Trace of Linear Transformations
up vote
1
down vote
favorite
I need to find the trace of the following powers of linear transformation which is already given.
My doubt is that Is there any better way than finding the matrix of linear transformation and then taking it's powers.
Because it will involve quite a lot of calculation as we will have a 10× 10 matrix.
Am I missing any trick??
Thanks and regards.
linear-algebra linear-transformations
asked Nov 13 at 13:31
Devendra Singh Rana
744216
add a comment |
up vote
1
down vote
favorite
I need to find the trace of the following powers of linear transformation which is already given.
My doubt is that Is there any better way than finding the matrix of linear transformation and then taking it's powers.
Because it will involve quite a lot of calculation as we will have a 10× 10 matrix.
Am I missing any trick??
Thanks and regards.
linear-algebra linear-transformations
asked Nov 13 at 13:31
Devendra Singh Rana
744216
Try computing eigenvalues. Note that there are some relatively obvious invariant subspaces, such as $operatorname{span}(1)$, $operatorname{span}(X, X^2)$, $operatorname{span}(X^3, X^4, X^5)$ and $operatorname{span}(X^6, X^7, X^8, X^9)$. You can break these subspaces down to eigenspaces without too much hassle.
– Theo Bendit
Nov 13 at 13:37
2
Or, given that this transformation just permutes monomials, computing powers is pretty easy to do.
– Theo Bendit
Nov 13 at 13:39
It will be helpful if you can elaborate a bit as an answer.
– Devendra Singh Rana
Nov 13 at 13:41
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to find the trace of the following powers of linear transformation which is already given.
My doubt is that Is there any better way than finding the matrix of linear transformation and then taking it's powers.
Because it will involve quite a lot of calculation as we will have a 10× 10 matrix.
Am I missing any trick??
Thanks and regards.
linear-algebra linear-transformations
asked Nov 13 at 13:31
Devendra Singh Rana
744216
I need to find the trace of the following powers of linear transformation which is already given.
My doubt is that Is there any better way than finding the matrix of linear transformation and then taking it's powers.
Because it will involve quite a lot of calculation as we will have a 10× 10 matrix.
Am I missing any trick??
Thanks and regards.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Nov 13 at 13:31
Devendra Singh Rana
744216
asked Nov 13 at 13:31
Devendra Singh Rana
744216
asked Nov 13 at 13:31
Devendra Singh Rana
744216
asked Nov 13 at 13:31
Devendra Singh Rana
744216
asked Nov 13 at 13:31
Devendra Singh Rana
744216
744216
Try computing eigenvalues. Note that there are some relatively obvious invariant subspaces, such as $operatorname{span}(1)$, $operatorname{span}(X, X^2)$, $operatorname{span}(X^3, X^4, X^5)$ and $operatorname{span}(X^6, X^7, X^8, X^9)$. You can break these subspaces down to eigenspaces without too much hassle.
– Theo Bendit
Nov 13 at 13:37
2
Or, given that this transformation just permutes monomials, computing powers is pretty easy to do.
– Theo Bendit
Nov 13 at 13:39
It will be helpful if you can elaborate a bit as an answer.
– Devendra Singh Rana
Nov 13 at 13:41
add a comment |
Try computing eigenvalues. Note that there are some relatively obvious invariant subspaces, such as $operatorname{span}(1)$, $operatorname{span}(X, X^2)$, $operatorname{span}(X^3, X^4, X^5)$ and $operatorname{span}(X^6, X^7, X^8, X^9)$. You can break these subspaces down to eigenspaces without too much hassle.
– Theo Bendit
Nov 13 at 13:37
2
Or, given that this transformation just permutes monomials, computing powers is pretty easy to do.
– Theo Bendit
Nov 13 at 13:39
It will be helpful if you can elaborate a bit as an answer.
– Devendra Singh Rana
Nov 13 at 13:41
Try computing eigenvalues. Note that there are some relatively obvious invariant subspaces, such as $operatorname{span}(1)$, $operatorname{span}(X, X^2)$, $operatorname{span}(X^3, X^4, X^5)$ and $operatorname{span}(X^6, X^7, X^8, X^9)$. You can break these subspaces down to eigenspaces without too much hassle.
– Theo Bendit
Nov 13 at 13:37
Try computing eigenvalues. Note that there are some relatively obvious invariant subspaces, such as $operatorname{span}(1)$, $operatorname{span}(X, X^2)$, $operatorname{span}(X^3, X^4, X^5)$ and $operatorname{span}(X^6, X^7, X^8, X^9)$. You can break these subspaces down to eigenspaces without too much hassle.
– Theo Bendit
Nov 13 at 13:37
2
2
Or, given that this transformation just permutes monomials, computing powers is pretty easy to do.
– Theo Bendit
Nov 13 at 13:39
Or, given that this transformation just permutes monomials, computing powers is pretty easy to do.
– Theo Bendit
Nov 13 at 13:39
It will be helpful if you can elaborate a bit as an answer.
– Devendra Singh Rana
Nov 13 at 13:41
It will be helpful if you can elaborate a bit as an answer.
