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Does $lVert B^TA^{-1}BrVert_2<1$ imply that...
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Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $leftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$, for $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$?
Or, at least, does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $rholeft(begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right)<1$, where $rho(A)$ is the spectral radius of $A$?
I'm trying to solve a task where the first implication seems to me to be the sufficient condition for the thesis of the task to hold while the second implication seems to me to be equivalent (sufficient and necessary) to the thesis of the task.
I can't say much about the spectral radius... but while I can't prove it, to my intuition the implication that $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1impliesleftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$ is actually false. This perplexes me.
Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply any of the above conditions?
matrices block-matrices spectral-radius matrix-norms
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user10354138
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gaazkam
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up vote
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Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $leftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$, for $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$?
Or, at least, does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $rholeft(begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right)<1$, where $rho(A)$ is the spectral radius of $A$?
I'm trying to solve a task where the first implication seems to me to be the sufficient condition for the thesis of the task to hold while the second implication seems to me to be equivalent (sufficient and necessary) to the thesis of the task.
I can't say much about the spectral radius... but while I can't prove it, to my intuition the implication that $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1impliesleftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$ is actually false. This perplexes me.
Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply any of the above conditions?
matrices block-matrices spectral-radius matrix-norms
edited yesterday
user10354138
6,284623
asked yesterday
gaazkam
422313
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $leftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$, for $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$?
Or, at least, does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $rholeft(begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right)<1$, where $rho(A)$ is the spectral radius of $A$?
I'm trying to solve a task where the first implication seems to me to be the sufficient condition for the thesis of the task to hold while the second implication seems to me to be equivalent (sufficient and necessary) to the thesis of the task.
I can't say much about the spectral radius... but while I can't prove it, to my intuition the implication that $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1impliesleftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$ is actually false. This perplexes me.
Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply any of the above conditions?
matrices block-matrices spectral-radius matrix-norms
edited yesterday
user10354138
6,284623
asked yesterday
gaazkam
422313
Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $leftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$, for $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$?
Or, at least, does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply that $rholeft(begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right)<1$, where $rho(A)$ is the spectral radius of $A$?
I'm trying to solve a task where the first implication seems to me to be the sufficient condition for the thesis of the task to hold while the second implication seems to me to be equivalent (sufficient and necessary) to the thesis of the task.
I can't say much about the spectral radius... but while I can't prove it, to my intuition the implication that $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1impliesleftlVertbegin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}rightrVert_2<1$ is actually false. This perplexes me.
Does $leftlVert B^{operatorname{T}}A^{-1}BrightrVert_2<1$ imply any of the above conditions?
matrices block-matrices spectral-radius matrix-norms
matrices block-matrices spectral-radius matrix-norms
edited yesterday
user10354138
6,284623
asked yesterday
gaazkam
422313
edited yesterday
user10354138
6,284623
asked yesterday
gaazkam
422313
edited yesterday
user10354138
6,284623
edited yesterday
user10354138
6,284623
edited yesterday
user10354138
6,284623
6,284623
asked yesterday
gaazkam
422313
asked yesterday
gaazkam
422313
asked yesterday
gaazkam
422313
422313
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The answer to your first question is no. The answer to your second question is yes.
If $N=1$ then this is equivalent to asking whether $b^2/a<1$ implies
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2<1,$$
where $binmathbb{R}$ and $ain(0,infty)$.
But this seems false. Take $a=1/2$ and $b=2/3$. Then $b^2/a=8/9<1$, however
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2=left|begin{bmatrix}0&-4/3\0&-8/9end{bmatrix}right|_2=frac{4sqrt{13}}{9}>1.$$
Probably this counterexample can be extended to all $Ninmathbb{N}$ by letting $A=frac{1}{2}I_N$ and $B=frac{2}{3}I_N$.
On the other hand, for $N=1$ the above matrix will always have spectrum ${-b^2/a,0}$. So the spectral radius will be $b^2/a<1$.
