prove the next algebras isomorphism
$begingroup$
Let V be a non trivial vector space s.t. $dim V = n$ over $mathbb{C}$. also let $T:Vto V$ be a linear transformation over $mathbb{C}$.
now we shall define A = $mathbb{C} [x]$ to be the polynomials algebra.
such that $V$ is a module of $A$ given by the following action:
for every $a = f(x) in A , vin V$ we define $a* v = f(T)(v)$. now suppose that $T$ is diagonalizable matrix and $lambda_1,...,lambda_k$ are it's different eigenvalues and $n_1,...,n_k$ are it's multiplicity.
prove that $End_A(V) cong bigoplus_{j=1}^{k} M_{n_j} (mathbb{C})$ as algebras (where $M_{n}(mathbb{C})$ is the $n times n$ matrix algebra)
abstract-algebra modules
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
$endgroup$
add a comment |
$begingroup$
Let V be a non trivial vector space s.t. $dim V = n$ over $mathbb{C}$. also let $T:Vto V$ be a linear transformation over $mathbb{C}$.
now we shall define A = $mathbb{C} [x]$ to be the polynomials algebra.
such that $V$ is a module of $A$ given by the following action:
for every $a = f(x) in A , vin V$ we define $a* v = f(T)(v)$. now suppose that $T$ is diagonalizable matrix and $lambda_1,...,lambda_k$ are it's different eigenvalues and $n_1,...,n_k$ are it's multiplicity.
prove that $End_A(V) cong bigoplus_{j=1}^{k} M_{n_j} (mathbb{C})$ as algebras (where $M_{n}(mathbb{C})$ is the $n times n$ matrix algebra)
abstract-algebra modules
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
$endgroup$
add a comment |
$begingroup$
Let V be a non trivial vector space s.t. $dim V = n$ over $mathbb{C}$. also let $T:Vto V$ be a linear transformation over $mathbb{C}$.
now we shall define A = $mathbb{C} [x]$ to be the polynomials algebra.
such that $V$ is a module of $A$ given by the following action:
for every $a = f(x) in A , vin V$ we define $a* v = f(T)(v)$. now suppose that $T$ is diagonalizable matrix and $lambda_1,...,lambda_k$ are it's different eigenvalues and $n_1,...,n_k$ are it's multiplicity.
prove that $End_A(V) cong bigoplus_{j=1}^{k} M_{n_j} (mathbb{C})$ as algebras (where $M_{n}(mathbb{C})$ is the $n times n$ matrix algebra)
abstract-algebra modules
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
$endgroup$
Let V be a non trivial vector space s.t. $dim V = n$ over $mathbb{C}$. also let $T:Vto V$ be a linear transformation over $mathbb{C}$.
now we shall define A = $mathbb{C} [x]$ to be the polynomials algebra.
such that $V$ is a module of $A$ given by the following action:
for every $a = f(x) in A , vin V$ we define $a* v = f(T)(v)$. now suppose that $T$ is diagonalizable matrix and $lambda_1,...,lambda_k$ are it's different eigenvalues and $n_1,...,n_k$ are it's multiplicity.
prove that $End_A(V) cong bigoplus_{j=1}^{k} M_{n_j} (mathbb{C})$ as algebras (where $M_{n}(mathbb{C})$ is the $n times n$ matrix algebra)
abstract-algebra modules
abstract-algebra modules
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
asked Dec 6 '18 at 10:50
ned grekerzbergned grekerzberg
482318
482318
add a comment |
add a comment |
1 Answer
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$begingroup$
Pick a basis of eigenvectors for $T$ : let's call it $B={v_1 dots v_n}$, such that $v_j$ has eigenvalue $lambda_i$ for every $n_i < i leq n_{i+1}$.
Now you can write $T$ in this basis ,getting $S in M(n,mathbb{C})$ and define a structure of $A$-module on $mathbb{C}^n$ as $xv=Sv$. I'm not gonna write down the complete proof, but you can chech that the "writing in $B$ basis" isomorphism between $End_{mathbb{C}}(V) $ and $M(n,mathbb{C})$ restricts to an isomorphism of algebras between $End_{A}(V) $ and $End_A(mathbb{C}^n) subset M(n,mathbb{C}).$
I've reduced to matrix case to make things clearer (at least I hope): now checking that a matrix $C in M(n,mathbb{C})$ defines an $A$-endomorphism means that $$C(xv)=x(Cv)$$ for every $v in mathbb{C}^n$ which turns out to be equivalent to $$CS=SC .$$
Now, $S$ is a block diagonal matrix with blocks given by scalar multiples of the identity of dimensions $n_1, dots n_k$. It's a linear algebra fact that the space of matrix that commutes with such a matrix is made by all block diagonal matrix with block given by scalar multiples of the identity of the previous dimensions and so you get the isomorphism .
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
$endgroup$
add a comment |
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$begingroup$
Pick a basis of eigenvectors for $T$ : let's call it $B={v_1 dots v_n}$, such that $v_j$ has eigenvalue $lambda_i$ for every $n_i < i leq n_{i+1}$.
