Mixing
A special type of similarity that “inherits multiplication†over addition.
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Let us investigate a special type of similarity.
$$A = S(C^-1+D)S^-1$$
$$B = S(D^-1-C)S^-1$$
$$AB = S(C^-1D^-1+I - I - DC)S^-1 \= S((DC)^-1-DC)S^-1$$
$$BA = S(D^-1C^-1-I + I - CD)S^-1 \= S((CD)^-1-CD)S^-1$$
- Multiplication does "transfer" to internals and result becomes a $B$-type matrix but with $D=C$, either equal to $CD$ or $DC$ in previous matrix.
Is that all there is to it or are there more things to discover?
linear-algebra abstract-algebra matrices soft-question linear-transformations
asked Jul 16 at 17:18
mathreadler
13.6k71857
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Let us investigate a special type of similarity.
$$A = S(C^-1+D)S^-1$$
$$B = S(D^-1-C)S^-1$$
$$AB = S(C^-1D^-1+I - I - DC)S^-1 \= S((DC)^-1-DC)S^-1$$
$$BA = S(D^-1C^-1-I + I - CD)S^-1 \= S((CD)^-1-CD)S^-1$$
- Multiplication does "transfer" to internals and result becomes a $B$-type matrix but with $D=C$, either equal to $CD$ or $DC$ in previous matrix.
Is that all there is to it or are there more things to discover?
linear-algebra abstract-algebra matrices soft-question linear-transformations
asked Jul 16 at 17:18
mathreadler
13.6k71857
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let us investigate a special type of similarity.
$$A = S(C^-1+D)S^-1$$
$$B = S(D^-1-C)S^-1$$
$$AB = S(C^-1D^-1+I - I - DC)S^-1 \= S((DC)^-1-DC)S^-1$$
$$BA = S(D^-1C^-1-I + I - CD)S^-1 \= S((CD)^-1-CD)S^-1$$
- Multiplication does "transfer" to internals and result becomes a $B$-type matrix but with $D=C$, either equal to $CD$ or $DC$ in previous matrix.
Is that all there is to it or are there more things to discover?
linear-algebra abstract-algebra matrices soft-question linear-transformations
asked Jul 16 at 17:18
mathreadler
13.6k71857
Let us investigate a special type of similarity.
$$A = S(C^-1+D)S^-1$$
$$B = S(D^-1-C)S^-1$$
$$AB = S(C^-1D^-1+I - I - DC)S^-1 \= S((DC)^-1-DC)S^-1$$
$$BA = S(D^-1C^-1-I + I - CD)S^-1 \= S((CD)^-1-CD)S^-1$$
- Multiplication does "transfer" to internals and result becomes a $B$-type matrix but with $D=C$, either equal to $CD$ or $DC$ in previous matrix.
Is that all there is to it or are there more things to discover?
linear-algebra abstract-algebra matrices soft-question linear-transformations
asked Jul 16 at 17:18
mathreadler
13.6k71857
edited Jul 16 at 17:33
asked Jul 16 at 17:18
mathreadler
13.6k71857
asked Jul 16 at 17:18
mathreadler
13.6k71857
asked Jul 16 at 17:18
mathreadler
13.6k71857
13.6k71857
add a comment |Â
add a comment |Â
1 Answer
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The layer of $S^-1$ really amounts to the observation that similarity preserves addition and multiplication (and consequently, inverses). That is, the map $X mapsto SXS^-1$ is a homomorphism of algebras. Since this map is invertible, it is also an isomorphism.
Stripping that away, you have observed that multiplying a "type-A" matrix with a "type-B" matrix produces another "type-B" matrix. I'm not really sure what to make of that, honestly. Maybe there's something to say in the case where $C,D$ belong to a class of matrices that is closed under multiplication. For instance, where both $C,D$ are orthogonal/unitary.
answered Jul 16 at 17:32
Omnomnomnom
121k784170
Homomorphism is a very algebraic word, I am not sure I understand it so well yet.. Yes I am also not sure what to make of it. That is why I ask.