– Devendra Singh Rana
Nov 13 at 13:41
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You have a 10x10 matrix indeed, but a really specific one.
Hint: Note that
$T(a_0)=a_0$
$X$ and $X^2$ coefficients are "mixed up"- So are $X^3$, $X^4$ and $X^5$ coefficients
- Same with $X^i, iin {6,ldots,9}$
Treat each "block" separatly, and arrange them to form a block diagonal matrix
answered Nov 13 at 13:42
F.Carette
1,19312
We need to evaluate powers of these diagonal blocks@F.Carette right??
– Devendra Singh Rana
Nov 13 at 14:01
You're right. And as theses are just permutations, a block to the right power will be nothing more than the identity matrix.
– F.Carette
Nov 13 at 14:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have a 10x10 matrix indeed, but a really specific one.
Hint: Note that
$T(a_0)=a_0$
$X$ and $X^2$ coefficients are "mixed up"- So are $X^3$, $X^4$ and $X^5$ coefficients
- Same with $X^i, iin {6,ldots,9}$
Treat each "block" separatly, and arrange them to form a block diagonal matrix
answered Nov 13 at 13:42
F.Carette
1,19312
We need to evaluate powers of these diagonal blocks@F.Carette right??
– Devendra Singh Rana
Nov 13 at 14:01
You're right. And as theses are just permutations, a block to the right power will be nothing more than the identity matrix.
– F.Carette
Nov 13 at 14:18
add a comment |
up vote
1
down vote
accepted
You have a 10x10 matrix indeed, but a really specific one.
Hint: Note that
$T(a_0)=a_0$
$X$ and $X^2$ coefficients are "mixed up"- So are $X^3$, $X^4$ and $X^5$ coefficients
- Same with $X^i, iin {6,ldots,9}$
Treat each "block" separatly, and arrange them to form a block diagonal matrix
answered Nov 13 at 13:42
F.Carette
1,19312
We need to evaluate powers of these diagonal blocks@F.Carette right??
– Devendra Singh Rana
Nov 13 at 14:01
You're right. And as theses are just permutations, a block to the right power will be nothing more than the identity matrix.
– F.Carette
Nov 13 at 14:18
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have a 10x10 matrix indeed, but a really specific one.
Hint: Note that
$T(a_0)=a_0$
$X$ and $X^2$ coefficients are "mixed up"- So are $X^3$, $X^4$ and $X^5$ coefficients
- Same with $X^i, iin {6,ldots,9}$
Treat each "block" separatly, and arrange them to form a block diagonal matrix
answered Nov 13 at 13:42
F.Carette
1,19312
You have a 10x10 matrix indeed, but a really specific one.
Hint: Note that
$T(a_0)=a_0$
$X$ and $X^2$ coefficients are "mixed up"- So are $X^3$, $X^4$ and $X^5$ coefficients
- Same with $X^i, iin {6,ldots,9}$
Treat each "block" separatly, and arrange them to form a block diagonal matrix
answered Nov 13 at 13:42
F.Carette
1,19312
edited Nov 13 at 14:19
answered Nov 13 at 13:42
F.Carette
1,19312
answered Nov 13 at 13:42
F.Carette
1,19312
answered Nov 13 at 13:42
F.Carette
1,19312
1,19312
We need to evaluate powers of these diagonal blocks@F.Carette right??
– Devendra Singh Rana
Nov 13 at 14:01
You're right. And as theses are just permutations, a block to the right power will be nothing more than the identity matrix.
– F.Carette
Nov 13 at 14:18
add a comment |
We need to evaluate powers of these diagonal blocks@F.Carette right??
– Devendra Singh Rana
Nov 13 at 14:01
You're right. And as theses are just permutations, a block to the right power will be nothing more than the identity matrix.
– F.Carette
Nov 13 at 14:18
We need to evaluate powers of these diagonal blocks@F.Carette right??
– Devendra Singh Rana
Nov 13 at 14:01
We need to evaluate powers of these diagonal blocks@F.Carette right??
– Devendra Singh Rana
Nov 13 at 14:01
You're right. And as theses are just permutations, a block to the right power will be nothing more than the identity matrix.
– F.Carette
Nov 13 at 14:18
You're right. And as theses are just permutations, a block to the right power will be nothing more than the identity matrix.
– F.Carette
Nov 13 at 14:18
add a comment |
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Try computing eigenvalues. Note that there are some relatively obvious invariant subspaces, such as $operatorname{span}(1)$, $operatorname{span}(X, X^2)$, $operatorname{span}(X^3, X^4, X^5)$ and $operatorname{span}(X^6, X^7, X^8, X^9)$. You can break these subspaces down to eigenspaces without too much hassle.
– Theo Bendit
Nov 13 at 13:37
2
Or, given that this transformation just permutes monomials, computing powers is pretty easy to do.
– Theo Bendit
Nov 13 at 13:39
It will be helpful if you can elaborate a bit as an answer.
– Devendra Singh Rana
Nov 13 at 13:41