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Let V is a complex vector space and if $A$ is a linear transformation such that spectar of $A=5,6$. Prove that $(A-5I)^n-1circ(A-6I)^n-1=O$. And $n=dim V$.
My teacher said that when you do not know on $n$-dimensional space you put down on 2 space that is easy, and I use that $n=2$. Now I know the spectrum of $(A-5I)=0,1$ and the spectrum of $(A-6I)=-1,0$ now I define linear transformation for some vector $(A-5I)(x,y)=(0,y)$ and $(A-6I)(x,y)=(-x,0)$ from here we can see that $operatornameim(A-6I)=ker(A-5I)$ so than $(A-5I)circ(A-6I)=O$, but what do you think is that enough?
linear-algebra
edited Aug 16 at 10:02
Bernard
111k635103
asked Aug 16 at 9:55
Marko Škorić
876
add a comment |Â
up vote
-2
down vote
favorite
Let V is a complex vector space and if $A$ is a linear transformation such that spectar of $A=5,6$. Prove that $(A-5I)^n-1circ(A-6I)^n-1=O$. And $n=dim V$.
My teacher said that when you do not know on $n$-dimensional space you put down on 2 space that is easy, and I use that $n=2$. Now I know the spectrum of $(A-5I)=0,1$ and the spectrum of $(A-6I)=-1,0$ now I define linear transformation for some vector $(A-5I)(x,y)=(0,y)$ and $(A-6I)(x,y)=(-x,0)$ from here we can see that $operatornameim(A-6I)=ker(A-5I)$ so than $(A-5I)circ(A-6I)=O$, but what do you think is that enough?
linear-algebra
edited Aug 16 at 10:02
Bernard
111k635103
asked Aug 16 at 9:55
Marko Škorić
876
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let V is a complex vector space and if $A$ is a linear transformation such that spectar of $A=5,6$. Prove that $(A-5I)^n-1circ(A-6I)^n-1=O$. And $n=dim V$.
My teacher said that when you do not know on $n$-dimensional space you put down on 2 space that is easy, and I use that $n=2$. Now I know the spectrum of $(A-5I)=0,1$ and the spectrum of $(A-6I)=-1,0$ now I define linear transformation for some vector $(A-5I)(x,y)=(0,y)$ and $(A-6I)(x,y)=(-x,0)$ from here we can see that $operatornameim(A-6I)=ker(A-5I)$ so than $(A-5I)circ(A-6I)=O$, but what do you think is that enough?
linear-algebra
edited Aug 16 at 10:02
Bernard
111k635103
asked Aug 16 at 9:55
Marko Škorić
876
Let V is a complex vector space and if $A$ is a linear transformation such that spectar of $A=5,6$. Prove that $(A-5I)^n-1circ(A-6I)^n-1=O$. And $n=dim V$.
My teacher said that when you do not know on $n$-dimensional space you put down on 2 space that is easy, and I use that $n=2$. Now I know the spectrum of $(A-5I)=0,1$ and the spectrum of $(A-6I)=-1,0$ now I define linear transformation for some vector $(A-5I)(x,y)=(0,y)$ and $(A-6I)(x,y)=(-x,0)$ from here we can see that $operatornameim(A-6I)=ker(A-5I)$ so than $(A-5I)circ(A-6I)=O$, but what do you think is that enough?
linear-algebra
edited Aug 16 at 10:02
Bernard
111k635103
asked Aug 16 at 9:55
Marko Škorić
876
edited Aug 16 at 10:02
Bernard
111k635103
edited Aug 16 at 10:02
Bernard
111k635103
edited Aug 16 at 10:02
Bernard
111k635103
111k635103
asked Aug 16 at 9:55
Marko Škorić
876
asked Aug 16 at 9:55
Marko Škorić
876
asked Aug 16 at 9:55
Marko Škorić
876
876
add a comment |Â
add a comment |Â
2 Answers
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No, you cannot define the linear transformation. Every square matrix satisfies its characteristic equaltion. Note that $5$ cannot be a root of order $n$ since $6$ is also a root. Hence $5$ is a root of order at most $n-1$. same thing is true of $6$. Hence $(lambda -5)^n-1 (lambda -6)^n-1$ is divisible by the characteristic polynomial. Hence $(A -5I)^n-1 (A-6I)^n-1=0$.
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
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Since the spectrum of $A$ is $5,6$, the char. polynomial $p$ of $A$ is given by
$p(z)=(z-5)^k(z-6)^l$,
with $k,l ge 1$ and $k+l=n$. By Cayley - Hamilton:
$0=p(A)=(A-5I)^k(A-6I)^l$.
