finding geometric multiplicity of a given matrix
$begingroup$
Let
$A =begin{bmatrix} a & 2f & 0 \ 2f & b & 3f \ 0 & 3f & c end{bmatrix}, tag 1$
where $a, b, c, f$ are real numbers and $f ne 0$. What is the geometric multiplicity of the largest eigenvalue of $A$?
For a symmetric matrix the geometric multiplicity equals algebraic multiplicity,so we can check the eigen values of the matrix but I find that difficult to solve. Is there any different approach to this problem ? Can anyone give some idea.
Thanks in advance.
linear-algebra
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
asked Dec 18 '18 at 17:01
math mathmath math
186
$endgroup$
add a comment |
$begingroup$
Let
$A =begin{bmatrix} a & 2f & 0 \ 2f & b & 3f \ 0 & 3f & c end{bmatrix}, tag 1$
where $a, b, c, f$ are real numbers and $f ne 0$. What is the geometric multiplicity of the largest eigenvalue of $A$?
For a symmetric matrix the geometric multiplicity equals algebraic multiplicity,so we can check the eigen values of the matrix but I find that difficult to solve. Is there any different approach to this problem ? Can anyone give some idea.
Thanks in advance.
linear-algebra
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
asked Dec 18 '18 at 17:01
math mathmath math
186
$endgroup$
$begingroup$
I edited your post to $LaTeX$ify it properly. Cheers!
$endgroup$
– Robert Lewis
Dec 18 '18 at 17:51
add a comment |
$begingroup$
Let
$A =begin{bmatrix} a & 2f & 0 \ 2f & b & 3f \ 0 & 3f & c end{bmatrix}, tag 1$
where $a, b, c, f$ are real numbers and $f ne 0$. What is the geometric multiplicity of the largest eigenvalue of $A$?
For a symmetric matrix the geometric multiplicity equals algebraic multiplicity,so we can check the eigen values of the matrix but I find that difficult to solve. Is there any different approach to this problem ? Can anyone give some idea.
Thanks in advance.
linear-algebra
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
asked Dec 18 '18 at 17:01
math mathmath math
186
$endgroup$
Let
$A =begin{bmatrix} a & 2f & 0 \ 2f & b & 3f \ 0 & 3f & c end{bmatrix}, tag 1$
where $a, b, c, f$ are real numbers and $f ne 0$. What is the geometric multiplicity of the largest eigenvalue of $A$?
For a symmetric matrix the geometric multiplicity equals algebraic multiplicity,so we can check the eigen values of the matrix but I find that difficult to solve. Is there any different approach to this problem ? Can anyone give some idea.
Thanks in advance.
linear-algebra
linear-algebra
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
asked Dec 18 '18 at 17:01
math mathmath math
186
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
asked Dec 18 '18 at 17:01
math mathmath math
186
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
edited Dec 18 '18 at 17:48
Robert Lewis
48.1k23167
48.1k23167
asked Dec 18 '18 at 17:01
math mathmath math
186
asked Dec 18 '18 at 17:01
math mathmath math
186
asked Dec 18 '18 at 17:01
math mathmath math
186
186
$begingroup$
I edited your post to $LaTeX$ify it properly. Cheers!
$endgroup$
– Robert Lewis
Dec 18 '18 at 17:51
add a comment |
$begingroup$
I edited your post to $LaTeX$ify it properly. Cheers!
$endgroup$
– Robert Lewis
Dec 18 '18 at 17:51
$begingroup$
I edited your post to $LaTeX$ify it properly. Cheers!
$endgroup$
– Robert Lewis
Dec 18 '18 at 17:51
$begingroup$
I edited your post to $LaTeX$ify it properly. Cheers!
$endgroup$
– Robert Lewis
Dec 18 '18 at 17:51
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: If we can show that the matrix has distinct eigenvalues then, we can say that the geometric multiplicity of the largest eigen value is 1.
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
$endgroup$
add a comment |
$begingroup$
Hint: If $lambda$ has the geometric multiplicity of $n,$ then
$$
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda
end{bmatrix}
$$
has the rank $3-n.$
A rank of $1$ means that each row (or each column) is a multiple of the same vector.
A rank of $0$ means that the matrix is the zero-matrix.
Can any of these happen?
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
$endgroup$
add a comment |
$begingroup$
As you claimed, the matrix is symmetric hence diagonalizable, and thus is the geometric multiplicity of any eigenvalue equal to its algebraic multiplicity.