– mathreadler
Jul 16 at 17:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The layer of $S^-1$ really amounts to the observation that similarity preserves addition and multiplication (and consequently, inverses). That is, the map $X mapsto SXS^-1$ is a homomorphism of algebras. Since this map is invertible, it is also an isomorphism.
Stripping that away, you have observed that multiplying a "type-A" matrix with a "type-B" matrix produces another "type-B" matrix. I'm not really sure what to make of that, honestly. Maybe there's something to say in the case where $C,D$ belong to a class of matrices that is closed under multiplication. For instance, where both $C,D$ are orthogonal/unitary.
answered Jul 16 at 17:32
Omnomnomnom
121k784170
Homomorphism is a very algebraic word, I am not sure I understand it so well yet.. Yes I am also not sure what to make of it. That is why I ask.
– mathreadler
Jul 16 at 17:39
add a comment |Â
up vote
1
down vote
The layer of $S^-1$ really amounts to the observation that similarity preserves addition and multiplication (and consequently, inverses). That is, the map $X mapsto SXS^-1$ is a homomorphism of algebras. Since this map is invertible, it is also an isomorphism.
Stripping that away, you have observed that multiplying a "type-A" matrix with a "type-B" matrix produces another "type-B" matrix. I'm not really sure what to make of that, honestly. Maybe there's something to say in the case where $C,D$ belong to a class of matrices that is closed under multiplication. For instance, where both $C,D$ are orthogonal/unitary.
answered Jul 16 at 17:32
Omnomnomnom
121k784170
Homomorphism is a very algebraic word, I am not sure I understand it so well yet.. Yes I am also not sure what to make of it. That is why I ask.
– mathreadler
Jul 16 at 17:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The layer of $S^-1$ really amounts to the observation that similarity preserves addition and multiplication (and consequently, inverses). That is, the map $X mapsto SXS^-1$ is a homomorphism of algebras. Since this map is invertible, it is also an isomorphism.
Stripping that away, you have observed that multiplying a "type-A" matrix with a "type-B" matrix produces another "type-B" matrix. I'm not really sure what to make of that, honestly. Maybe there's something to say in the case where $C,D$ belong to a class of matrices that is closed under multiplication. For instance, where both $C,D$ are orthogonal/unitary.
answered Jul 16 at 17:32
Omnomnomnom
121k784170
The layer of $S^-1$ really amounts to the observation that similarity preserves addition and multiplication (and consequently, inverses). That is, the map $X mapsto SXS^-1$ is a homomorphism of algebras. Since this map is invertible, it is also an isomorphism.
Stripping that away, you have observed that multiplying a "type-A" matrix with a "type-B" matrix produces another "type-B" matrix. I'm not really sure what to make of that, honestly. Maybe there's something to say in the case where $C,D$ belong to a class of matrices that is closed under multiplication. For instance, where both $C,D$ are orthogonal/unitary.
answered Jul 16 at 17:32
Omnomnomnom
121k784170
answered Jul 16 at 17:32
Omnomnomnom
121k784170
answered Jul 16 at 17:32
Omnomnomnom
121k784170
answered Jul 16 at 17:32
Omnomnomnom
121k784170
121k784170
Homomorphism is a very algebraic word, I am not sure I understand it so well yet.. Yes I am also not sure what to make of it. That is why I ask.
– mathreadler
Jul 16 at 17:39
add a comment |Â
Homomorphism is a very algebraic word, I am not sure I understand it so well yet.. Yes I am also not sure what to make of it. That is why I ask.
– mathreadler
Jul 16 at 17:39
Homomorphism is a very algebraic word, I am not sure I understand it so well yet.. Yes I am also not sure what to make of it. That is why I ask.
– mathreadler
Jul 16 at 17:39
Homomorphism is a very algebraic word, I am not sure I understand it so well yet.. Yes I am also not sure what to make of it. That is why I ask.
– mathreadler
Jul 16 at 17:39
add a comment |Â
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