 16:39
$begingroup$
From wikipedia - Nullity, the dimension of the kernel of a mathematical operator or null space of a matrix. Nullity is just =1 in this case, rank =2.
$endgroup$
– Widawensen
Jan 28 at 16:39
$begingroup$
Yes got your point. Thanks for sharing.
$endgroup$
– Pradeep Bihani
Jan 28 at 16:40
$begingroup$
Yes got your point. Thanks for sharing.
$endgroup$
– Pradeep Bihani
Jan 28 at 16:40
$begingroup$
O.k. Now answer is accurate.
$endgroup$
– Widawensen
Jan 28 at 16:42
$begingroup$
O.k. Now answer is accurate.
$endgroup$
– Widawensen
Jan 28 at 16:42
add a comment |
$begingroup$
We can check the matrix for some values (matrix is symmetric so it has only real eigenvalues)
Substitute for example $a=b=c=f=1$.
For this you have $λ_1 = 1 + sqrt{13}$, $λ_2 = 1 - sqrt{13}$ , $λ_3=1$.
More generally for $a=b=c=n$ and $f $
(matrix of the form $nI+fB$ where $B= begin{bmatrix}
0 & 2 & 0 \
2 & 0 & 3 \
0 & 3 & 0
end{bmatrix}$)
you have $λ_1 = n+fsqrt{13}$, $λ_2 = n -fsqrt{13}$ , $λ_3=n$.
Evidently at least for some pattern of values multiplicity is just $1$.
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
$endgroup$
$begingroup$
See also for eigendecomposition of sym. matrices en.wikipedia.org/wiki/…
$endgroup$
– Widawensen
Jan 25 at 16:42
add a comment |
$begingroup$
We can check the matrix for some values (matrix is symmetric so it has only real eigenvalues)
Substitute for example $a=b=c=f=1$.
For this you have $λ_1 = 1 + sqrt{13}$, $λ_2 = 1 - sqrt{13}$ , $λ_3=1$.
More generally for $a=b=c=n$ and $f $
(matrix of the form $nI+fB$ where $B= begin{bmatrix}
0 & 2 & 0 \
2 & 0 & 3 \
0 & 3 & 0
end{bmatrix}$)
you have $λ_1 = n+fsqrt{13}$, $λ_2 = n -fsqrt{13}$ , $λ_3=n$.
Evidently at least for some pattern of values multiplicity is just $1$.
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
$endgroup$
$begingroup$
See also for eigendecomposition of sym. matrices en.wikipedia.org/wiki/…
$endgroup$
– Widawensen
Jan 25 at 16:42
add a comment |
$begingroup$
We can check the matrix for some values (matrix is symmetric so it has only real eigenvalues)
Substitute for example $a=b=c=f=1$.
For this you have $λ_1 = 1 + sqrt{13}$, $λ_2 = 1 - sqrt{13}$ , $λ_3=1$.
More generally for $a=b=c=n$ and $f $
(matrix of the form $nI+fB$ where $B= begin{bmatrix}
0 & 2 & 0 \
2 & 0 & 3 \
0 & 3 & 0
end{bmatrix}$)
you have $λ_1 = n+fsqrt{13}$, $λ_2 = n -fsqrt{13}$ , $λ_3=n$.
Evidently at least for some pattern of values multiplicity is just $1$.
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
$endgroup$
We can check the matrix for some values (matrix is symmetric so it has only real eigenvalues)
Substitute for example $a=b=c=f=1$.
For this you have $λ_1 = 1 + sqrt{13}$, $λ_2 = 1 - sqrt{13}$ , $λ_3=1$.
More generally for $a=b=c=n$ and $f $
(matrix of the form $nI+fB$ where $B= begin{bmatrix}
0 & 2 & 0 \
2 & 0 & 3 \
0 & 3 & 0
end{bmatrix}$)
you have $λ_1 = n+fsqrt{13}$, $λ_2 = n -fsqrt{13}$ , $λ_3=n$.
Evidently at least for some pattern of values multiplicity is just $1$.
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
edited Jan 25 at 17:15
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
answered Jan 25 at 16:21
WidawensenWidawensen
4,51321446
4,51321446
$begingroup$
See also for eigendecomposition of sym. matrices en.wikipedia.org/wiki/…
$endgroup$
– Widawensen
Jan 25 at 16:42
add a comment |
$begingroup$
See also for eigendecomposition of sym. matrices en.wikipedia.org/wiki/…
$endgroup$
– Widawensen
Jan 25 at 16:42
$begingroup$
See also for eigendecomposition of sym. matrices en.wikipedia.org/wiki/…
$endgroup$
– Widawensen
Jan 25 at 16:42
$begingroup$
See also for eigendecomposition of sym. matrices en.wikipedia.org/wiki/…
$endgroup$
– Widawensen
Jan 25 at 16:42
add a comment |
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$begingroup$
Welcome to MSE. You do have a math editor. It is Latex. See math.meta.stackexchange.com/questions/5020/….
$endgroup$
– mathcounterexamples.net
Jan 17 at 9:13
$begingroup$
This same question was asked a couple of weeks ago (math.stackexchange.com/q/3063378/265466) and got no response.
$endgroup$
– amd
Jan 17 at 22:50