$A^3 = A^2$ How can $A$'s minimal polynomial look like?
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Let $K$ be a field and $A in K^n times n$ a matrix with $A^3 = A^2$.
How can $A$'s minimal polynomial $mu_A$ look like?
The only possibilities I could think of are
- $A = 0$. Then the characteristic polynomial is $P_A(t) = -t^n$.
- $A = E$, where $E$ is the $n times n$ identity matrix. Then the characteristic polynomial is $P_A(t) = (1 -t)^n$.
I am surely missing some possibilities.
How can I draw a conclusion about the minimal polynomial from characteristic polynomial, knowing that the exponents in the first are the sizes of the largest Jordan blocks.
linear-algebra matrices proof-verification jordan-normal-form minimal-polynomials
asked Aug 28 at 14:38
Viktor Glombik
193219
add a comment |Â
up vote
1
down vote
favorite
Let $K$ be a field and $A in K^n times n$ a matrix with $A^3 = A^2$.
How can $A$'s minimal polynomial $mu_A$ look like?
The only possibilities I could think of are
- $A = 0$. Then the characteristic polynomial is $P_A(t) = -t^n$.
- $A = E$, where $E$ is the $n times n$ identity matrix. Then the characteristic polynomial is $P_A(t) = (1 -t)^n$.
I am surely missing some possibilities.
How can I draw a conclusion about the minimal polynomial from characteristic polynomial, knowing that the exponents in the first are the sizes of the largest Jordan blocks.
linear-algebra matrices proof-verification jordan-normal-form minimal-polynomials
asked Aug 28 at 14:38
Viktor Glombik
193219
Note that $A^3-A^2=0$. $mu_A$ must be a divisor of $x^3-x^2$.
– Cave Johnson
Aug 28 at 14:41
Any divisor of $X^3-X^2$ can be the minimal polynomial. Don't forget rings of matrices are not integral domains.
– Bernard
Aug 28 at 14:41
Note that the minimal polynomial must divide the annhilitaor polynomial (any polynomial f such that $f(A)=0$ ).
– Panchal Shamsundar
Aug 28 at 14:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $K$ be a field and $A in K^n times n$ a matrix with $A^3 = A^2$.
How can $A$'s minimal polynomial $mu_A$ look like?
The only possibilities I could think of are
- $A = 0$. Then the characteristic polynomial is $P_A(t) = -t^n$.
- $A = E$, where $E$ is the $n times n$ identity matrix. Then the characteristic polynomial is $P_A(t) = (1 -t)^n$.
I am surely missing some possibilities.
How can I draw a conclusion about the minimal polynomial from characteristic polynomial, knowing that the exponents in the first are the sizes of the largest Jordan blocks.
linear-algebra matrices proof-verification jordan-normal-form minimal-polynomials
asked Aug 28 at 14:38
Viktor Glombik
193219
Let $K$ be a field and $A in K^n times n$ a matrix with $A^3 = A^2$.
How can $A$'s minimal polynomial $mu_A$ look like?
The only possibilities I could think of are
- $A = 0$. Then the characteristic polynomial is $P_A(t) = -t^n$.
- $A = E$, where $E$ is the $n times n$ identity matrix. Then the characteristic polynomial is $P_A(t) = (1 -t)^n$.
I am surely missing some possibilities.
How can I draw a conclusion about the minimal polynomial from characteristic polynomial, knowing that the exponents in the first are the sizes of the largest Jordan blocks.
linear-algebra matrices proof-verification jordan-normal-form minimal-polynomials
asked Aug 28 at 14:38
Viktor Glombik
193219
asked Aug 28 at 14:38
Viktor Glombik
193219
asked Aug 28 at 14:38
Viktor Glombik
193219
asked Aug 28 at 14:38
Viktor Glombik
193219
193219
Note that $A^3-A^2=0$. $mu_A$ must be a divisor of $x^3-x^2$.
– Cave Johnson
Aug 28 at 14:41
Any divisor of $X^3-X^2$ can be the minimal polynomial. Don't forget rings of matrices are not integral domains.
– Bernard
Aug 28 at 14:41
Note that the minimal polynomial must divide the annhilitaor polynomial (any polynomial f such that $f(A)=0$ ).
– Panchal Shamsundar
Aug 28 at 14:43
add a comment |Â
Note that $A^3-A^2=0$. $mu_A$ must be a divisor of $x^3-x^2$.
– Cave Johnson
Aug 28 at 14:41
Any divisor of $X^3-X^2$ can be the minimal polynomial. Don't forget rings of matrices are not integral domains.
– Bernard
Aug 28 at 14:41
Note that the minimal polynomial must divide the annhilitaor polynomial (any polynomial f such that $f(A)=0$ ).
– Panchal Shamsundar
Aug 28 at 14:43
Note that $A^3-A^2=0$. $mu_A$ must be a divisor of $x^3-x^2$.
