Mixing
Does a matrix of minimum norm in an affine subspace of $M_n(mathbb R)$ have minimum spectral radius?
Let $mathcal U in M_n(mathbb R)$ be a subspace defined by declaring certain entries to be $0$. More precisely, let $Lambda, Theta subset {1, dots, n}$, $mathcal U$ is defined as
begin{align*}
mathcal U = { A in M_n(mathbb R): a_{ij} = 0 text{ for } (i, j) in Lambda times Theta text{ when } i neq j text{ and } a_{ii} = 0 text{ for } i = {1, dots, n} }.
end{align*}
I hope this is clear. Essentially $mathcal U$ is a subspace with certain zero patterns on its entries and in particular, the diagonal entries are $0$.
Now let $A in mathcal U$ and we consider the affine subspace $mathcal S := A + mathcal U^{perp}$ where $mathcal U^{perp}$ would be the matrix with zero pattern opposite to $mathcal U$. It is clear with respect to the inner product defined by $langle M, N rangle = text{tr}(M^TN)$, $A$ is of minimum norm in $mathcal S$.
My question: suppose $rho(A) ge a$ where $a$ is some scalar in $mathbb R$ and $rho$ denotes spectral radius, is it possible for any $B in mathcal S$, we have $rho(B) ge a$. In general, I know matrix norm is an upper bound of spectral radius and we should not expect such inequality. But I failed to construct a counterexample or prove it.
linear-algebra matrices spectral-radius matrix-norms
asked Nov 20 '18 at 0:52
user1101010
7551630
add a comment |
Let $mathcal U in M_n(mathbb R)$ be a subspace defined by declaring certain entries to be $0$. More precisely, let $Lambda, Theta subset {1, dots, n}$, $mathcal U$ is defined as
begin{align*}
mathcal U = { A in M_n(mathbb R): a_{ij} = 0 text{ for } (i, j) in Lambda times Theta text{ when } i neq j text{ and } a_{ii} = 0 text{ for } i = {1, dots, n} }.
end{align*}
I hope this is clear. Essentially $mathcal U$ is a subspace with certain zero patterns on its entries and in particular, the diagonal entries are $0$.
Now let $A in mathcal U$ and we consider the affine subspace $mathcal S := A + mathcal U^{perp}$ where $mathcal U^{perp}$ would be the matrix with zero pattern opposite to $mathcal U$. It is clear with respect to the inner product defined by $langle M, N rangle = text{tr}(M^TN)$, $A$ is of minimum norm in $mathcal S$.
My question: suppose $rho(A) ge a$ where $a$ is some scalar in $mathbb R$ and $rho$ denotes spectral radius, is it possible for any $B in mathcal S$, we have $rho(B) ge a$. In general, I know matrix norm is an upper bound of spectral radius and we should not expect such inequality. But I failed to construct a counterexample or prove it.
linear-algebra matrices spectral-radius matrix-norms
asked Nov 20 '18 at 0:52
user1101010
7551630
This may not be essential, but just to be clear: does "$iinLambda, jinTheta$" mean "$(i,j)inLambdatimesTheta$" or "$(i,j)in(Lambdatimes{1,2,ldots,n})cup({1,2,ldots,n}timesTheta)$"?
– user1551
Nov 21 '18 at 10:13
@user1551: Yes. I have a problem in formulating formally. Ideally, I want to guarantee the diagonal entries are $0$. In addition, the other entries can be zeros at will. In writing down the question, I was thinking that $Lambda cap Theta$ might not be empty so I chose the way in the question. Thanks.
– user1101010
Nov 21 '18 at 16:50
add a comment |
Let $mathcal U in M_n(mathbb R)$ be a subspace defined by declaring certain entries to be $0$. More precisely, let $Lambda, Theta subset {1, dots, n}$, $mathcal U$ is defined as
begin{align*}
mathcal U = { A in M_n(mathbb R): a_{ij} = 0 text{ for } (i, j) in Lambda times Theta text{ when } i neq j text{ and } a_{ii} = 0 text{ for } i = {1, dots, n} }.
end{align*}
I hope this is clear. Essentially $mathcal U$ is a subspace with certain zero patterns on its entries and in particular, the diagonal entries are $0$.
