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Sums of cubes of two complex matrices are closed under matrix multiplication
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Show that the set of matrices $M^3+N^3: M,Nin M_n(mathbb C)$ is closed under matrix multiplication.
Thanks for your help.
abstract-algebra matrices proof-writing
edited Jul 29 at 8:29
user1551
66.9k564122
asked Jul 17 at 12:44
Morpheus
336
 |Â
show 15 more comments
up vote
1
down vote
favorite
Show that the set of matrices $M^3+N^3: M,Nin M_n(mathbb C)$ is closed under matrix multiplication.
Thanks for your help.
abstract-algebra matrices proof-writing
edited Jul 29 at 8:29
user1551
66.9k564122
asked Jul 17 at 12:44
Morpheus
336
2
What does it mean for an operation to be "intern"? That word is not in common use in mathematical English.
– Gerry Myerson
Jul 17 at 13:10
It means that for A,B,C,D,M,N in Mn(C), (A^3+B^3)(C^3+D^3) can be written as M^3+N^3.
– Morpheus
Jul 17 at 20:17
In other words, the set is closed under multiplication. But maybe every complex matrix is a cube?
– Gerry Myerson
Jul 17 at 21:54
Ignore that – not every complex matrix is a cube.
– Gerry Myerson
Jul 19 at 2:54
1
It's an exercise given by my teacher, so it has to be true.
– Morpheus
Jul 20 at 11:16
 |Â
show 15 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that the set of matrices $M^3+N^3: M,Nin M_n(mathbb C)$ is closed under matrix multiplication.
Thanks for your help.
abstract-algebra matrices proof-writing
edited Jul 29 at 8:29
user1551
66.9k564122
asked Jul 17 at 12:44
Morpheus
336
Show that the set of matrices $M^3+N^3: M,Nin M_n(mathbb C)$ is closed under matrix multiplication.
Thanks for your help.
abstract-algebra matrices proof-writing
edited Jul 29 at 8:29
user1551
66.9k564122
asked Jul 17 at 12:44
Morpheus
336
edited Jul 29 at 8:29
user1551
66.9k564122
edited Jul 29 at 8:29
user1551
66.9k564122
edited Jul 29 at 8:29
user1551
66.9k564122
66.9k564122
asked Jul 17 at 12:44
Morpheus
336
asked Jul 17 at 12:44
Morpheus
336
asked Jul 17 at 12:44
Morpheus
336
336
2
What does it mean for an operation to be "intern"? That word is not in common use in mathematical English.
– Gerry Myerson
Jul 17 at 13:10
It means that for A,B,C,D,M,N in Mn(C), (A^3+B^3)(C^3+D^3) can be written as M^3+N^3.
– Morpheus
Jul 17 at 20:17
In other words, the set is closed under multiplication. But maybe every complex matrix is a cube?
– Gerry Myerson
Jul 17 at 21:54
Ignore that – not every complex matrix is a cube.
– Gerry Myerson
Jul 19 at 2:54
1
It's an exercise given by my teacher, so it has to be true.
– Morpheus
Jul 20 at 11:16
 |Â
show 15 more comments
2
What does it mean for an operation to be "intern"? That word is not in common use in mathematical English.
– Gerry Myerson
Jul 17 at 13:10
It means that for A,B,C,D,M,N in Mn(C), (A^3+B^3)(C^3+D^3) can be written as M^3+N^3.
– Morpheus
Jul 17 at 20:17
In other words, the set is closed under multiplication. But maybe every complex matrix is a cube?
– Gerry Myerson
Jul 17 at 21:54
Ignore that – not every complex matrix is a cube.
– Gerry Myerson
Jul 19 at 2:54
1
It's an exercise given by my teacher, so it has to be true.
– Morpheus
Jul 20 at 11:16
2
2
What does it mean for an operation to be "intern"? That word is not in common use in mathematical English.
– Gerry Myerson
Jul 17 at 13:10
What does it mean for an operation to be "intern"? That word is not in common use in mathematical English.
– Gerry Myerson
Jul 17 at 13:10
It means that for A,B,C,D,M,N in Mn(C), (A^3+B^3)(C^3+D^3) can be written as M^3+N^3.
– Morpheus
Jul 17 at 20:17
It means that for A,B,C,D,M,N in Mn(C), (A^3+B^3)(C^3+D^3) can be written as M^3+N^3.
– Morpheus
Jul 17 at 20:17
In other words, the set is closed under multiplication. But maybe every complex matrix is a cube?
– Gerry Myerson
Jul 17 at 21:54
In other words, the set is closed under multiplication. But maybe every complex matrix is a cube?
– Gerry Myerson
Jul 17 at 21:54
Ignore that – not every complex matrix is a cube.