The case for $N>1$ requires the block matrix property of determinants:
$$text{det}begin{pmatrix}A&B\0&Dend{pmatrix}=text{det}(A)text{det}(D).$$
Observe that the eigenvalues to your matrix are the solutions to
$$0=text{det}begin{pmatrix}-lambda&-A^{-1}B\0&-B^TA^{-1}B-lambdaend{pmatrix}=text{det}(-lambda)text{det}(-B^TA^{-1}B-lambda)$$
which is true precisely when $text{det}(-lambda)=0$ or else $text{det}(-B^TA^{-1}B-lambda)=0$. The former implies $lambda=0$, while the latter gives the negative eigenvalues to $B^TA^{-1}B$. It is well-known that the spectral radius is bounded above by the 2-norm. Hence
$$rhobegin{pmatrix}0&-A^{-1}B\0&-B^TA^{-1}Bend{pmatrix}=rho(B^TA^{-1}B)leq|B^TA^{-1}B|_2<1.$$
answered yesterday
Ben W
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The answer to your first question is no. The answer to your second question is yes.
If $N=1$ then this is equivalent to asking whether $b^2/a<1$ implies
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2<1,$$
where $binmathbb{R}$ and $ain(0,infty)$.
But this seems false. Take $a=1/2$ and $b=2/3$. Then $b^2/a=8/9<1$, however
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2=left|begin{bmatrix}0&-4/3\0&-8/9end{bmatrix}right|_2=frac{4sqrt{13}}{9}>1.$$
Probably this counterexample can be extended to all $Ninmathbb{N}$ by letting $A=frac{1}{2}I_N$ and $B=frac{2}{3}I_N$.
On the other hand, for $N=1$ the above matrix will always have spectrum ${-b^2/a,0}$. So the spectral radius will be $b^2/a<1$.
The case for $N>1$ requires the block matrix property of determinants:
$$text{det}begin{pmatrix}A&B\0&Dend{pmatrix}=text{det}(A)text{det}(D).$$
Observe that the eigenvalues to your matrix are the solutions to
$$0=text{det}begin{pmatrix}-lambda&-A^{-1}B\0&-B^TA^{-1}B-lambdaend{pmatrix}=text{det}(-lambda)text{det}(-B^TA^{-1}B-lambda)$$
which is true precisely when $text{det}(-lambda)=0$ or else $text{det}(-B^TA^{-1}B-lambda)=0$. The former implies $lambda=0$, while the latter gives the negative eigenvalues to $B^TA^{-1}B$. It is well-known that the spectral radius is bounded above by the 2-norm. Hence
$$rhobegin{pmatrix}0&-A^{-1}B\0&-B^TA^{-1}Bend{pmatrix}=rho(B^TA^{-1}B)leq|B^TA^{-1}B|_2<1.$$
answered yesterday
Ben W
1,079510
add a comment |
up vote
2
down vote
accepted
The answer to your first question is no. The answer to your second question is yes.
If $N=1$ then this is equivalent to asking whether $b^2/a<1$ implies
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2<1,$$
where $binmathbb{R}$ and $ain(0,infty)$.
But this seems false. Take $a=1/2$ and $b=2/3$. Then $b^2/a=8/9<1$, however
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2=left|begin{bmatrix}0&-4/3\0&-8/9end{bmatrix}right|_2=frac{4sqrt{13}}{9}>1.$$
Probably this counterexample can be extended to all $Ninmathbb{N}$ by letting $A=frac{1}{2}I_N$ and $B=frac{2}{3}I_N$.
On the other hand, for $N=1$ the above matrix will always have spectrum ${-b^2/a,0}$. So the spectral radius will be $b^2/a<1$.
The case for $N>1$ requires the block matrix property of determinants:
$$text{det}begin{pmatrix}A&B\0&Dend{pmatrix}=text{det}(A)text{det}(D).$$
Observe that the eigenvalues to your matrix are the solutions to
$$0=text{det}begin{pmatrix}-lambda&-A^{-1}B\0&-B^TA^{-1}B-lambdaend{pmatrix}=text{det}(-lambda)text{det}(-B^TA^{-1}B-lambda)$$
which is true precisely when $text{det}(-lambda)=0$ or else $text{det}(-B^TA^{-1}B-lambda)=0$. The former implies $lambda=0$, while the latter gives the negative eigenvalues to $B^TA^{-1}B$. It is well-known that the spectral radius is bounded above by the 2-norm. Hence
$$rhobegin{pmatrix}0&-A^{-1}B\0&-B^TA^{-1}Bend{pmatrix}=rho(B^TA^{-1}B)leq|B^TA^{-1}B|_2<1.$$
answered yesterday
Ben W
1,079510
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The answer to your first question is no. The answer to your second question is yes.
