Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4...
$begingroup$
Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.
So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.
So my first question is: Can you have a submodule of rationals in a module over the integers?
If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.
(Is this correct?)
linear-algebra
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
$endgroup$
add a comment |
$begingroup$
Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.
So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.
So my first question is: Can you have a submodule of rationals in a module over the integers?
If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.
(Is this correct?)
linear-algebra
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
$endgroup$
$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35
$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41
1
$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56
add a comment |
$begingroup$
Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.
So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.
So my first question is: Can you have a submodule of rationals in a module over the integers?
If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.
(Is this correct?)
linear-algebra
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
$endgroup$
Let $V=mathbb{Z}^2$ and let $L$ be the submodule of $V$ spanned by the columns of $A=begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}$. Find a basis $(overrightarrow{alpha_1}, overrightarrow{alpha_2})$ of $V$ and integers $c_1,c_2$ so that $c_1overrightarrow{alpha_1}$, $c_2overrightarrow{alpha_2}$ is a basis for $L$.
So I find that the characteristic polynomial for this matrix is $x^2-18x+40$. Hence it's roots are $x=9pm sqrt{41}$.
So my first question is: Can you have a submodule of rationals in a module over the integers?
If so, then I start to find the eigenvectors
$$
begin{bmatrix} 3pm sqrt{41} & -4 \ -8 & -3pm sqrt{41} end{bmatrix} = begin{bmatrix} 3 & -4 \ -8 & -3 end{bmatrix} pm sqrt{41} cdot I
$$
Since the since term in the sum is a linear combination of the basis in $mathbb{Z}$, I disregarded it and try to find the eigenvector using the first term.
(Is this correct?)
linear-algebra
linear-algebra
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
edited Dec 24 '18 at 14:45
Username Unknown
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
asked Dec 24 '18 at 2:02
Username UnknownUsername Unknown
1,26442158
1,26442158
$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35
$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41
1
$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56
add a comment |
$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35
$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41
1
$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56
$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35
$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35
$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41
$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41
1
1
$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56
$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Perform elementary column operations:
$$
begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
$$
Compare this with
$$
begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
$$
whose columns are a basis for $V$.
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Perform elementary column operations:
$$
begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
$$
Compare this with
$$
begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
$$
whose columns are a basis for $V$.
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
Perform elementary column operations:
$$
begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
$$
Compare this with
$$
begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
$$
whose columns are a basis for $V$.
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
Perform elementary column operations:
$$
begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
$$
Compare this with
$$
begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
$$
whose columns are a basis for $V$.
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
$endgroup$
Perform elementary column operations:
$$
begin{bmatrix} 6 & 4 \8 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 4 \-4 & 12 end{bmatrix}
to
begin{bmatrix} 2 & 0 \-4 & 20 end{bmatrix}
$$
Compare this with
$$
begin{bmatrix} 1 & 0 \-2 & 1 end{bmatrix}
$$
whose columns are a basis for $V$.
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
edited Dec 24 '18 at 15:00
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
answered Dec 24 '18 at 14:52
lhflhf
165k10171396
165k10171396
add a comment |
add a comment |
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$begingroup$
Why are you trying to find the eigenvalues of $A$? For your other question: yes, you can have a submodule of rations in a module over the integers (let the module be $mathbb{Q}$?). I suspect you meant to ask something else, and if so, please phrase your question more clearly.
$endgroup$
– Aniruddh Agarwal
Dec 24 '18 at 8:35
$begingroup$
Then how do I find the basis for $L$?
$endgroup$
– Username Unknown
Dec 24 '18 at 14:41
1
$begingroup$
See en.wikipedia.org/wiki/Smith_normal_form
$endgroup$
– lhf
Dec 24 '18 at 14:56