– Gerry Myerson
Jul 19 at 2:54
Ignore that – not every complex matrix is a cube.
– Gerry Myerson
Jul 19 at 2:54
1
1
It's an exercise given by my teacher, so it has to be true.
– Morpheus
Jul 20 at 11:16
It's an exercise given by my teacher, so it has to be true.
– Morpheus
Jul 20 at 11:16
 |Â
show 15 more comments
1 Answer
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I think you may be able to produce a proof by fleshing out the following sketch:
Every diagonalizable matrix is a cube of a matrix.
Every matrix is a sum of two diagonalizable matrices.
It follows that every matrix is a sum of two cubes.
Fleshing this out:
Let $A$ be a diagonalizable matrix. That means there's a diagonal matrix $D$, and an invertible matrix $P$, such that $A=P^-1DP$. There's a diagonal matrix $E$ such that $E^3=D$; just let each entry of $E$ be a cube root of the corresponding entry of $D$. Let $B=P^-1EP$. Then $$B^3=P^-1EPP^-1EPP^-1EP=P^-1E^3P=P^-1DP=A$$ Thus, every diagonalizable matrix is a cube.
Now every matrix $C$ is similar to a matrix in Jordan form, that is, there is a matrix $J$ in Jordan form and an invertible matrix $P$ such that $C=P^-1JP$. A matrix in Jordan form is composed of Jordan blocks $J_i$, $i=1,dots,m$, $$J=pmatrixJ_1&0&dots&0cr0&J_2&dots&0crvdots&&ddots&vdotscr0&0&dots&J_mcr$$ where each block $J_i$ is of the form $$J_i=pmatrixa_i&1&0&dots&0cr0&a_i&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i&1cr0&dots&dots&0&a_icr$$ Each $J_i$ can be written as a sum of two diagonlizable matrices, $J_i=R_i+S_i$, $$R_i=pmatrix1&1&0&dots&0cr0&2&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&m-1&1cr0&dots&dots&0&mcr$$ and $$S_i=pmatrixa_i-1&0&0&dots&0cr0&a_i-2&0&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i-(m-1)&0cr0&dots&dots&0&a_i-mcr$$ $R_i$ is diagonalizable since it has distinct eigenvalues, and $S_i$ is diagonal. Then $J=R+S$, where $R$ and $S$ are composed of the blocks $R_i$ and $S_i$, respectively, and are diagonalizable, say $R=Q^-1FQ$ with $F$ diagonal ($S$ is already diagonal). So $$C=P^-1JP=P^-1(R+S)P=P^-1RP+P^-1SP=P^-1Q^-1FQP+P^-1SP=(QP)^-1F(QP)+P^-1SP$$ is a sum of two diagonal matrices.
So, every matrix is a sum of two diagonalizable matrices, and every diagonalizable matrix is a cube, hence, every matrix is a sum of two cubes, hence, the set of all sums of two cubes, being the same as the set of all matrices, is closed under multiplication.
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I think you may be able to produce a proof by fleshing out the following sketch:
Every diagonalizable matrix is a cube of a matrix.
Every matrix is a sum of two diagonalizable matrices.
It follows that every matrix is a sum of two cubes.
Fleshing this out:
Let $A$ be a diagonalizable matrix. That means there's a diagonal matrix $D$, and an invertible matrix $P$, such that $A=P^-1DP$. There's a diagonal matrix $E$ such that $E^3=D$; just let each entry of $E$ be a cube root of the corresponding entry of $D$. Let $B=P^-1EP$. Then $$B^3=P^-1EPP^-1EPP^-1EP=P^-1E^3P=P^-1DP=A$$ Thus, every diagonalizable matrix is a cube.
Now every matrix $C$ is similar to a matrix in Jordan form, that is, there is a matrix $J$ in Jordan form and an invertible matrix $P$ such that $C=P^-1JP$. A matrix in Jordan form is composed of Jordan blocks $J_i$, $i=1,dots,m$, $$J=pmatrixJ_1&0&dots&0cr0&J_2&dots&0crvdots&&ddots&vdotscr0&0&dots&J_mcr$$ where each block $J_i$ is of the form $$J_i=pmatrixa_i&1&0&dots&0cr0&a_i&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i&1cr0&dots&dots&0&a_icr$$ Each $J_i$ can be written as a sum of two diagonlizable matrices, $J_i=R_i+S_i$, $$R_i=pmatrix1&1&0&dots&0cr0&2&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&m-1&1cr0&dots&dots&0&mcr$$ and $$S_i=pmatrixa_i-1&0&0&dots&0cr0&a_i-2&0&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i-(m-1)&0cr0&dots&dots&0&a_i-mcr$$ $R_i$ is diagonalizable since it has distinct eigenvalues, and $S_i$ is diagonal. Then $J=R+S$, where $R$ and $S$ are composed of the blocks $R_i$ and $S_i$, respectively, and are diagonalizable, say $R=Q^-1FQ$ with $F$ diagonal ($S$ is already diagonal). So $$C=P^-1JP=P^-1(R+S)P=P^-1RP+P^-1SP=P^-1Q^-1FQP+P^-1SP=(QP)^-1F(QP)+P^-1SP$$ is a sum of two diagonal matrices.