If $N=1$ then this is equivalent to asking whether $b^2/a<1$ implies
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2<1,$$
where $binmathbb{R}$ and $ain(0,infty)$.
But this seems false. Take $a=1/2$ and $b=2/3$. Then $b^2/a=8/9<1$, however
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2=left|begin{bmatrix}0&-4/3\0&-8/9end{bmatrix}right|_2=frac{4sqrt{13}}{9}>1.$$
Probably this counterexample can be extended to all $Ninmathbb{N}$ by letting $A=frac{1}{2}I_N$ and $B=frac{2}{3}I_N$.
On the other hand, for $N=1$ the above matrix will always have spectrum ${-b^2/a,0}$. So the spectral radius will be $b^2/a<1$.
The case for $N>1$ requires the block matrix property of determinants:
$$text{det}begin{pmatrix}A&B\0&Dend{pmatrix}=text{det}(A)text{det}(D).$$
Observe that the eigenvalues to your matrix are the solutions to
$$0=text{det}begin{pmatrix}-lambda&-A^{-1}B\0&-B^TA^{-1}B-lambdaend{pmatrix}=text{det}(-lambda)text{det}(-B^TA^{-1}B-lambda)$$
which is true precisely when $text{det}(-lambda)=0$ or else $text{det}(-B^TA^{-1}B-lambda)=0$. The former implies $lambda=0$, while the latter gives the negative eigenvalues to $B^TA^{-1}B$. It is well-known that the spectral radius is bounded above by the 2-norm. Hence
$$rhobegin{pmatrix}0&-A^{-1}B\0&-B^TA^{-1}Bend{pmatrix}=rho(B^TA^{-1}B)leq|B^TA^{-1}B|_2<1.$$
answered yesterday
Ben W
1,079510
The answer to your first question is no. The answer to your second question is yes.
If $N=1$ then this is equivalent to asking whether $b^2/a<1$ implies
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2<1,$$
where $binmathbb{R}$ and $ain(0,infty)$.
But this seems false. Take $a=1/2$ and $b=2/3$. Then $b^2/a=8/9<1$, however
$$left|begin{bmatrix}0&-b/a\0&-b^2/aend{bmatrix}right|_2=left|begin{bmatrix}0&-4/3\0&-8/9end{bmatrix}right|_2=frac{4sqrt{13}}{9}>1.$$
Probably this counterexample can be extended to all $Ninmathbb{N}$ by letting $A=frac{1}{2}I_N$ and $B=frac{2}{3}I_N$.
On the other hand, for $N=1$ the above matrix will always have spectrum ${-b^2/a,0}$. So the spectral radius will be $b^2/a<1$.
The case for $N>1$ requires the block matrix property of determinants:
$$text{det}begin{pmatrix}A&B\0&Dend{pmatrix}=text{det}(A)text{det}(D).$$
Observe that the eigenvalues to your matrix are the solutions to
$$0=text{det}begin{pmatrix}-lambda&-A^{-1}B\0&-B^TA^{-1}B-lambdaend{pmatrix}=text{det}(-lambda)text{det}(-B^TA^{-1}B-lambda)$$
which is true precisely when $text{det}(-lambda)=0$ or else $text{det}(-B^TA^{-1}B-lambda)=0$. The former implies $lambda=0$, while the latter gives the negative eigenvalues to $B^TA^{-1}B$. It is well-known that the spectral radius is bounded above by the 2-norm. Hence
$$rhobegin{pmatrix}0&-A^{-1}B\0&-B^TA^{-1}Bend{pmatrix}=rho(B^TA^{-1}B)leq|B^TA^{-1}B|_2<1.$$
answered yesterday
Ben W
1,079510
edited yesterday
answered yesterday
Ben W
1,079510
answered yesterday
Ben W
1,079510
answered yesterday
Ben W
1,079510
1,079510
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