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For $AinmathbbR^ntimes n$ and $BinmathbbR^ntimes k$, does $(I_n-aA)^-1Bb$ determine $ainmathbbR$ and $binmathbbR^k$?
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Let $A$ be a real $ntimes n$ matrix, and $B$ be a real $ntimes k$
matrix of rank $k<n$, with $mathrmcol(B)$ (the column space of $B$) not an
invariant subspace of $A$. We assume $Aneq0,I_n$. Consider $$f(a,b)=(I_n-aA)^-1Bb,$$ for real
scalar $ain S$ and $binmathbbR^k$. $S$ is the set of values of $a$ such that
$(I_n-aA)$ is invertible.
We say that $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $f(a,b)=f(a^ast
,b^ast)$ implies $(a,b)=(a^ast,b^ast)$ for any two pairs
$(a,b),(a^ast,b^ast)inOmega$.
The problem is to describe the set over which $f(a,b)$ does not identify $a$ and $b$.
What I have done so far:
Rewrite $f(a,b)=f(a^ast,b^ast)$ as
$$
B(b-b^ast)+AB(ab^ast-a^astb)=0.tag1label1
$$
Thus, $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $eqref1$ implies
$(a,b)=(a^ast,b^ast)$ for all $(a,b),(a^ast,b^ast)inOmega$.
So far I understand only two particular cases:
- If $mathrmrank(B,AB)=2k$, then $eqref1$ is satisfied iff $b-b^ast=0$ and
$ab^ast-a^astb=0$, from which $(a,b)=(a^ast,b^ast)$ provided that
$bneq0$. Hence, in this case $f(a,b)$ identifies $a$ and $b$ over
$StimesmathbbR^k/0$. - Partition $B$ as $(B_1,B_2)$ where $B_1$ is $ntimes k_1$ and
$B_2$ is $ntimes k_2$, with $0<k_1<k$, and suppose that $AB_1=B_1L$
for some $k_1times k_1$ matrix $L$ ($mathrmcol(B_1)$ is an invariant
subspace of $A$), and that $mathrmrank(B,AB)=k+k_2.$ Then, letting
$left( b_1^prime,b_2^primeright) $ be the partition of
$b^prime$ conformable with that of $B,$ $eqref1$ becomes
$$
B_1(b_1-b_1^ast+L(ab_1^ast-a^astb_1))+B_2(b_2%
-b_2^ast)+AB_2(ab_2^ast-a^astb_2)=0.
$$
Since the columns of $(B_1,B_2,AB_2)$ are linearly independent,
$eqref1$ is satisfied if and only if $b_1-b_1^ast+L(ab_1^ast
-a^astb_1)=0$, $b_2-b_2^ast=0$, and $ab_2^ast-a^ast%
b_2=0$. Combining the second and third equalities, gives $b_2=b_2^ast
$ and $a=a^ast$, provided that $b_2neq0.$ Hence the first equality
becomes $left( I_k_1-aLright) left( b_1-b_1^astright) =0$,
which is equivalent to $b_1-b_1^ast$ because $I_k_1-aL$ is
invertible for any $ain S$. This means that $f(a,b)$ identifies $(a,b)$
provided that not all entries of $b$ associated to columns of $B$ not in
$mathrmcol(A)$ are zero (if they were we would essentially be in the
case when $mathrmcol(B)$ is an invariant subspace of $A$--in that case it is clear that
$left( a,bright) $ cannot be identified from $f(a,b)$)
Update:
Let's look at the case $k=2.$ Write $B=(B_1,B_2)$, so $B_1$ and $B_2$ are vectors. Assume
$mathrmrank(B,AB)=mathrmrank(B,AB_2)=k+1$ (the case $mathrmrank(B,AB)=2k$ is trivial, see case 1 above, and so is the case $mathrmrank(B,AB)=k$), and write $$AB_1%
=l_1B_1+l_2B_2+l_3AB_2,$$ for $l_1,l_2,l_3inmathbbR$.