– Cave Johnson
Aug 28 at 14:41
Note that $A^3-A^2=0$. $mu_A$ must be a divisor of $x^3-x^2$.
– Cave Johnson
Aug 28 at 14:41
Any divisor of $X^3-X^2$ can be the minimal polynomial. Don't forget rings of matrices are not integral domains.
– Bernard
Aug 28 at 14:41
Any divisor of $X^3-X^2$ can be the minimal polynomial. Don't forget rings of matrices are not integral domains.
– Bernard
Aug 28 at 14:41
Note that the minimal polynomial must divide the annhilitaor polynomial (any polynomial f such that $f(A)=0$ ).
– Panchal Shamsundar
Aug 28 at 14:43
Note that the minimal polynomial must divide the annhilitaor polynomial (any polynomial f such that $f(A)=0$ ).
– Panchal Shamsundar
Aug 28 at 14:43
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
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accepted
Let $m(x)$ be the minimal polynomial. Because of Hamilton-Cayley's theorem, we have $m(x) | x^3-x^2$.
So $m(x)$ can be $x,x-1$ (as you said) , but also
- $x(x-1)$ for example in $$beginbmatrix
1 & 0 \
0 & 0
endbmatrix$$ - $x^2$ for example in $$beginbmatrix
0 & 1 \
0 & 0
endbmatrix$$ - $x^2(x-1)$ for example in $$beginbmatrix
1 & 0 & 0 \
0 & 0 & 1 \
0 & 0 & 0
endbmatrix$$
In general, given a polynomial $p$ such that $p(A)=0$ any $m(x) | p(x)$ can be $A$'s minimal polynomial.
answered Aug 28 at 14:44
LucaMac
1,24614
So the conditions for $p$ being the minimal polynomial are $p(A) = 0$ and $p(t) vert t^3 - t^2$ ?
– Viktor Glombik
Aug 28 at 16:10
No! A polynomial of the form you said CAN be a minimal polynomial, given $A^3=A^2$. Now, of course there is only one minimal polynomial, which depends on $A$!
– LucaMac
Aug 28 at 20:16
But we're not given A in the exercise.
– Viktor Glombik
10 hours ago
Indeed, the polynomial can assume that values, but we do not know which one it is, it depends on $A$.
– LucaMac
10 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $m(x)$ be the minimal polynomial. Because of Hamilton-Cayley's theorem, we have $m(x) | x^3-x^2$.
So $m(x)$ can be $x,x-1$ (as you said) , but also
- $x(x-1)$ for example in $$beginbmatrix
1 & 0 \
0 & 0
endbmatrix$$ - $x^2$ for example in $$beginbmatrix
0 & 1 \
0 & 0
endbmatrix$$ - $x^2(x-1)$ for example in $$beginbmatrix
1 & 0 & 0 \
0 & 0 & 1 \
0 & 0 & 0
endbmatrix$$
In general, given a polynomial $p$ such that $p(A)=0$ any $m(x) | p(x)$ can be $A$'s minimal polynomial.
answered Aug 28 at 14:44
LucaMac
1,24614
So the conditions for $p$ being the minimal polynomial are $p(A) = 0$ and $p(t) vert t^3 - t^2$ ?
– Viktor Glombik
Aug 28 at 16:10
No! A polynomial of the form you said CAN be a minimal polynomial, given $A^3=A^2$. Now, of course there is only one minimal polynomial, which depends on $A$!
– LucaMac
Aug 28 at 20:16
But we're not given A in the exercise.
– Viktor Glombik
10 hours ago
Indeed, the polynomial can assume that values, but we do not know which one it is, it depends on $A$.
– LucaMac
10 hours ago
add a comment |Â
up vote
4
down vote
accepted
Let $m(x)$ be the minimal polynomial. Because of Hamilton-Cayley's theorem, we have $m(x) | x^3-x^2$.
So $m(x)$ can be $x,x-1$ (as you said) , but also
- $x(x-1)$ for example in $$beginbmatrix
1 & 0 \
0 & 0
endbmatrix$$ - $x^2$ for example in $$beginbmatrix
0 & 1 \
0 & 0
endbmatrix$$ - $x^2(x-1)$ for example in $$beginbmatrix
1 & 0 & 0 \
0 & 0 & 1 \
0 & 0 & 0
endbmatrix$$
In general, given a polynomial $p$ such that $p(A)=0$ any $m(x) | p(x)$ can be $A$'s minimal polynomial.
answered Aug 28 at 14:44
LucaMac
1,24614
So the conditions for $p$ being the minimal polynomial are $p(A) = 0$ and $p(t) vert t^3 - t^2$ ?
– Viktor Glombik
Aug 28 at 16:10
No! A polynomial of the form you said CAN be a minimal polynomial, given $A^3=A^2$. Now, of course there is only one minimal polynomial, which depends on $A$!
