If $D$ is a diagonal matrix, when is the commutator $DA - AD$ full rank?Relationship between $|a_{i,j}|$ and...
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If $D$ is a diagonal matrix, when is the commutator $DA - AD$ full rank?
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$begingroup$
Suppose $A in mathbb{C}^{n times n}$ is full rank. I'm looking for a sufficient condition on $A$ such that for some diagonal matrix $D in mathbb{C}^{n times n}$, the commutator $D A - A D$ is full rank.
I've worked out some necessary conditions- $A$ and $D$ cannot share an eigenvector, so no column of $D$ can have $n-1$ zero indices (ie, no column of $A$ is a scaled column of the identity matrix).
For $n=2$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} \
(d_2 - d_1) a_{2, 1} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)^2 a_{1,2} a_{2,1},
end{align}
so a $a_{1,2}, a_{2,1} neq 0$ is both necessary and sufficient.
But for $n=3$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} & (d_1 - d_3) a_{1, 3} \
(d_2 - d_1) a_{2, 1} & 0 & (d_2 - d_3) a_{2, 3} \
(d_3 - d_1) a_{3, 1} & (d_3 - d_2) a_{3, 2} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)(d_1-d_3)(d_2-d_3) (a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}).
end{align}
So if $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, $DA - AD$ is rank deficient regardless of the choice of $D$.
Q1: Is there a geometric interpretation for the constraint $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}) neq 0$?
Q2: I don't get a factored form of the determinant (one term depending on $D$, one term on $A$) for $n > 3$. Is $n=3$ a special case?
Q3: Is there a sufficient condition on $A$ such that $DA - AD$ is full rank for $n>3$?
I'd prefer to not assume $A$ is positive definite.
Edit: If $A$ is symmetric, $DA - AD$ is skew-symmetric and thus rank deficient if $n$ is odd.
Q4: $A$ symmetric is one way to make $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, but clearly other choices result in a rank-deficient commutator. Is there a clean way to express this for $n > 3$?
linear-algebra matrices
asked 2 days ago
lp333lp333
1336
$endgroup$
add a comment |
$begingroup$
Suppose $A in mathbb{C}^{n times n}$ is full rank. I'm looking for a sufficient condition on $A$ such that for some diagonal matrix $D in mathbb{C}^{n times n}$, the commutator $D A - A D$ is full rank.
I've worked out some necessary conditions- $A$ and $D$ cannot share an eigenvector, so no column of $D$ can have $n-1$ zero indices (ie, no column of $A$ is a scaled column of the identity matrix).
For $n=2$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} \
(d_2 - d_1) a_{2, 1} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)^2 a_{1,2} a_{2,1},
end{align}
so a $a_{1,2}, a_{2,1} neq 0$ is both necessary and sufficient.
But for $n=3$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} & (d_1 - d_3) a_{1, 3} \
(d_2 - d_1) a_{2, 1} & 0 & (d_2 - d_3) a_{2, 3} \
(d_3 - d_1) a_{3, 1} & (d_3 - d_2) a_{3, 2} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)(d_1-d_3)(d_2-d_3) (a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}).
end{align}
So if $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, $DA - AD$ is rank deficient regardless of the choice of $D$.
Q1: Is there a geometric interpretation for the constraint $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}) neq 0$?
Q2: I don't get a factored form of the determinant (one term depending on $D$, one term on $A$) for $n > 3$. Is $n=3$ a special case?
Q3: Is there a sufficient condition on $A$ such that $DA - AD$ is full rank for $n>3$?
I'd prefer to not assume $A$ is positive definite.
Edit: If $A$ is symmetric, $DA - AD$ is skew-symmetric and thus rank deficient if $n$ is odd.
Q4: $A$ symmetric is one way to make $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, but clearly other choices result in a rank-deficient commutator. Is there a clean way to express this for $n > 3$?
linear-algebra matrices
asked 2 days ago
lp333lp333
1336
$endgroup$
$begingroup$
Very interesting problem! The only thing I see for now is a generalization of what you get in Q4: if $n$ is odd and $A$ is symmetric, then the commutator $[D,A]$ is never of full rank--regardless of the choice of $D$--due to $$det([D,A])=det([D,A]^T)=det(A^TD^T-D^TA^T)=det(-[D,A])=(-1)^ndet([D,A])=-det([D,A])$$ so $det([D,A])=0$.
$endgroup$
– Frederik vom Ende
yesterday
add a comment |
$begingroup$
Suppose $A in mathbb{C}^{n times n}$ is full rank. I'm looking for a sufficient condition on $A$ such that for some diagonal matrix $D in mathbb{C}^{n times n}$, the commutator $D A - A D$ is full rank.
I've worked out some necessary conditions- $A$ and $D$ cannot share an eigenvector, so no column of $D$ can have $n-1$ zero indices (ie, no column of $A$ is a scaled column of the identity matrix).
For $n=2$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} \
(d_2 - d_1) a_{2, 1} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)^2 a_{1,2} a_{2,1},
end{align}
so a $a_{1,2}, a_{2,1} neq 0$ is both necessary and sufficient.
