Minimal polynomial linear operator $A$ in finite dimension complex vector space ptoblem
Let $ m_{A} ( lambda) = (lambda -2)^2$ be a minimal polynomial of linear operator $A$ in finite dimension complex vector space. Find the mimimal polynomial of the operator $A^2$ .
This is what I tried:
There exists a vector v which is not zero so that
$ Av=2v$ . Then we have $ A^2(v)=A(2v)=2A(v)=4v$. So $lambda -4 $ is a factor in the minimal polynomial of $A^2$ . But how do I know if there are any other factors?
I got the same result when I tried looking at the determinants. From $ (A-2)^2 =0$ I got $A^2= 4A-4I$ and then I looked at the deterimants minus $lambda I $ but got nowhere.
linear-algebra
asked Nov 28 '18 at 15:35
user15269
1608
add a comment |
Let $ m_{A} ( lambda) = (lambda -2)^2$ be a minimal polynomial of linear operator $A$ in finite dimension complex vector space. Find the mimimal polynomial of the operator $A^2$ .
This is what I tried:
There exists a vector v which is not zero so that
$ Av=2v$ . Then we have $ A^2(v)=A(2v)=2A(v)=4v$. So $lambda -4 $ is a factor in the minimal polynomial of $A^2$ . But how do I know if there are any other factors?
I got the same result when I tried looking at the determinants. From $ (A-2)^2 =0$ I got $A^2= 4A-4I$ and then I looked at the deterimants minus $lambda I $ but got nowhere.
linear-algebra
asked Nov 28 '18 at 15:35
user15269
1608
Possibly relevant fact: If $A,B$ are two commuting operators then every eigenspace of $A$ is an eigenspace of $B$, and vice versa.
– user25959
Nov 28 '18 at 15:48
add a comment |
Let $ m_{A} ( lambda) = (lambda -2)^2$ be a minimal polynomial of linear operator $A$ in finite dimension complex vector space. Find the mimimal polynomial of the operator $A^2$ .
This is what I tried:
There exists a vector v which is not zero so that
$ Av=2v$ . Then we have $ A^2(v)=A(2v)=2A(v)=4v$. So $lambda -4 $ is a factor in the minimal polynomial of $A^2$ . But how do I know if there are any other factors?
I got the same result when I tried looking at the determinants. From $ (A-2)^2 =0$ I got $A^2= 4A-4I$ and then I looked at the deterimants minus $lambda I $ but got nowhere.
linear-algebra
asked Nov 28 '18 at 15:35
user15269
1608
Let $ m_{A} ( lambda) = (lambda -2)^2$ be a minimal polynomial of linear operator $A$ in finite dimension complex vector space. Find the mimimal polynomial of the operator $A^2$ .
This is what I tried:
There exists a vector v which is not zero so that
$ Av=2v$ . Then we have $ A^2(v)=A(2v)=2A(v)=4v$. So $lambda -4 $ is a factor in the minimal polynomial of $A^2$ . But how do I know if there are any other factors?
I got the same result when I tried looking at the determinants. From $ (A-2)^2 =0$ I got $A^2= 4A-4I$ and then I looked at the deterimants minus $lambda I $ but got nowhere.
linear-algebra
linear-algebra
asked Nov 28 '18 at 15:35
user15269
1608
asked Nov 28 '18 at 15:35
user15269
1608
asked Nov 28 '18 at 15:35
user15269
1608
asked Nov 28 '18 at 15:35
user15269
1608
asked Nov 28 '18 at 15:35
user15269
1608
1608
Possibly relevant fact: If $A,B$ are two commuting operators then every eigenspace of $A$ is an eigenspace of $B$, and vice versa.
– user25959
Nov 28 '18 at 15:48
add a comment |
Possibly relevant fact: If $A,B$ are two commuting operators then every eigenspace of $A$ is an eigenspace of $B$, and vice versa.
– user25959
Nov 28 '18 at 15:48
Possibly relevant fact: If $A,B$ are two commuting operators then every eigenspace of $A$ is an eigenspace of $B$, and vice versa.
– user25959
Nov 28 '18 at 15:48
Possibly relevant fact: If $A,B$ are two commuting operators then every eigenspace of $A$ is an eigenspace of $B$, and vice versa.
– user25959
Nov 28 '18 at 15:48
add a comment |
1 Answer
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You've made a good start; the fact that $A^2=4A-4I$ implies that
$$(A^2+4I)^2-16(A^2+4I)+64I=(4A)^2-16(4A)+64I=16(A^2-4A+4I)=0,$$
and so the minimal polynomial of $A^2$ divides
$$(lambda+4)^2-16(lambda+4)+64=((lambda+4)-8)^2=(lambda-4)^2,$$
hence there are no other factors.
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
You've made a good start; the fact that $A^2=4A-4I$ implies that
$$(A^2+4I)^2-16(A^2+4I)+64I=(4A)^2-16(4A)+64I=16(A^2-4A+4I)=0,$$
and so the minimal polynomial of $A^2$ divides
$$(lambda+4)^2-16(lambda+4)+64=((lambda+4)-8)^2=(lambda-4)^2,$$
hence there are no other factors.
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
add a comment |
You've made a good start; the fact that $A^2=4A-4I$ implies that
$$(A^2+4I)^2-16(A^2+4I)+64I=(4A)^2-16(4A)+64I=16(A^2-4A+4I)=0,$$
and so the minimal polynomial of $A^2$ divides
$$(lambda+4)^2-16(lambda+4)+64=((lambda+4)-8)^2=(lambda-4)^2,$$
hence there are no other factors.
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
add a comment |
You've made a good start; the fact that $A^2=4A-4I$ implies that
$$(A^2+4I)^2-16(A^2+4I)+64I=(4A)^2-16(4A)+64I=16(A^2-4A+4I)=0,$$
and so the minimal polynomial of $A^2$ divides
$$(lambda+4)^2-16(lambda+4)+64=((lambda+4)-8)^2=(lambda-4)^2,$$
hence there are no other factors.
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
You've made a good start; the fact that $A^2=4A-4I$ implies that
$$(A^2+4I)^2-16(A^2+4I)+64I=(4A)^2-16(4A)+64I=16(A^2-4A+4I)=0,$$
and so the minimal polynomial of $A^2$ divides
$$(lambda+4)^2-16(lambda+4)+64=((lambda+4)-8)^2=(lambda-4)^2,$$
hence there are no other factors.
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
answered Nov 28 '18 at 17:58
Servaes
22.4k33793
22.4k33793
add a comment |
add a comment |
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Possibly relevant fact: If $A,B$ are two commuting operators then every eigenspace of $A$ is an eigenspace of $B$, and vice versa.
– user25959
Nov 28 '18 at 15:48