On approximate simultaneous diagonalizationFor what values of $k$ is matrix $k A - B$ positive...
On approximate simultaneous diagonalization
For what values of $k$ is matrix $k A - B$ positive semidefinite?Bandwidth reduction of multiple matricesNearby matrices have nearby leading eigenvectors?On the solvability of a matrix equation« Generalized simultaneous diagonalization » of a pair of symmetric, non-commuting, positive semi-definite matricesError in Hoffman-Kunze (normal operators on finite-dimensional inner product space with a cyclic vector)
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It is well known that two $ntimes n$ symmetric positive semidefinite matrices $A$, $B$ such that $AB=0$ are simultaneously diagonalizable.
My question is related to the existence of a specific simultaneous diagonalization in the following sense: Let ${A_k}$, ${B_k}$ be two sequences of symmetric matrices converging to positive semidefinite matrices $A$ and $B$, respectively, such that $AB=0$. Is it the case that there exist a basis ${v_i^k}$ of eigenvectors of $A_k$ and a basis ${w_i^k}$ of eigenvectors of $B_k$, for all k, such that each $v_i^k$ and $w_i^k$ converge to some $c_i$ such that ${c_1,dots, c_n}$ form a simultaneous basis of eigenvectors for $A$ and $B$?
linear-algebra matrix-analysis
asked 10 hours ago
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$begingroup$
It is well known that two $ntimes n$ symmetric positive semidefinite matrices $A$, $B$ such that $AB=0$ are simultaneously diagonalizable.
My question is related to the existence of a specific simultaneous diagonalization in the following sense: Let ${A_k}$, ${B_k}$ be two sequences of symmetric matrices converging to positive semidefinite matrices $A$ and $B$, respectively, such that $AB=0$. Is it the case that there exist a basis ${v_i^k}$ of eigenvectors of $A_k$ and a basis ${w_i^k}$ of eigenvectors of $B_k$, for all k, such that each $v_i^k$ and $w_i^k$ converge to some $c_i$ such that ${c_1,dots, c_n}$ form a simultaneous basis of eigenvectors for $A$ and $B$?
linear-algebra matrix-analysis
asked 10 hours ago
Shake BabyShake Baby
8531 gold badge9 silver badges24 bronze badges
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1
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Just for the record: two normal matrices $A$ and $B$ are simultaneously diagonalizable if they commute, and $AB=0$ is a sufficient condition for normal matrices $A$ and $B$ to commute.
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– Jochen Glueck
10 hours ago
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@ChristianRemling: I'm not sure whether it's really that simple: what if $A$ has a double eigenvalue that splits into two simple eigenvalues for each $A_k$? Then the eigenvectors of $A_k$ for those eigenvalues are fixed up to scalar multiples, but those eigenvectors might not be the right ones to diagonalize $B$ as well.
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– Jochen Glueck
9 hours ago
add a comment
|
$begingroup$
It is well known that two $ntimes n$ symmetric positive semidefinite matrices $A$, $B$ such that $AB=0$ are simultaneously diagonalizable.
My question is related to the existence of a specific simultaneous diagonalization in the following sense: Let ${A_k}$, ${B_k}$ be two sequences of symmetric matrices converging to positive semidefinite matrices $A$ and $B$, respectively, such that $AB=0$. Is it the case that there exist a basis ${v_i^k}$ of eigenvectors of $A_k$ and a basis ${w_i^k}$ of eigenvectors of $B_k$, for all k, such that each $v_i^k$ and $w_i^k$ converge to some $c_i$ such that ${c_1,dots, c_n}$ form a simultaneous basis of eigenvectors for $A$ and $B$?
linear-algebra matrix-analysis
asked 10 hours ago
Shake BabyShake Baby
8531 gold badge9 silver badges24 bronze badges
$endgroup$
It is well known that two $ntimes n$ symmetric positive semidefinite matrices $A$, $B$ such that $AB=0$ are simultaneously diagonalizable.