– LucaMac
Aug 28 at 20:16
But we're not given A in the exercise.
– Viktor Glombik
10 hours ago
Indeed, the polynomial can assume that values, but we do not know which one it is, it depends on $A$.
– LucaMac
10 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $m(x)$ be the minimal polynomial. Because of Hamilton-Cayley's theorem, we have $m(x) | x^3-x^2$.
So $m(x)$ can be $x,x-1$ (as you said) , but also
- $x(x-1)$ for example in $$beginbmatrix
1 & 0 \
0 & 0
endbmatrix$$ - $x^2$ for example in $$beginbmatrix
0 & 1 \
0 & 0
endbmatrix$$ - $x^2(x-1)$ for example in $$beginbmatrix
1 & 0 & 0 \
0 & 0 & 1 \
0 & 0 & 0
endbmatrix$$
In general, given a polynomial $p$ such that $p(A)=0$ any $m(x) | p(x)$ can be $A$'s minimal polynomial.
answered Aug 28 at 14:44
LucaMac
1,24614
Let $m(x)$ be the minimal polynomial. Because of Hamilton-Cayley's theorem, we have $m(x) | x^3-x^2$.
So $m(x)$ can be $x,x-1$ (as you said) , but also
- $x(x-1)$ for example in $$beginbmatrix
1 & 0 \
0 & 0
endbmatrix$$ - $x^2$ for example in $$beginbmatrix
0 & 1 \
0 & 0
endbmatrix$$ - $x^2(x-1)$ for example in $$beginbmatrix
1 & 0 & 0 \
0 & 0 & 1 \
0 & 0 & 0
endbmatrix$$
In general, given a polynomial $p$ such that $p(A)=0$ any $m(x) | p(x)$ can be $A$'s minimal polynomial.
answered Aug 28 at 14:44
LucaMac
1,24614
answered Aug 28 at 14:44
LucaMac
1,24614
answered Aug 28 at 14:44
LucaMac
1,24614
answered Aug 28 at 14:44
LucaMac
1,24614
1,24614
So the conditions for $p$ being the minimal polynomial are $p(A) = 0$ and $p(t) vert t^3 - t^2$ ?
– Viktor Glombik
Aug 28 at 16:10
No! A polynomial of the form you said CAN be a minimal polynomial, given $A^3=A^2$. Now, of course there is only one minimal polynomial, which depends on $A$!
– LucaMac
Aug 28 at 20:16
But we're not given A in the exercise.
– Viktor Glombik
10 hours ago
Indeed, the polynomial can assume that values, but we do not know which one it is, it depends on $A$.
– LucaMac
10 hours ago
add a comment |Â
So the conditions for $p$ being the minimal polynomial are $p(A) = 0$ and $p(t) vert t^3 - t^2$ ?
– Viktor Glombik
Aug 28 at 16:10
No! A polynomial of the form you said CAN be a minimal polynomial, given $A^3=A^2$. Now, of course there is only one minimal polynomial, which depends on $A$!
– LucaMac
Aug 28 at 20:16
But we're not given A in the exercise.
– Viktor Glombik
10 hours ago
Indeed, the polynomial can assume that values, but we do not know which one it is, it depends on $A$.
– LucaMac
10 hours ago
So the conditions for $p$ being the minimal polynomial are $p(A) = 0$ and $p(t) vert t^3 - t^2$ ?
– Viktor Glombik
Aug 28 at 16:10
So the conditions for $p$ being the minimal polynomial are $p(A) = 0$ and $p(t) vert t^3 - t^2$ ?
– Viktor Glombik
Aug 28 at 16:10
No! A polynomial of the form you said CAN be a minimal polynomial, given $A^3=A^2$. Now, of course there is only one minimal polynomial, which depends on $A$!
– LucaMac
Aug 28 at 20:16
No! A polynomial of the form you said CAN be a minimal polynomial, given $A^3=A^2$. Now, of course there is only one minimal polynomial, which depends on $A$!
– LucaMac
Aug 28 at 20:16
But we're not given A in the exercise.
– Viktor Glombik
10 hours ago
But we're not given A in the exercise.
– Viktor Glombik
10 hours ago
Indeed, the polynomial can assume that values, but we do not know which one it is, it depends on $A$.
– LucaMac
10 hours ago
Indeed, the polynomial can assume that values, but we do not know which one it is, it depends on $A$.
– LucaMac
10 hours ago
add a comment |Â
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Note that $A^3-A^2=0$. $mu_A$ must be a divisor of $x^3-x^2$.
– Cave Johnson
Aug 28 at 14:41
Any divisor of $X^3-X^2$ can be the minimal polynomial. Don't forget rings of matrices are not integral domains.
– Bernard
Aug 28 at 14:41
Note that the minimal polynomial must divide the annhilitaor polynomial (any polynomial f such that $f(A)=0$ ).
– Panchal Shamsundar
Aug 28 at 14:43