Dimension of span of powers of a given matrix Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Cayley-Hamilton theorem on square matricesFind $A^1000$ by using Cayley-Hamilton TheoremThe minimal polynomial divides the characteristic polynomialShow that the minimal polynomial over $mathbbC$ of a real matrix has real coefficientsAvoiding the Cayley–Hamilton theoremCayley Hamilton TheoremA positive definite matrix A can be express as a polynomial of $ A^2 $Generalization of Cayley-Hamilton for non-diagonal coefficients related to other properties than eigenvalues?Cayley-Hamilton and linear dependenceDimension on span of End(V) composition
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Dimension of span of powers of a given matrix
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Cayley-Hamilton theorem on square matricesFind $A^1000$ by using Cayley-Hamilton TheoremThe minimal polynomial divides the characteristic polynomialShow that the minimal polynomial over $mathbbC$ of a real matrix has real coefficientsAvoiding the Cayley–Hamilton theoremCayley Hamilton TheoremA positive definite matrix A can be express as a polynomial of $ A^2 $Generalization of Cayley-Hamilton for non-diagonal coefficients related to other properties than eigenvalues?Cayley-Hamilton and linear dependenceDimension on span of End(V) composition
$begingroup$
Let $A$ be a matrix $ntimes n$ of rank n, then Cayley–Hamilton theorem states that powers of A up to $n-1$ are linearly depend, more preciously $rho_A(A)=0$ for characteristic polynomial of $A$.
I've stuck with an elementary question, what is the dimension of linear span of powers of A up to k-th power? It seems to be related to the theorem above, but it's not clear for me how understand it and the dimension of the span.
linear-algebra matrices
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
$endgroup$
add a comment |
$begingroup$
Let $A$ be a matrix $ntimes n$ of rank n, then Cayley–Hamilton theorem states that powers of A up to $n-1$ are linearly depend, more preciously $rho_A(A)=0$ for characteristic polynomial of $A$.
I've stuck with an elementary question, what is the dimension of linear span of powers of A up to k-th power? It seems to be related to the theorem above, but it's not clear for me how understand it and the dimension of the span.
linear-algebra matrices
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
$endgroup$
add a comment |
$begingroup$
Let $A$ be a matrix $ntimes n$ of rank n, then Cayley–Hamilton theorem states that powers of A up to $n-1$ are linearly depend, more preciously $rho_A(A)=0$ for characteristic polynomial of $A$.
I've stuck with an elementary question, what is the dimension of linear span of powers of A up to k-th power? It seems to be related to the theorem above, but it's not clear for me how understand it and the dimension of the span.
linear-algebra matrices
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
$endgroup$
Let $A$ be a matrix $ntimes n$ of rank n, then Cayley–Hamilton theorem states that powers of A up to $n-1$ are linearly depend, more preciously $rho_A(A)=0$ for characteristic polynomial of $A$.
I've stuck with an elementary question, what is the dimension of linear span of powers of A up to k-th power? It seems to be related to the theorem above, but it's not clear for me how understand it and the dimension of the span.
linear-algebra matrices
linear-algebra matrices
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
asked Apr 2 at 17:17
Grog SmithGrog Smith
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
The space of $n times n$ matrices is a vector space (e.g., just view each matrix as a vector with $n^2$ components).
$A^n-1, A^n-2, ldots, A, I$ is a set of elements of this vector space, so you can consider its span (i.e. all matrices of the form $c_n-1 A^n-1 + cdots + c_1 A + c_0 I$) or ask about whether the set is linearly independent or not, etc. This is not anything different than what you know about working with vector spaces of the form $mathbbR^n$.
If you know that, say, $A^3 + 2 A^2 + 5 A - 2I = 0$ (this is an arbitrary example), then you know $A^3, A^2, A, I$ is a linearly dependent set. (Why?) Can you now figure out what the Cayley-Hamilton theorem implies about linear dependence of powers of $A$?
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
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add a comment |
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$begingroup$
The space of $n times n$ matrices is a vector space (e.g., just view each matrix as a vector with $n^2$ components).
$A^n-1, A^n-2, ldots, A, I$ is a set of elements of this vector space, so you can consider its span (i.e. all matrices of the form $c_n-1 A^n-1 + cdots + c_1 A + c_0 I$) or ask about whether the set is linearly independent or not, etc. This is not anything different than what you know about working with vector spaces of the form $mathbbR^n$.
If you know that, say, $A^3 + 2 A^2 + 5 A - 2I = 0$ (this is an arbitrary example), then you know $A^3, A^2, A, I$ is a linearly dependent set. (Why?) Can you now figure out what the Cayley-Hamilton theorem implies about linear dependence of powers of $A$?
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
$endgroup$
add a comment |
$begingroup$
The space of $n times n$ matrices is a vector space (e.g., just view each matrix as a vector with $n^2$ components).
$A^n-1, A^n-2, ldots, A, I$ is a set of elements of this vector space, so you can consider its span (i.e. all matrices of the form $c_n-1 A^n-1 + cdots + c_1 A + c_0 I$) or ask about whether the set is linearly independent or not, etc. This is not anything different than what you know about working with vector spaces of the form $mathbbR^n$.
If you know that, say, $A^3 + 2 A^2 + 5 A - 2I = 0$ (this is an arbitrary example), then you know $A^3, A^2, A, I$ is a linearly dependent set. (Why?) Can you now figure out what the Cayley-Hamilton theorem implies about linear dependence of powers of $A$?
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
$endgroup$
add a comment |
$begingroup$
The space of $n times n$ matrices is a vector space (e.g., just view each matrix as a vector with $n^2$ components).
$A^n-1, A^n-2, ldots, A, I$ is a set of elements of this vector space, so you can consider its span (i.e. all matrices of the form $c_n-1 A^n-1 + cdots + c_1 A + c_0 I$) or ask about whether the set is linearly independent or not, etc. This is not anything different than what you know about working with vector spaces of the form $mathbbR^n$.
If you know that, say, $A^3 + 2 A^2 + 5 A - 2I = 0$ (this is an arbitrary example), then you know $A^3, A^2, A, I$ is a linearly dependent set. (Why?) Can you now figure out what the Cayley-Hamilton theorem implies about linear dependence of powers of $A$?
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
$endgroup$
The space of $n times n$ matrices is a vector space (e.g., just view each matrix as a vector with $n^2$ components).
$A^n-1, A^n-2, ldots, A, I$ is a set of elements of this vector space, so you can consider its span (i.e. all matrices of the form $c_n-1 A^n-1 + cdots + c_1 A + c_0 I$) or ask about whether the set is linearly independent or not, etc. This is not anything different than what you know about working with vector spaces of the form $mathbbR^n$.
If you know that, say, $A^3 + 2 A^2 + 5 A - 2I = 0$ (this is an arbitrary example), then you know $A^3, A^2, A, I$ is a linearly dependent set. (Why?) Can you now figure out what the Cayley-Hamilton theorem implies about linear dependence of powers of $A$?
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
answered Apr 2 at 17:26
angryavianangryavian
42.9k23482
42.9k23482
add a comment |
add a comment |
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