If all $a,b,c,f$ are positive, by Perron-Frobenius theorem is the eigenvalue with the largest absolute value real and positive, and is simple (multiplicity 1). In other words, the spectral radius of the matrix is also an eigenvalue.
answered Jan 2 at 23:24
user376343user376343
3,9584829
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: If we can show that the matrix has distinct eigenvalues then, we can say that the geometric multiplicity of the largest eigen value is 1.
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
$endgroup$
add a comment |
$begingroup$
Hint: If we can show that the matrix has distinct eigenvalues then, we can say that the geometric multiplicity of the largest eigen value is 1.
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
$endgroup$
add a comment |
$begingroup$
Hint: If we can show that the matrix has distinct eigenvalues then, we can say that the geometric multiplicity of the largest eigen value is 1.
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
$endgroup$
Hint: If we can show that the matrix has distinct eigenvalues then, we can say that the geometric multiplicity of the largest eigen value is 1.
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
answered Jan 2 at 12:29
shreyas s rshreyas s r
1
1
add a comment |
add a comment |
$begingroup$
Hint: If $lambda$ has the geometric multiplicity of $n,$ then
$$
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda
end{bmatrix}
$$
has the rank $3-n.$
A rank of $1$ means that each row (or each column) is a multiple of the same vector.
A rank of $0$ means that the matrix is the zero-matrix.
Can any of these happen?
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
$endgroup$
add a comment |
$begingroup$
Hint: If $lambda$ has the geometric multiplicity of $n,$ then
$$
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda
end{bmatrix}
$$
has the rank $3-n.$
A rank of $1$ means that each row (or each column) is a multiple of the same vector.
A rank of $0$ means that the matrix is the zero-matrix.
Can any of these happen?
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
$endgroup$
add a comment |
$begingroup$
Hint: If $lambda$ has the geometric multiplicity of $n,$ then
$$
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda
end{bmatrix}
$$
has the rank $3-n.$
A rank of $1$ means that each row (or each column) is a multiple of the same vector.
A rank of $0$ means that the matrix is the zero-matrix.
Can any of these happen?
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
$endgroup$
Hint: If $lambda$ has the geometric multiplicity of $n,$ then
$$
begin{bmatrix}
a-lambda & 2f & 0 \
2f & b-lambda & 3f \
0 & 3f & c-lambda
end{bmatrix}
$$
has the rank $3-n.$
A rank of $1$ means that each row (or each column) is a multiple of the same vector.
A rank of $0$ means that the matrix is the zero-matrix.
Can any of these happen?
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
answered Jan 2 at 15:35
Reinhard MeierReinhard Meier
2,912310
2,912310
add a comment |
add a comment |
$begingroup$
As you claimed, the matrix is symmetric hence diagonalizable, and thus is the geometric multiplicity of any eigenvalue equal to its algebraic multiplicity.
If all $a,b,c,f$ are positive, by Perron-Frobenius theorem is the eigenvalue with the largest absolute value real and positive, and is simple (multiplicity 1). In other words, the spectral radius of the matrix is also an eigenvalue.
answered Jan 2 at 23:24
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
As you claimed, the matrix is symmetric hence diagonalizable, and thus is the geometric multiplicity of any eigenvalue equal to its algebraic multiplicity.
If all $a,b,c,f$ are positive, by Perron-Frobenius theorem is the eigenvalue with the largest absolute value real and positive, and is simple (multiplicity 1). In other words, the spectral radius of the matrix is also an eigenvalue.
answered Jan 2 at 23:24
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
As you claimed, the matrix is symmetric hence diagonalizable, and thus is the geometric multiplicity of any eigenvalue equal to its algebraic multiplicity.
If all $a,b,c,f$ are positive, by Perron-Frobenius theorem is the eigenvalue with the largest absolute value real and positive, and is simple (multiplicity 1). In other words, the spectral radius of the matrix is also an eigenvalue.
answered Jan 2 at 23:24
user376343user376343
3,9584829
$endgroup$
As you claimed, the matrix is symmetric hence diagonalizable, and thus is the geometric multiplicity of any eigenvalue equal to its algebraic multiplicity.
If all $a,b,c,f$ are positive, by Perron-Frobenius theorem is the eigenvalue with the largest absolute value real and positive, and is simple (multiplicity 1). In other words, the spectral radius of the matrix is also an eigenvalue.
answered Jan 2 at 23:24
user376343user376343
3,9584829
answered Jan 2 at 23:24
user376343user376343
3,9584829
answered Jan 2 at 23:24
user376343user376343
3,9584829
answered Jan 2 at 23:24
user376343user376343
3,9584829
3,9584829
add a comment |
add a comment |
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$begingroup$
I edited your post to $LaTeX$ify it properly. Cheers!
$endgroup$
– Robert Lewis
Dec 18 '18 at 17:51