Now let $A in mathcal U$ and we consider the affine subspace $mathcal S := A + mathcal U^{perp}$ where $mathcal U^{perp}$ would be the matrix with zero pattern opposite to $mathcal U$. It is clear with respect to the inner product defined by $langle M, N rangle = text{tr}(M^TN)$, $A$ is of minimum norm in $mathcal S$.
My question: suppose $rho(A) ge a$ where $a$ is some scalar in $mathbb R$ and $rho$ denotes spectral radius, is it possible for any $B in mathcal S$, we have $rho(B) ge a$. In general, I know matrix norm is an upper bound of spectral radius and we should not expect such inequality. But I failed to construct a counterexample or prove it.
linear-algebra matrices spectral-radius matrix-norms
asked Nov 20 '18 at 0:52
user1101010
7551630
Let $mathcal U in M_n(mathbb R)$ be a subspace defined by declaring certain entries to be $0$. More precisely, let $Lambda, Theta subset {1, dots, n}$, $mathcal U$ is defined as
begin{align*}
mathcal U = { A in M_n(mathbb R): a_{ij} = 0 text{ for } (i, j) in Lambda times Theta text{ when } i neq j text{ and } a_{ii} = 0 text{ for } i = {1, dots, n} }.
end{align*}
I hope this is clear. Essentially $mathcal U$ is a subspace with certain zero patterns on its entries and in particular, the diagonal entries are $0$.
Now let $A in mathcal U$ and we consider the affine subspace $mathcal S := A + mathcal U^{perp}$ where $mathcal U^{perp}$ would be the matrix with zero pattern opposite to $mathcal U$. It is clear with respect to the inner product defined by $langle M, N rangle = text{tr}(M^TN)$, $A$ is of minimum norm in $mathcal S$.
My question: suppose $rho(A) ge a$ where $a$ is some scalar in $mathbb R$ and $rho$ denotes spectral radius, is it possible for any $B in mathcal S$, we have $rho(B) ge a$. In general, I know matrix norm is an upper bound of spectral radius and we should not expect such inequality. But I failed to construct a counterexample or prove it.
linear-algebra matrices spectral-radius matrix-norms
linear-algebra matrices spectral-radius matrix-norms
asked Nov 20 '18 at 0:52
user1101010
7551630
asked Nov 20 '18 at 0:52
user1101010
7551630
edited Nov 21 '18 at 16:51
asked Nov 20 '18 at 0:52
user1101010
7551630
asked Nov 20 '18 at 0:52
user1101010
7551630
asked Nov 20 '18 at 0:52
user1101010
7551630
7551630
This may not be essential, but just to be clear: does "$iinLambda, jinTheta$" mean "$(i,j)inLambdatimesTheta$" or "$(i,j)in(Lambdatimes{1,2,ldots,n})cup({1,2,ldots,n}timesTheta)$"?
– user1551
Nov 21 '18 at 10:13
@user1551: Yes. I have a problem in formulating formally. Ideally, I want to guarantee the diagonal entries are $0$. In addition, the other entries can be zeros at will. In writing down the question, I was thinking that $Lambda cap Theta$ might not be empty so I chose the way in the question. Thanks.
– user1101010
Nov 21 '18 at 16:50
add a comment |
This may not be essential, but just to be clear: does "$iinLambda, jinTheta$" mean "$(i,j)inLambdatimesTheta$" or "$(i,j)in(Lambdatimes{1,2,ldots,n})cup({1,2,ldots,n}timesTheta)$"?
– user1551
Nov 21 '18 at 10:13
@user1551: Yes. I have a problem in formulating formally. Ideally, I want to guarantee the diagonal entries are $0$. In addition, the other entries can be zeros at will. In writing down the question, I was thinking that $Lambda cap Theta$ might not be empty so I chose the way in the question. Thanks.
