Calculate minimal polynomial of a matrix
$begingroup$
begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve
linear-algebra matrices minimal-polynomials
edited May 11 '17 at 4:19
Parcly Taxel
1
asked May 11 '17 at 4:08
user439852user439852
27113
$endgroup$
add a comment |
$begingroup$
begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve
linear-algebra matrices minimal-polynomials
edited May 11 '17 at 4:19
Parcly Taxel
1
asked May 11 '17 at 4:08
user439852user439852
27113
$endgroup$
add a comment |
$begingroup$
begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve
linear-algebra matrices minimal-polynomials
edited May 11 '17 at 4:19
Parcly Taxel
1
asked May 11 '17 at 4:08
user439852user439852
27113
$endgroup$
begin{bmatrix}0&1&0&1\1&0&1&0\0&1&0&1\1&0&1&0end{bmatrix}
I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve
linear-algebra matrices minimal-polynomials
linear-algebra matrices minimal-polynomials
edited May 11 '17 at 4:19
Parcly Taxel
1
asked May 11 '17 at 4:08
user439852user439852
27113
edited May 11 '17 at 4:19
Parcly Taxel
1
asked May 11 '17 at 4:08
user439852user439852
27113
edited May 11 '17 at 4:19
Parcly Taxel
1
edited May 11 '17 at 4:19
Parcly Taxel
1
edited May 11 '17 at 4:19
Parcly Taxel
1
1
asked May 11 '17 at 4:08
user439852user439852
27113
asked May 11 '17 at 4:08
user439852user439852
27113
asked May 11 '17 at 4:08
user439852user439852
27113
27113
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.
Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
$$pmatrix{1\0\0\0}overset Amapsto
pmatrix{0\1\0\1}overset Amapsto
pmatrix{2\0\2\0}overset Amapsto
pmatrix{0\4\0\4}
$$
with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.
As you see, one can do entirely without the characteristic polynomial.
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
add a comment |
$begingroup$
All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$
Hence $x(x^2-4)$ divides the minimal polynomial,
Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.
Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.
Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.
Another way to decide on the last part:
The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
$endgroup$
1
$begingroup$
Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 15:57
add a comment |
$begingroup$
As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)
answered Jan 19 '18 at 5:22
user454200user454200
412
$endgroup$
$begingroup$
@ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
$endgroup$
– Jose Brox
Dec 5 '18 at 20:08
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.
Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
$$pmatrix{1\0\0\0}overset Amapsto
pmatrix{0\1\0\1}overset Amapsto
pmatrix{2\0\2\0}overset Amapsto
pmatrix{0\4\0\4}
$$
with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.
As you see, one can do entirely without the characteristic polynomial.
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
add a comment |
$begingroup$
These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.
Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
$$pmatrix{1\0\0\0}overset Amapsto
pmatrix{0\1\0\1}overset Amapsto
pmatrix{2\0\2\0}overset Amapsto
pmatrix{0\4\0\4}
$$
with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.
As you see, one can do entirely without the characteristic polynomial.
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
add a comment |
$begingroup$
These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.
Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
$$pmatrix{1\0\0\0}overset Amapsto
pmatrix{0\1\0\1}overset Amapsto
pmatrix{2\0\2\0}overset Amapsto
pmatrix{0\4\0\4}
$$
with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.
As you see, one can do entirely without the characteristic polynomial.
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
$endgroup$
These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $lambda=0$), which is a huge give-away. But I'll apply a general method instead.
Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
$$pmatrix{1\0\0\0}overset Amapsto
pmatrix{0\1\0\1}overset Amapsto
pmatrix{2\0\2\0}overset Amapsto
pmatrix{0\4\0\4}
$$
with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.
As you see, one can do entirely without the characteristic polynomial.
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
edited May 11 '17 at 9:25
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
answered May 11 '17 at 4:42
Marc van LeeuwenMarc van Leeuwen
87.6k5110225
87.6k5110225
add a comment |
add a comment |
$begingroup$
All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$
Hence $x(x^2-4)$ divides the minimal polynomial,
Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.
Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.
Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.
Another way to decide on the last part:
The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
$endgroup$
1
$begingroup$
Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 15:57
add a comment |
$begingroup$
All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$
Hence $x(x^2-4)$ divides the minimal polynomial,
Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.
Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.
Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.
Another way to decide on the last part:
The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
$endgroup$
1
$begingroup$
Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 15:57
add a comment |
$begingroup$
All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$
Hence $x(x^2-4)$ divides the minimal polynomial,
Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.
Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.
Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.
Another way to decide on the last part:
The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
$endgroup$
All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$
Hence $x(x^2-4)$ divides the minimal polynomial,
Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.
Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.
Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.
Another way to decide on the last part:
The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be $x^2(x^2-4)$, hence the answer is $x(x^2-4)$.
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
edited May 11 '17 at 4:43
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
answered May 11 '17 at 4:33
Arpan1729Arpan1729
2,7871320
2,7871320
1
$begingroup$
Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 15:57
add a comment |
1
$begingroup$
Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 15:57
1
1
$begingroup$
Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 15:57
$begingroup$
Alternatives to 'hence': 'this means that', 'so', 'it follows that', 'thus', 'so we see that', 'from where', etc.
$endgroup$
– Pedro Tamaroff♦
Dec 5 '18 at 15:57
add a comment |
$begingroup$
As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)
answered Jan 19 '18 at 5:22
user454200user454200
412
$endgroup$
$begingroup$
@ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
$endgroup$
– Jose Brox
Dec 5 '18 at 20:08
add a comment |
$begingroup$
As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)
answered Jan 19 '18 at 5:22
user454200user454200
412
$endgroup$
$begingroup$
@ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
$endgroup$
– Jose Brox
Dec 5 '18 at 20:08
add a comment |
$begingroup$
As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)
answered Jan 19 '18 at 5:22
user454200user454200
412
$endgroup$
As the given matrix is symmetric it is diagonalizable => its minimal polynomial has distinct roots => min polynomial = x(x-2)(x+2)
answered Jan 19 '18 at 5:22
user454200user454200
412
answered Jan 19 '18 at 5:22
user454200user454200
412
answered Jan 19 '18 at 5:22
user454200user454200
412
answered Jan 19 '18 at 5:22
user454200user454200
412
412
$begingroup$
@ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
$endgroup$
– Jose Brox
Dec 5 '18 at 20:08
add a comment |
$begingroup$
@ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
$endgroup$
– Jose Brox
Dec 5 '18 at 20:08
$begingroup$
@ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
$endgroup$
– Jose Brox
Dec 5 '18 at 20:08
$begingroup$
@ancientmathematician Yes, of course! I suppose I wrongly read characteristic polynomial or something. I'm deleting my previous misleading comment, and this one in a while. Thanks!
$endgroup$
– Jose Brox
Dec 5 '18 at 20:08
add a comment |
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