So, every matrix is a sum of two diagonalizable matrices, and every diagonalizable matrix is a cube, hence, every matrix is a sum of two cubes, hence, the set of all sums of two cubes, being the same as the set of all matrices, is closed under multiplication.
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
add a comment |Â
up vote
2
down vote
accepted
I think you may be able to produce a proof by fleshing out the following sketch:
Every diagonalizable matrix is a cube of a matrix.
Every matrix is a sum of two diagonalizable matrices.
It follows that every matrix is a sum of two cubes.
Fleshing this out:
Let $A$ be a diagonalizable matrix. That means there's a diagonal matrix $D$, and an invertible matrix $P$, such that $A=P^-1DP$. There's a diagonal matrix $E$ such that $E^3=D$; just let each entry of $E$ be a cube root of the corresponding entry of $D$. Let $B=P^-1EP$. Then $$B^3=P^-1EPP^-1EPP^-1EP=P^-1E^3P=P^-1DP=A$$ Thus, every diagonalizable matrix is a cube.
Now every matrix $C$ is similar to a matrix in Jordan form, that is, there is a matrix $J$ in Jordan form and an invertible matrix $P$ such that $C=P^-1JP$. A matrix in Jordan form is composed of Jordan blocks $J_i$, $i=1,dots,m$, $$J=pmatrixJ_1&0&dots&0cr0&J_2&dots&0crvdots&&ddots&vdotscr0&0&dots&J_mcr$$ where each block $J_i$ is of the form $$J_i=pmatrixa_i&1&0&dots&0cr0&a_i&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i&1cr0&dots&dots&0&a_icr$$ Each $J_i$ can be written as a sum of two diagonlizable matrices, $J_i=R_i+S_i$, $$R_i=pmatrix1&1&0&dots&0cr0&2&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&m-1&1cr0&dots&dots&0&mcr$$ and $$S_i=pmatrixa_i-1&0&0&dots&0cr0&a_i-2&0&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i-(m-1)&0cr0&dots&dots&0&a_i-mcr$$ $R_i$ is diagonalizable since it has distinct eigenvalues, and $S_i$ is diagonal. Then $J=R+S$, where $R$ and $S$ are composed of the blocks $R_i$ and $S_i$, respectively, and are diagonalizable, say $R=Q^-1FQ$ with $F$ diagonal ($S$ is already diagonal). So $$C=P^-1JP=P^-1(R+S)P=P^-1RP+P^-1SP=P^-1Q^-1FQP+P^-1SP=(QP)^-1F(QP)+P^-1SP$$ is a sum of two diagonal matrices.
So, every matrix is a sum of two diagonalizable matrices, and every diagonalizable matrix is a cube, hence, every matrix is a sum of two cubes, hence, the set of all sums of two cubes, being the same as the set of all matrices, is closed under multiplication.
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I think you may be able to produce a proof by fleshing out the following sketch:
Every diagonalizable matrix is a cube of a matrix.
Every matrix is a sum of two diagonalizable matrices.
It follows that every matrix is a sum of two cubes.
Fleshing this out:
Let $A$ be a diagonalizable matrix. That means there's a diagonal matrix $D$, and an invertible matrix $P$, such that $A=P^-1DP$. There's a diagonal matrix $E$ such that $E^3=D$; just let each entry of $E$ be a cube root of the corresponding entry of $D$. Let $B=P^-1EP$. Then $$B^3=P^-1EPP^-1EPP^-1EP=P^-1E^3P=P^-1DP=A$$ Thus, every diagonalizable matrix is a cube.