From $eqref1$,
beginequation
left[ beta_1-beta_1^ast+l_1(abeta_1^ast-a^astbeta
_1)right] B_1+left[ beta_2-beta_2^ast+l_2(abeta_1^ast
-a^astbeta_1)right] B_2+left[ abeta_2^ast-a^astbeta
_2+l_3(abeta_1^ast-a^astbeta_1)right] AB_2=0.
endequation
Since $mathrmrank(B,AB_2)=k+1,$ this equality is satisfied iff
$$left{
beginarray
[c]c%
b_1-b_1^ast+l_1(ab_1^ast-a^astb_1)=0\
b_2-b_2^ast+l_2(ab_1^ast-a^astb_1)=0\
ab_2^ast-a^astb_2+l_3(ab_1^ast-a^astb_1)=0
endarray
right.$$
As a linear system in the unknowns $b_1^ast,b_2^ast,a^ast$, this
is
$$Cleft[
beginarray
[c]c%
b_1^ast\
b_2^ast\
a^ast%
endarray
right] =left[
beginarray
[c]c%
b_1\
b_2\
0
endarray
right],tag1label2$$
where
$$C=left[
beginarray
[c]ccc%
1-l_1a & 0 & l_1b_1\
-l_2a & 1 & l_2b_1\
-l_3a & a & l_3b_1+b_2%
endarray
right] .$$
If $mathrmrank(C)=3$, the only solution to the system is $left( b^ast
,a^astright) =left( b,aright) $. If $mathrmrank(C)<3$, there may
be other solutions.
Note that
$$
det(C)=left( l_1a-1right) b_2-left( l_2a+l_3right) b_1.
$$
Note that for fixed $a$, the set of $(b_1,b_2)$ such that $det(C)=0$ is a line in $mathbbR^2$.
Incidentally note that case 2. above (with $k=2$) obtains when $l_2=l_3=0.$ In that case,
$det(C)=left( l_1a-1right) b_2$, so $det(C)=0$ only if $b_2=0$
($l_1a-1=0$ is impossible as $l_1$ must be an eigenvalue of $A$, so $l_1a-1=0$ would contradict invertibility of $(I_n-aA)$). That is,
in that case, we find again the result that $b,a$ is identifiable provided
that $b_2neq0$.
linear-algebra matrices invariant-subspace
asked Jul 15 at 23:15
mark
244214
add a comment |Â
up vote
8
down vote
favorite
Let $A$ be a real $ntimes n$ matrix, and $B$ be a real $ntimes k$
matrix of rank $k<n$, with $mathrmcol(B)$ (the column space of $B$) not an
invariant subspace of $A$. We assume $Aneq0,I_n$. Consider $$f(a,b)=(I_n-aA)^-1Bb,$$ for real
scalar $ain S$ and $binmathbbR^k$. $S$ is the set of values of $a$ such that
$(I_n-aA)$ is invertible.
We say that $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $f(a,b)=f(a^ast
,b^ast)$ implies $(a,b)=(a^ast,b^ast)$ for any two pairs
$(a,b),(a^ast,b^ast)inOmega$.
The problem is to describe the set over which $f(a,b)$ does not identify $a$ and $b$.
What I have done so far:
Rewrite $f(a,b)=f(a^ast,b^ast)$ as
$$
B(b-b^ast)+AB(ab^ast-a^astb)=0.tag1label1
$$
Thus, $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $eqref1$ implies
$(a,b)=(a^ast,b^ast)$ for all $(a,b),(a^ast,b^ast)inOmega$.
So far I understand only two particular cases:
- If $mathrmrank(B,AB)=2k$, then $eqref1$ is satisfied iff $b-b^ast=0$ and
$ab^ast-a^astb=0$, from which $(a,b)=(a^ast,b^ast)$ provided that
$bneq0$. Hence, in this case $f(a,b)$ identifies $a$ and $b$ over
$StimesmathbbR^k/0$. - Partition $B$ as $(B_1,B_2)$ where $B_1$ is $ntimes k_1$ and
$B_2$ is $ntimes k_2$, with $0<k_1<k$, and suppose that $AB_1=B_1L$
for some $k_1times k_1$ matrix $L$ ($mathrmcol(B_1)$ is an invariant
subspace of $A$), and that $mathrmrank(B,AB)=k+k_2.$ Then, letting
$left( b_1^prime,b_2^primeright) $ be the partition of
$b^prime$ conformable with that of $B,$ $eqref1$ becomes
$$
B_1(b_1-b_1^ast+L(ab_1^ast-a^astb_1))+B_2(b_2%
-b_2^ast)+AB_2(ab_2^ast-a^astb_2)=0.