My question is related to the existence of a specific simultaneous diagonalization in the following sense: Let ${A_k}$, ${B_k}$ be two sequences of symmetric matrices converging to positive semidefinite matrices $A$ and $B$, respectively, such that $AB=0$. Is it the case that there exist a basis ${v_i^k}$ of eigenvectors of $A_k$ and a basis ${w_i^k}$ of eigenvectors of $B_k$, for all k, such that each $v_i^k$ and $w_i^k$ converge to some $c_i$ such that ${c_1,dots, c_n}$ form a simultaneous basis of eigenvectors for $A$ and $B$?
linear-algebra matrix-analysis
linear-algebra matrix-analysis
asked 10 hours ago
Shake BabyShake Baby
8531 gold badge9 silver badges24 bronze badges
asked 10 hours ago
Shake BabyShake Baby
8531 gold badge9 silver badges24 bronze badges
asked 10 hours ago
Shake BabyShake Baby
8531 gold badge9 silver badges24 bronze badges
asked 10 hours ago
Shake BabyShake Baby
8531 gold badge9 silver badges24 bronze badges
asked 10 hours ago
Shake BabyShake Baby
8531 gold badge9 silver badges24 bronze badges
8531 gold badge9 silver badges24 bronze badges
1
$begingroup$
Just for the record: two normal matrices $A$ and $B$ are simultaneously diagonalizable if they commute, and $AB=0$ is a sufficient condition for normal matrices $A$ and $B$ to commute.
$endgroup$
– Jochen Glueck
10 hours ago
$begingroup$
@ChristianRemling: I'm not sure whether it's really that simple: what if $A$ has a double eigenvalue that splits into two simple eigenvalues for each $A_k$? Then the eigenvectors of $A_k$ for those eigenvalues are fixed up to scalar multiples, but those eigenvectors might not be the right ones to diagonalize $B$ as well.
$endgroup$
– Jochen Glueck
9 hours ago
add a comment
|
1
$begingroup$
Just for the record: two normal matrices $A$ and $B$ are simultaneously diagonalizable if they commute, and $AB=0$ is a sufficient condition for normal matrices $A$ and $B$ to commute.
$endgroup$
– Jochen Glueck
10 hours ago
$begingroup$
@ChristianRemling: I'm not sure whether it's really that simple: what if $A$ has a double eigenvalue that splits into two simple eigenvalues for each $A_k$? Then the eigenvectors of $A_k$ for those eigenvalues are fixed up to scalar multiples, but those eigenvectors might not be the right ones to diagonalize $B$ as well.
$endgroup$
– Jochen Glueck
9 hours ago
1
1
$begingroup$
Just for the record: two normal matrices $A$ and $B$ are simultaneously diagonalizable if they commute, and $AB=0$ is a sufficient condition for normal matrices $A$ and $B$ to commute.
$endgroup$
– Jochen Glueck
10 hours ago
$begingroup$
Just for the record: two normal matrices $A$ and $B$ are simultaneously diagonalizable if they commute, and $AB=0$ is a sufficient condition for normal matrices $A$ and $B$ to commute.
$endgroup$
– Jochen Glueck
10 hours ago
$begingroup$
@ChristianRemling: I'm not sure whether it's really that simple: what if $A$ has a double eigenvalue that splits into two simple eigenvalues for each $A_k$? Then the eigenvectors of $A_k$ for those eigenvalues are fixed up to scalar multiples, but those eigenvectors might not be the right ones to diagonalize $B$ as well.
$endgroup$
– Jochen Glueck
9 hours ago
$begingroup$
@ChristianRemling: I'm not sure whether it's really that simple: what if $A$ has a double eigenvalue that splits into two simple eigenvalues for each $A_k$? Then the eigenvectors of $A_k$ for those eigenvalues are fixed up to scalar multiples, but those eigenvectors might not be the right ones to diagonalize $B$ as well.
$endgroup$
– Jochen Glueck
9 hours ago
add a comment
|
1 Answer
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The answer is no in general.
For a $2times 2$-counterexample, let $A = 0$, let $B$ be the diagonal matrix with diagonal entries $1$ and $0$ (i.e. $B$ is the projection onto the first component), choose $B_k = B$ for each $k in mathbb{N}$ and
$$
A_k =
frac{1}{k}
begin{pmatrix}
1 & 1 \
1 & 1
end{pmatrix}
$$
for each $k in mathbb{N}$. Then each sequence $(v_k) subseteq mathbb{R}^2$ of eigenvectors of $A_k$ can only converge to a scalar multiple of $(1,1)$ or to a scalar multiple of $(1,-1)$.
However, only (scalar multiples of) the canonical unit vectors $(1,0)$ and $(0,1)$ simultaneously diagonalize $A$ and $B$.
answered 9 hours ago
Jochen GlueckJochen Glueck
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1 Answer
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1 Answer
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$begingroup$
The answer is no in general.