– user1101010
Nov 21 '18 at 16:50
This may not be essential, but just to be clear: does "$iinLambda, jinTheta$" mean "$(i,j)inLambdatimesTheta$" or "$(i,j)in(Lambdatimes{1,2,ldots,n})cup({1,2,ldots,n}timesTheta)$"?
– user1551
Nov 21 '18 at 10:13
This may not be essential, but just to be clear: does "$iinLambda, jinTheta$" mean "$(i,j)inLambdatimesTheta$" or "$(i,j)in(Lambdatimes{1,2,ldots,n})cup({1,2,ldots,n}timesTheta)$"?
– user1551
Nov 21 '18 at 10:13
@user1551: Yes. I have a problem in formulating formally. Ideally, I want to guarantee the diagonal entries are $0$. In addition, the other entries can be zeros at will. In writing down the question, I was thinking that $Lambda cap Theta$ might not be empty so I chose the way in the question. Thanks.
– user1101010
Nov 21 '18 at 16:50
@user1551: Yes. I have a problem in formulating formally. Ideally, I want to guarantee the diagonal entries are $0$. In addition, the other entries can be zeros at will. In writing down the question, I was thinking that $Lambda cap Theta$ might not be empty so I chose the way in the question. Thanks.
– user1101010
Nov 21 '18 at 16:50
add a comment |
1 Answer
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If $A$ is nilpotent, $a$ has to be non-positive. Hence $rho(B)$ is always $ge a$ and the answer to your question is yes in this case.
If $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ is not an eigenvalue of $A$, then provided that $a$ is sufficiently close to $rho(A)$, there always exists some scalar matrix $tIin U^perp$ such that $rho(A-tI)<a$ and hence the answer to your question is no in this case.
E.g. suppose $n=3, Lambda={1}$ and $Theta={3}$. Then $Ainmathcal U$ if and only if $a_{13}=0$ and $A$ has a zero diagonal. Let
$$
A=pmatrix{0&1&0\ 2&0&2\ 2&2&0}.
$$
Its spectrum is ${1+sqrt{3}, -2, 1-sqrt{3}}$ and $rho(A)=1+sqrt{3}approx2.732$. Let $ain(2, rho(A))$. Then for any $tin(rho(A)-a, rho(A)-2)$, we have $tIinmathcal U^perp$ but $rho(A-tI)=rho(A)-t<a$.
So, the only interesting case is when both $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ are eigenvalues of $A$, for which I haven't any answer yet.
answered Nov 21 '18 at 18:20
user1551
71.3k566125
add a comment |
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If $A$ is nilpotent, $a$ has to be non-positive. Hence $rho(B)$ is always $ge a$ and the answer to your question is yes in this case.
If $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ is not an eigenvalue of $A$, then provided that $a$ is sufficiently close to $rho(A)$, there always exists some scalar matrix $tIin U^perp$ such that $rho(A-tI)<a$ and hence the answer to your question is no in this case.
E.g. suppose $n=3, Lambda={1}$ and $Theta={3}$. Then $Ainmathcal U$ if and only if $a_{13}=0$ and $A$ has a zero diagonal. Let
$$
A=pmatrix{0&1&0\ 2&0&2\ 2&2&0}.
$$
Its spectrum is ${1+sqrt{3}, -2, 1-sqrt{3}}$ and $rho(A)=1+sqrt{3}approx2.732$. Let $ain(2, rho(A))$. Then for any $tin(rho(A)-a, rho(A)-2)$, we have $tIinmathcal U^perp$ but $rho(A-tI)=rho(A)-t<a$.
So, the only interesting case is when both $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ are eigenvalues of $A$, for which I haven't any answer yet.
answered Nov 21 '18 at 18:20
user1551
71.3k566125
add a comment |
If $A$ is nilpotent, $a$ has to be non-positive. Hence $rho(B)$ is always $ge a$ and the answer to your question is yes in this case.
If $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ is not an eigenvalue of $A$, then provided that $a$ is sufficiently close to $rho(A)$, there always exists some scalar matrix $tIin U^perp$ such that $rho(A-tI)<a$ and hence the answer to your question is no in this case.