Now you can write $T$ in this basis ,getting $S in M(n,mathbb{C})$ and define a structure of $A$-module on $mathbb{C}^n$ as $xv=Sv$. I'm not gonna write down the complete proof, but you can chech that the "writing in $B$ basis" isomorphism between $End_{mathbb{C}}(V) $ and $M(n,mathbb{C})$ restricts to an isomorphism of algebras between $End_{A}(V) $ and $End_A(mathbb{C}^n) subset M(n,mathbb{C}).$
I've reduced to matrix case to make things clearer (at least I hope): now checking that a matrix $C in M(n,mathbb{C})$ defines an $A$-endomorphism means that $$C(xv)=x(Cv)$$ for every $v in mathbb{C}^n$ which turns out to be equivalent to $$CS=SC .$$
Now, $S$ is a block diagonal matrix with blocks given by scalar multiples of the identity of dimensions $n_1, dots n_k$. It's a linear algebra fact that the space of matrix that commutes with such a matrix is made by all block diagonal matrix with block given by scalar multiples of the identity of the previous dimensions and so you get the isomorphism .
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
$endgroup$
add a comment |
$begingroup$
Pick a basis of eigenvectors for $T$ : let's call it $B={v_1 dots v_n}$, such that $v_j$ has eigenvalue $lambda_i$ for every $n_i < i leq n_{i+1}$.
Now you can write $T$ in this basis ,getting $S in M(n,mathbb{C})$ and define a structure of $A$-module on $mathbb{C}^n$ as $xv=Sv$. I'm not gonna write down the complete proof, but you can chech that the "writing in $B$ basis" isomorphism between $End_{mathbb{C}}(V) $ and $M(n,mathbb{C})$ restricts to an isomorphism of algebras between $End_{A}(V) $ and $End_A(mathbb{C}^n) subset M(n,mathbb{C}).$
I've reduced to matrix case to make things clearer (at least I hope): now checking that a matrix $C in M(n,mathbb{C})$ defines an $A$-endomorphism means that $$C(xv)=x(Cv)$$ for every $v in mathbb{C}^n$ which turns out to be equivalent to $$CS=SC .$$
Now, $S$ is a block diagonal matrix with blocks given by scalar multiples of the identity of dimensions $n_1, dots n_k$. It's a linear algebra fact that the space of matrix that commutes with such a matrix is made by all block diagonal matrix with block given by scalar multiples of the identity of the previous dimensions and so you get the isomorphism .
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
$endgroup$
add a comment |
$begingroup$
Pick a basis of eigenvectors for $T$ : let's call it $B={v_1 dots v_n}$, such that $v_j$ has eigenvalue $lambda_i$ for every $n_i < i leq n_{i+1}$.
Now you can write $T$ in this basis ,getting $S in M(n,mathbb{C})$ and define a structure of $A$-module on $mathbb{C}^n$ as $xv=Sv$. I'm not gonna write down the complete proof, but you can chech that the "writing in $B$ basis" isomorphism between $End_{mathbb{C}}(V) $ and $M(n,mathbb{C})$ restricts to an isomorphism of algebras between $End_{A}(V) $ and $End_A(mathbb{C}^n) subset M(n,mathbb{C}).$
I've reduced to matrix case to make things clearer (at least I hope): now checking that a matrix $C in M(n,mathbb{C})$ defines an $A$-endomorphism means that $$C(xv)=x(Cv)$$ for every $v in mathbb{C}^n$ which turns out to be equivalent to $$CS=SC .$$
Now, $S$ is a block diagonal matrix with blocks given by scalar multiples of the identity of dimensions $n_1, dots n_k$. It's a linear algebra fact that the space of matrix that commutes with such a matrix is made by all block diagonal matrix with block given by scalar multiples of the identity of the previous dimensions and so you get the isomorphism .
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
$endgroup$
Pick a basis of eigenvectors for $T$ : let's call it $B={v_1 dots v_n}$, such that $v_j$ has eigenvalue $lambda_i$ for every $n_i < i leq n_{i+1}$.
Now you can write $T$ in this basis ,getting $S in M(n,mathbb{C})$ and define a structure of $A$-module on $mathbb{C}^n$ as $xv=Sv$. I'm not gonna write down the complete proof, but you can chech that the "writing in $B$ basis" isomorphism between $End_{mathbb{C}}(V) $ and $M(n,mathbb{C})$ restricts to an isomorphism of algebras between $End_{A}(V) $ and $End_A(mathbb{C}^n) subset M(n,mathbb{C}).$
I've reduced to matrix case to make things clearer (at least I hope): now checking that a matrix $C in M(n,mathbb{C})$ defines an $A$-endomorphism means that $$C(xv)=x(Cv)$$ for every $v in mathbb{C}^n$ which turns out to be equivalent to $$CS=SC .$$
Now, $S$ is a block diagonal matrix with blocks given by scalar multiples of the identity of dimensions $n_1, dots n_k$. It's a linear algebra fact that the space of matrix that commutes with such a matrix is made by all block diagonal matrix with block given by scalar multiples of the identity of the previous dimensions and so you get the isomorphism .
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
answered Dec 6 '18 at 13:50
Tommaso ScognamiglioTommaso Scognamiglio
487312
487312
add a comment |
add a comment |
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