For a $2times 2$-counterexample, let $A = 0$, let $B$ be the diagonal matrix with diagonal entries $1$ and $0$ (i.e. $B$ is the projection onto the first component), choose $B_k = B$ for each $k in mathbb{N}$ and
$$
A_k =
frac{1}{k}
begin{pmatrix}
1 & 1 \
1 & 1
end{pmatrix}
$$
for each $k in mathbb{N}$. Then each sequence $(v_k) subseteq mathbb{R}^2$ of eigenvectors of $A_k$ can only converge to a scalar multiple of $(1,1)$ or to a scalar multiple of $(1,-1)$.
However, only (scalar multiples of) the canonical unit vectors $(1,0)$ and $(0,1)$ simultaneously diagonalize $A$ and $B$.
answered 9 hours ago
Jochen GlueckJochen Glueck
3,9641 gold badge15 silver badges31 bronze badges
$endgroup$
add a comment
|
$begingroup$
The answer is no in general.
For a $2times 2$-counterexample, let $A = 0$, let $B$ be the diagonal matrix with diagonal entries $1$ and $0$ (i.e. $B$ is the projection onto the first component), choose $B_k = B$ for each $k in mathbb{N}$ and
$$
A_k =
frac{1}{k}
begin{pmatrix}
1 & 1 \
1 & 1
end{pmatrix}
$$
for each $k in mathbb{N}$. Then each sequence $(v_k) subseteq mathbb{R}^2$ of eigenvectors of $A_k$ can only converge to a scalar multiple of $(1,1)$ or to a scalar multiple of $(1,-1)$.
However, only (scalar multiples of) the canonical unit vectors $(1,0)$ and $(0,1)$ simultaneously diagonalize $A$ and $B$.
answered 9 hours ago
Jochen GlueckJochen Glueck
3,9641 gold badge15 silver badges31 bronze badges
$endgroup$
add a comment
|
$begingroup$
The answer is no in general.
For a $2times 2$-counterexample, let $A = 0$, let $B$ be the diagonal matrix with diagonal entries $1$ and $0$ (i.e. $B$ is the projection onto the first component), choose $B_k = B$ for each $k in mathbb{N}$ and
$$
A_k =
frac{1}{k}
begin{pmatrix}
1 & 1 \
1 & 1
end{pmatrix}
$$
for each $k in mathbb{N}$. Then each sequence $(v_k) subseteq mathbb{R}^2$ of eigenvectors of $A_k$ can only converge to a scalar multiple of $(1,1)$ or to a scalar multiple of $(1,-1)$.
However, only (scalar multiples of) the canonical unit vectors $(1,0)$ and $(0,1)$ simultaneously diagonalize $A$ and $B$.
answered 9 hours ago
Jochen GlueckJochen Glueck
3,9641 gold badge15 silver badges31 bronze badges
$endgroup$
The answer is no in general.
For a $2times 2$-counterexample, let $A = 0$, let $B$ be the diagonal matrix with diagonal entries $1$ and $0$ (i.e. $B$ is the projection onto the first component), choose $B_k = B$ for each $k in mathbb{N}$ and
$$
A_k =
frac{1}{k}
begin{pmatrix}
1 & 1 \
1 & 1
end{pmatrix}
$$
for each $k in mathbb{N}$. Then each sequence $(v_k) subseteq mathbb{R}^2$ of eigenvectors of $A_k$ can only converge to a scalar multiple of $(1,1)$ or to a scalar multiple of $(1,-1)$.
However, only (scalar multiples of) the canonical unit vectors $(1,0)$ and $(0,1)$ simultaneously diagonalize $A$ and $B$.
answered 9 hours ago
Jochen GlueckJochen Glueck
3,9641 gold badge15 silver badges31 bronze badges
answered 9 hours ago
Jochen GlueckJochen Glueck
3,9641 gold badge15 silver badges31 bronze badges
answered 9 hours ago
Jochen GlueckJochen Glueck
3,9641 gold badge15 silver badges31 bronze badges
answered 9 hours ago
Jochen GlueckJochen Glueck
3,9641 gold badge15 silver badges31 bronze badges
3,9641 gold badge15 silver badges31 bronze badges
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$begingroup$
Just for the record: two normal matrices $A$ and $B$ are simultaneously diagonalizable if they commute, and $AB=0$ is a sufficient condition for normal matrices $A$ and $B$ to commute.
$endgroup$
– Jochen Glueck
10 hours ago
$begingroup$
@ChristianRemling: I'm not sure whether it's really that simple: what if $A$ has a double eigenvalue that splits into two simple eigenvalues for each $A_k$? Then the eigenvectors of $A_k$ for those eigenvalues are fixed up to scalar multiples, but those eigenvectors might not be the right ones to diagonalize $B$ as well.
$endgroup$
– Jochen Glueck
9 hours ago