E.g. suppose $n=3, Lambda={1}$ and $Theta={3}$. Then $Ainmathcal U$ if and only if $a_{13}=0$ and $A$ has a zero diagonal. Let
$$
A=pmatrix{0&1&0\ 2&0&2\ 2&2&0}.
$$
Its spectrum is ${1+sqrt{3}, -2, 1-sqrt{3}}$ and $rho(A)=1+sqrt{3}approx2.732$. Let $ain(2, rho(A))$. Then for any $tin(rho(A)-a, rho(A)-2)$, we have $tIinmathcal U^perp$ but $rho(A-tI)=rho(A)-t<a$.
So, the only interesting case is when both $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ are eigenvalues of $A$, for which I haven't any answer yet.
answered Nov 21 '18 at 18:20
user1551
71.3k566125
add a comment |
If $A$ is nilpotent, $a$ has to be non-positive. Hence $rho(B)$ is always $ge a$ and the answer to your question is yes in this case.
If $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ is not an eigenvalue of $A$, then provided that $a$ is sufficiently close to $rho(A)$, there always exists some scalar matrix $tIin U^perp$ such that $rho(A-tI)<a$ and hence the answer to your question is no in this case.
E.g. suppose $n=3, Lambda={1}$ and $Theta={3}$. Then $Ainmathcal U$ if and only if $a_{13}=0$ and $A$ has a zero diagonal. Let
$$
A=pmatrix{0&1&0\ 2&0&2\ 2&2&0}.
$$
Its spectrum is ${1+sqrt{3}, -2, 1-sqrt{3}}$ and $rho(A)=1+sqrt{3}approx2.732$. Let $ain(2, rho(A))$. Then for any $tin(rho(A)-a, rho(A)-2)$, we have $tIinmathcal U^perp$ but $rho(A-tI)=rho(A)-t<a$.
So, the only interesting case is when both $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ are eigenvalues of $A$, for which I haven't any answer yet.
answered Nov 21 '18 at 18:20
user1551
71.3k566125
If $A$ is nilpotent, $a$ has to be non-positive. Hence $rho(B)$ is always $ge a$ and the answer to your question is yes in this case.
If $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ is not an eigenvalue of $A$, then provided that $a$ is sufficiently close to $rho(A)$, there always exists some scalar matrix $tIin U^perp$ such that $rho(A-tI)<a$ and hence the answer to your question is no in this case.
E.g. suppose $n=3, Lambda={1}$ and $Theta={3}$. Then $Ainmathcal U$ if and only if $a_{13}=0$ and $A$ has a zero diagonal. Let
$$
A=pmatrix{0&1&0\ 2&0&2\ 2&2&0}.
$$
Its spectrum is ${1+sqrt{3}, -2, 1-sqrt{3}}$ and $rho(A)=1+sqrt{3}approx2.732$. Let $ain(2, rho(A))$. Then for any $tin(rho(A)-a, rho(A)-2)$, we have $tIinmathcal U^perp$ but $rho(A-tI)=rho(A)-t<a$.
So, the only interesting case is when both $lambda=argmax_{|lambda_i(A)|=rho(A)}|Re(lambda_i(A))|$ and $-lambda$ are eigenvalues of $A$, for which I haven't any answer yet.
answered Nov 21 '18 at 18:20
user1551
71.3k566125
answered Nov 21 '18 at 18:20
user1551
71.3k566125
answered Nov 21 '18 at 18:20
user1551
71.3k566125
answered Nov 21 '18 at 18:20
user1551
71.3k566125
71.3k566125
add a comment |
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This may not be essential, but just to be clear: does "$iinLambda, jinTheta$" mean "$(i,j)inLambdatimesTheta$" or "$(i,j)in(Lambdatimes{1,2,ldots,n})cup({1,2,ldots,n}timesTheta)$"?
– user1551
Nov 21 '18 at 10:13
@user1551: Yes. I have a problem in formulating formally. Ideally, I want to guarantee the diagonal entries are $0$. In addition, the other entries can be zeros at will. In writing down the question, I was thinking that $Lambda cap Theta$ might not be empty so I chose the way in the question. Thanks.
– user1101010
Nov 21 '18 at 16:50