But for $n=3$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} & (d_1 - d_3) a_{1, 3} \
(d_2 - d_1) a_{2, 1} & 0 & (d_2 - d_3) a_{2, 3} \
(d_3 - d_1) a_{3, 1} & (d_3 - d_2) a_{3, 2} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)(d_1-d_3)(d_2-d_3) (a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}).
end{align}
So if $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, $DA - AD$ is rank deficient regardless of the choice of $D$.
Q1: Is there a geometric interpretation for the constraint $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}) neq 0$?
Q2: I don't get a factored form of the determinant (one term depending on $D$, one term on $A$) for $n > 3$. Is $n=3$ a special case?
Q3: Is there a sufficient condition on $A$ such that $DA - AD$ is full rank for $n>3$?
I'd prefer to not assume $A$ is positive definite.
Edit: If $A$ is symmetric, $DA - AD$ is skew-symmetric and thus rank deficient if $n$ is odd.
Q4: $A$ symmetric is one way to make $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, but clearly other choices result in a rank-deficient commutator. Is there a clean way to express this for $n > 3$?
linear-algebra matrices
asked 2 days ago
lp333lp333
1336
$endgroup$
Suppose $A in mathbb{C}^{n times n}$ is full rank. I'm looking for a sufficient condition on $A$ such that for some diagonal matrix $D in mathbb{C}^{n times n}$, the commutator $D A - A D$ is full rank.
I've worked out some necessary conditions- $A$ and $D$ cannot share an eigenvector, so no column of $D$ can have $n-1$ zero indices (ie, no column of $A$ is a scaled column of the identity matrix).
For $n=2$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} \
(d_2 - d_1) a_{2, 1} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)^2 a_{1,2} a_{2,1},
end{align}
so a $a_{1,2}, a_{2,1} neq 0$ is both necessary and sufficient.
But for $n=3$,
begin{align}
DA - A D &=
begin{bmatrix}
0 & (d_1 - d_2) a_{1,2} & (d_1 - d_3) a_{1, 3} \
(d_2 - d_1) a_{2, 1} & 0 & (d_2 - d_3) a_{2, 3} \
(d_3 - d_1) a_{3, 1} & (d_3 - d_2) a_{3, 2} & 0
end{bmatrix}\
det(DA - AD) &= (d_1-d_2)(d_1-d_3)(d_2-d_3) (a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}).
end{align}
So if $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, $DA - AD$ is rank deficient regardless of the choice of $D$.
Q1: Is there a geometric interpretation for the constraint $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2}) neq 0$?
Q2: I don't get a factored form of the determinant (one term depending on $D$, one term on $A$) for $n > 3$. Is $n=3$ a special case?
Q3: Is there a sufficient condition on $A$ such that $DA - AD$ is full rank for $n>3$?
I'd prefer to not assume $A$ is positive definite.
Edit: If $A$ is symmetric, $DA - AD$ is skew-symmetric and thus rank deficient if $n$ is odd.
Q4: $A$ symmetric is one way to make $(a_{1,2} a_{2,3} a_{3,1} - a_{1,3}a_{2,1}a_{3,2})=0$, but clearly other choices result in a rank-deficient commutator. Is there a clean way to express this for $n > 3$?
linear-algebra matrices
linear-algebra matrices
asked 2 days ago
lp333lp333
1336
asked 2 days ago
lp333lp333
1336
edited yesterday
lp333
asked 2 days ago
lp333lp333
1336
asked 2 days ago
lp333lp333
1336
asked 2 days ago
lp333lp333
1336
1336
$begingroup$
Very interesting problem! The only thing I see for now is a generalization of what you get in Q4: if $n$ is odd and $A$ is symmetric, then the commutator $[D,A]$ is never of full rank--regardless of the choice of $D$--due to $$det([D,A])=det([D,A]^T)=det(A^TD^T-D^TA^T)=det(-[D,A])=(-1)^ndet([D,A])=-det([D,A])$$ so $det([D,A])=0$.
$endgroup$
– Frederik vom Ende
yesterday
add a comment |
$begingroup$
Very interesting problem! The only thing I see for now is a generalization of what you get in Q4: if $n$ is odd and $A$ is symmetric, then the commutator $[D,A]$ is never of full rank--regardless of the choice of $D$--due to $$det([D,A])=det([D,A]^T)=det(A^TD^T-D^TA^T)=det(-[D,A])=(-1)^ndet([D,A])=-det([D,A])$$ so $det([D,A])=0$.
$endgroup$
– Frederik vom Ende
yesterday
$begingroup$
Very interesting problem! The only thing I see for now is a generalization of what you get in Q4: if $n$ is odd and $A$ is symmetric, then the commutator $[D,A]$ is never of full rank--regardless of the choice of $D$--due to $$det([D,A])=det([D,A]^T)=det(A^TD^T-D^TA^T)=det(-[D,A])=(-1)^ndet([D,A])=-det([D,A])$$ so $det([D,A])=0$.