Can you proceed ?
answered Aug 16 at 10:00
Fred
38k1238
I see that, then for every number of k and l will be zero in my case k=n-1 and l=n-1
– Marko Å korić
Aug 16 at 10:07
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
No, you cannot define the linear transformation. Every square matrix satisfies its characteristic equaltion. Note that $5$ cannot be a root of order $n$ since $6$ is also a root. Hence $5$ is a root of order at most $n-1$. same thing is true of $6$. Hence $(lambda -5)^n-1 (lambda -6)^n-1$ is divisible by the characteristic polynomial. Hence $(A -5I)^n-1 (A-6I)^n-1=0$.
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
add a comment |Â
up vote
0
down vote
accepted
No, you cannot define the linear transformation. Every square matrix satisfies its characteristic equaltion. Note that $5$ cannot be a root of order $n$ since $6$ is also a root. Hence $5$ is a root of order at most $n-1$. same thing is true of $6$. Hence $(lambda -5)^n-1 (lambda -6)^n-1$ is divisible by the characteristic polynomial. Hence $(A -5I)^n-1 (A-6I)^n-1=0$.
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
No, you cannot define the linear transformation. Every square matrix satisfies its characteristic equaltion. Note that $5$ cannot be a root of order $n$ since $6$ is also a root. Hence $5$ is a root of order at most $n-1$. same thing is true of $6$. Hence $(lambda -5)^n-1 (lambda -6)^n-1$ is divisible by the characteristic polynomial. Hence $(A -5I)^n-1 (A-6I)^n-1=0$.
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
No, you cannot define the linear transformation. Every square matrix satisfies its characteristic equaltion. Note that $5$ cannot be a root of order $n$ since $6$ is also a root. Hence $5$ is a root of order at most $n-1$. same thing is true of $6$. Hence $(lambda -5)^n-1 (lambda -6)^n-1$ is divisible by the characteristic polynomial. Hence $(A -5I)^n-1 (A-6I)^n-1=0$.
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
answered Aug 16 at 10:04
Kavi Rama Murthy
22.6k2933
22.6k2933
add a comment |Â
add a comment |Â
up vote
0
down vote
Since the spectrum of $A$ is $5,6$, the char. polynomial $p$ of $A$ is given by
$p(z)=(z-5)^k(z-6)^l$,
with $k,l ge 1$ and $k+l=n$. By Cayley - Hamilton:
$0=p(A)=(A-5I)^k(A-6I)^l$.
Can you proceed ?
answered Aug 16 at 10:00
Fred
38k1238
I see that, then for every number of k and l will be zero in my case k=n-1 and l=n-1
– Marko Å korić
Aug 16 at 10:07
add a comment |Â
up vote
0
down vote
Since the spectrum of $A$ is $5,6$, the char. polynomial $p$ of $A$ is given by
$p(z)=(z-5)^k(z-6)^l$,
with $k,l ge 1$ and $k+l=n$. By Cayley - Hamilton:
$0=p(A)=(A-5I)^k(A-6I)^l$.
Can you proceed ?
answered Aug 16 at 10:00
Fred
38k1238
I see that, then for every number of k and l will be zero in my case k=n-1 and l=n-1
– Marko Å korić
Aug 16 at 10:07
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since the spectrum of $A$ is $5,6$, the char. polynomial $p$ of $A$ is given by
$p(z)=(z-5)^k(z-6)^l$,
with $k,l ge 1$ and $k+l=n$. By Cayley - Hamilton:
$0=p(A)=(A-5I)^k(A-6I)^l$.
Can you proceed ?
answered Aug 16 at 10:00
Fred
38k1238
Since the spectrum of $A$ is $5,6$, the char. polynomial $p$ of $A$ is given by
$p(z)=(z-5)^k(z-6)^l$,
with $k,l ge 1$ and $k+l=n$. By Cayley - Hamilton:
$0=p(A)=(A-5I)^k(A-6I)^l$.
Can you proceed ?
answered Aug 16 at 10:00
Fred
38k1238
answered Aug 16 at 10:00
Fred
38k1238
answered Aug 16 at 10:00
Fred
38k1238
answered Aug 16 at 10:00
Fred
38k1238
38k1238
I see that, then for every number of k and l will be zero in my case k=n-1 and l=n-1
– Marko Å korić
Aug 16 at 10:07
add a comment |Â
I see that, then for every number of k and l will be zero in my case k=n-1 and l=n-1
– Marko Å korić
Aug 16 at 10:07
I see that, then for every number of k and l will be zero in my case k=n-1 and l=n-1
– Marko Å korić
Aug 16 at 10:07
I see that, then for every number of k and l will be zero in my case k=n-1 and l=n-1
– Marko Å korić
Aug 16 at 10:07
add a comment |Â
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