Now every matrix $C$ is similar to a matrix in Jordan form, that is, there is a matrix $J$ in Jordan form and an invertible matrix $P$ such that $C=P^-1JP$. A matrix in Jordan form is composed of Jordan blocks $J_i$, $i=1,dots,m$, $$J=pmatrixJ_1&0&dots&0cr0&J_2&dots&0crvdots&&ddots&vdotscr0&0&dots&J_mcr$$ where each block $J_i$ is of the form $$J_i=pmatrixa_i&1&0&dots&0cr0&a_i&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i&1cr0&dots&dots&0&a_icr$$ Each $J_i$ can be written as a sum of two diagonlizable matrices, $J_i=R_i+S_i$, $$R_i=pmatrix1&1&0&dots&0cr0&2&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&m-1&1cr0&dots&dots&0&mcr$$ and $$S_i=pmatrixa_i-1&0&0&dots&0cr0&a_i-2&0&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i-(m-1)&0cr0&dots&dots&0&a_i-mcr$$ $R_i$ is diagonalizable since it has distinct eigenvalues, and $S_i$ is diagonal. Then $J=R+S$, where $R$ and $S$ are composed of the blocks $R_i$ and $S_i$, respectively, and are diagonalizable, say $R=Q^-1FQ$ with $F$ diagonal ($S$ is already diagonal). So $$C=P^-1JP=P^-1(R+S)P=P^-1RP+P^-1SP=P^-1Q^-1FQP+P^-1SP=(QP)^-1F(QP)+P^-1SP$$ is a sum of two diagonal matrices.
So, every matrix is a sum of two diagonalizable matrices, and every diagonalizable matrix is a cube, hence, every matrix is a sum of two cubes, hence, the set of all sums of two cubes, being the same as the set of all matrices, is closed under multiplication.
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
I think you may be able to produce a proof by fleshing out the following sketch:
Every diagonalizable matrix is a cube of a matrix.
Every matrix is a sum of two diagonalizable matrices.
It follows that every matrix is a sum of two cubes.
Fleshing this out:
Let $A$ be a diagonalizable matrix. That means there's a diagonal matrix $D$, and an invertible matrix $P$, such that $A=P^-1DP$. There's a diagonal matrix $E$ such that $E^3=D$; just let each entry of $E$ be a cube root of the corresponding entry of $D$. Let $B=P^-1EP$. Then $$B^3=P^-1EPP^-1EPP^-1EP=P^-1E^3P=P^-1DP=A$$ Thus, every diagonalizable matrix is a cube.
Now every matrix $C$ is similar to a matrix in Jordan form, that is, there is a matrix $J$ in Jordan form and an invertible matrix $P$ such that $C=P^-1JP$. A matrix in Jordan form is composed of Jordan blocks $J_i$, $i=1,dots,m$, $$J=pmatrixJ_1&0&dots&0cr0&J_2&dots&0crvdots&&ddots&vdotscr0&0&dots&J_mcr$$ where each block $J_i$ is of the form $$J_i=pmatrixa_i&1&0&dots&0cr0&a_i&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i&1cr0&dots&dots&0&a_icr$$ Each $J_i$ can be written as a sum of two diagonlizable matrices, $J_i=R_i+S_i$, $$R_i=pmatrix1&1&0&dots&0cr0&2&1&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&m-1&1cr0&dots&dots&0&mcr$$ and $$S_i=pmatrixa_i-1&0&0&dots&0cr0&a_i-2&0&dots&0crvdots&&ddots&ddots&vdotscrvdots&&&a_i-(m-1)&0cr0&dots&dots&0&a_i-mcr$$ $R_i$ is diagonalizable since it has distinct eigenvalues, and $S_i$ is diagonal. Then $J=R+S$, where $R$ and $S$ are composed of the blocks $R_i$ and $S_i$, respectively, and are diagonalizable, say $R=Q^-1FQ$ with $F$ diagonal ($S$ is already diagonal). So $$C=P^-1JP=P^-1(R+S)P=P^-1RP+P^-1SP=P^-1Q^-1FQP+P^-1SP=(QP)^-1F(QP)+P^-1SP$$ is a sum of two diagonal matrices.
So, every matrix is a sum of two diagonalizable matrices, and every diagonalizable matrix is a cube, hence, every matrix is a sum of two cubes, hence, the set of all sums of two cubes, being the same as the set of all matrices, is closed under multiplication.
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
edited Jul 28 at 22:37
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
answered Jul 24 at 23:12
Gerry Myerson
143k7145294
143k7145294
add a comment |Â
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2
What does it mean for an operation to be "intern"? That word is not in common use in mathematical English.
– Gerry Myerson
Jul 17 at 13:10
It means that for A,B,C,D,M,N in Mn(C), (A^3+B^3)(C^3+D^3) can be written as M^3+N^3.
– Morpheus
Jul 17 at 20:17
In other words, the set is closed under multiplication. But maybe every complex matrix is a cube?
– Gerry Myerson
Jul 17 at 21:54
Ignore that – not every complex matrix is a cube.
– Gerry Myerson
Jul 19 at 2:54
1
It's an exercise given by my teacher, so it has to be true.
– Morpheus
Jul 20 at 11:16