$endgroup$
– Frederik vom Ende
yesterday
$begingroup$
Very interesting problem! The only thing I see for now is a generalization of what you get in Q4: if $n$ is odd and $A$ is symmetric, then the commutator $[D,A]$ is never of full rank--regardless of the choice of $D$--due to $$det([D,A])=det([D,A]^T)=det(A^TD^T-D^TA^T)=det(-[D,A])=(-1)^ndet([D,A])=-det([D,A])$$ so $det([D,A])=0$.
$endgroup$
– Frederik vom Ende
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One answer for Q3:
Sufficient condition: $n$ is even and $A$ can be partitioned as
begin{equation}
A = begin{bmatrix}
A_{11} & A_{12} \
A_{21} & A_{22}
end{bmatrix}
end{equation}
where the $n/2 times n/2$ blocks $A_{12}$ and $A_{21}$ are each full rank.
Proof: Construct $D$ as
begin{equation}
D = begin{bmatrix}
I_{n/2} & 0 \
0 & 0_{n/2}
end{bmatrix},
end{equation}
and the commutator is
begin{equation}
DA - AD = begin{bmatrix}
0_{n/2} & A_{12} \
-A_{21} & 0_{n/2}
end{bmatrix}.
end{equation}
The nonzero blocks are full rank by assumption.
answered yesterday
lp333lp333
1336
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
One answer for Q3:
Sufficient condition: $n$ is even and $A$ can be partitioned as
begin{equation}
A = begin{bmatrix}
A_{11} & A_{12} \
A_{21} & A_{22}
end{bmatrix}
end{equation}
where the $n/2 times n/2$ blocks $A_{12}$ and $A_{21}$ are each full rank.
Proof: Construct $D$ as
begin{equation}
D = begin{bmatrix}
I_{n/2} & 0 \
0 & 0_{n/2}
end{bmatrix},
end{equation}
and the commutator is
begin{equation}
DA - AD = begin{bmatrix}
0_{n/2} & A_{12} \
-A_{21} & 0_{n/2}
end{bmatrix}.
end{equation}
The nonzero blocks are full rank by assumption.
answered yesterday
lp333lp333
1336
$endgroup$
add a comment |
$begingroup$
One answer for Q3:
Sufficient condition: $n$ is even and $A$ can be partitioned as
begin{equation}
A = begin{bmatrix}
A_{11} & A_{12} \
A_{21} & A_{22}
end{bmatrix}
end{equation}
where the $n/2 times n/2$ blocks $A_{12}$ and $A_{21}$ are each full rank.
Proof: Construct $D$ as
begin{equation}
D = begin{bmatrix}
I_{n/2} & 0 \
0 & 0_{n/2}
end{bmatrix},
end{equation}
and the commutator is
begin{equation}
DA - AD = begin{bmatrix}
0_{n/2} & A_{12} \
-A_{21} & 0_{n/2}
end{bmatrix}.
end{equation}
The nonzero blocks are full rank by assumption.
answered yesterday
lp333lp333
1336
$endgroup$
add a comment |
$begingroup$
One answer for Q3:
Sufficient condition: $n$ is even and $A$ can be partitioned as
begin{equation}
A = begin{bmatrix}
A_{11} & A_{12} \
A_{21} & A_{22}
end{bmatrix}
end{equation}
where the $n/2 times n/2$ blocks $A_{12}$ and $A_{21}$ are each full rank.
Proof: Construct $D$ as
begin{equation}
D = begin{bmatrix}
I_{n/2} & 0 \
0 & 0_{n/2}
end{bmatrix},
end{equation}
and the commutator is
begin{equation}
DA - AD = begin{bmatrix}
0_{n/2} & A_{12} \
-A_{21} & 0_{n/2}
end{bmatrix}.
end{equation}
The nonzero blocks are full rank by assumption.
answered yesterday
lp333lp333
1336
$endgroup$
One answer for Q3:
Sufficient condition: $n$ is even and $A$ can be partitioned as
begin{equation}
A = begin{bmatrix}
A_{11} & A_{12} \
A_{21} & A_{22}
end{bmatrix}
end{equation}
where the $n/2 times n/2$ blocks $A_{12}$ and $A_{21}$ are each full rank.
Proof: Construct $D$ as
begin{equation}
D = begin{bmatrix}
I_{n/2} & 0 \
0 & 0_{n/2}
end{bmatrix},
end{equation}
and the commutator is
begin{equation}
DA - AD = begin{bmatrix}
0_{n/2} & A_{12} \
-A_{21} & 0_{n/2}
end{bmatrix}.
end{equation}
The nonzero blocks are full rank by assumption.
answered yesterday
lp333lp333
1336
answered yesterday
lp333lp333
1336
answered yesterday
lp333lp333
1336
answered yesterday
lp333lp333
1336
1336
add a comment |
add a comment |
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$begingroup$
Very interesting problem! The only thing I see for now is a generalization of what you get in Q4: if $n$ is odd and $A$ is symmetric, then the commutator $[D,A]$ is never of full rank--regardless of the choice of $D$--due to $$det([D,A])=det([D,A]^T)=det(A^TD^T-D^TA^T)=det(-[D,A])=(-1)^ndet([D,A])=-det([D,A])$$ so $det([D,A])=0$.
$endgroup$
– Frederik vom Ende
yesterday