$$
Since the columns of $(B_1,B_2,AB_2)$ are linearly independent,
$eqref1$ is satisfied if and only if $b_1-b_1^ast+L(ab_1^ast
-a^astb_1)=0$, $b_2-b_2^ast=0$, and $ab_2^ast-a^ast%
b_2=0$. Combining the second and third equalities, gives $b_2=b_2^ast
$ and $a=a^ast$, provided that $b_2neq0.$ Hence the first equality
becomes $left( I_k_1-aLright) left( b_1-b_1^astright) =0$,
which is equivalent to $b_1-b_1^ast$ because $I_k_1-aL$ is
invertible for any $ain S$. This means that $f(a,b)$ identifies $(a,b)$
provided that not all entries of $b$ associated to columns of $B$ not in
$mathrmcol(A)$ are zero (if they were we would essentially be in the
case when $mathrmcol(B)$ is an invariant subspace of $A$--in that case it is clear that
$left( a,bright) $ cannot be identified from $f(a,b)$)
Update:
Let's look at the case $k=2.$ Write $B=(B_1,B_2)$, so $B_1$ and $B_2$ are vectors. Assume
$mathrmrank(B,AB)=mathrmrank(B,AB_2)=k+1$ (the case $mathrmrank(B,AB)=2k$ is trivial, see case 1 above, and so is the case $mathrmrank(B,AB)=k$), and write $$AB_1%
=l_1B_1+l_2B_2+l_3AB_2,$$ for $l_1,l_2,l_3inmathbbR$.
From $eqref1$,
beginequation
left[ beta_1-beta_1^ast+l_1(abeta_1^ast-a^astbeta
_1)right] B_1+left[ beta_2-beta_2^ast+l_2(abeta_1^ast
-a^astbeta_1)right] B_2+left[ abeta_2^ast-a^astbeta
_2+l_3(abeta_1^ast-a^astbeta_1)right] AB_2=0.
endequation
Since $mathrmrank(B,AB_2)=k+1,$ this equality is satisfied iff
$$left{
beginarray
[c]c%
b_1-b_1^ast+l_1(ab_1^ast-a^astb_1)=0\
b_2-b_2^ast+l_2(ab_1^ast-a^astb_1)=0\
ab_2^ast-a^astb_2+l_3(ab_1^ast-a^astb_1)=0
endarray
right.$$
As a linear system in the unknowns $b_1^ast,b_2^ast,a^ast$, this
is
$$Cleft[
beginarray
[c]c%
b_1^ast\
b_2^ast\
a^ast%
endarray
right] =left[
beginarray
[c]c%
b_1\
b_2\
0
endarray
right],tag1label2$$
where
$$C=left[
beginarray
[c]ccc%
1-l_1a & 0 & l_1b_1\
-l_2a & 1 & l_2b_1\
-l_3a & a & l_3b_1+b_2%
endarray
right] .$$
If $mathrmrank(C)=3$, the only solution to the system is $left( b^ast
,a^astright) =left( b,aright) $. If $mathrmrank(C)<3$, there may
be other solutions.
Note that
$$
det(C)=left( l_1a-1right) b_2-left( l_2a+l_3right) b_1.
$$
Note that for fixed $a$, the set of $(b_1,b_2)$ such that $det(C)=0$ is a line in $mathbbR^2$.
Incidentally note that case 2. above (with $k=2$) obtains when $l_2=l_3=0.$ In that case,
$det(C)=left( l_1a-1right) b_2$, so $det(C)=0$ only if $b_2=0$
($l_1a-1=0$ is impossible as $l_1$ must be an eigenvalue of $A$, so $l_1a-1=0$ would contradict invertibility of $(I_n-aA)$). That is,
in that case, we find again the result that $b,a$ is identifiable provided
that $b_2neq0$.
linear-algebra matrices invariant-subspace
asked Jul 15 at 23:15
mark
244214
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Let $A$ be a real $ntimes n$ matrix, and $B$ be a real $ntimes k$
matrix of rank $k<n$, with $mathrmcol(B)$ (the column space of $B$) not an
invariant subspace of $A$. We assume $Aneq0,I_n$. Consider $$f(a,b)=(I_n-aA)^-1Bb,$$ for real
scalar $ain S$ and $binmathbbR^k$. $S$ is the set of values of $a$ such that
$(I_n-aA)$ is invertible.
We say that $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $f(a,b)=f(a^ast
,b^ast)$ implies $(a,b)=(a^ast,b^ast)$ for any two pairs
$(a,b),(a^ast,b^ast)inOmega$.
The problem is to describe the set over which $f(a,b)$ does not identify $a$ and $b$.
What I have done so far:
Rewrite $f(a,b)=f(a^ast,b^ast)$ as
$$
B(b-b^ast)+AB(ab^ast-a^astb)=0.tag1label1
$$
Thus, $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $eqref1$ implies
$(a,b)=(a^ast,b^ast)$ for all $(a,b),(a^ast,b^ast)inOmega$.
So far I understand only two particular cases:
- If $mathrmrank(B,AB)=2k$, then $eqref1$ is satisfied iff $b-b^ast=0$ and
$ab^ast-a^astb=0$, from which $(a,b)=(a^ast,b^ast)$ provided that
$bneq0$. Hence, in this case $f(a,b)$ identifies $a$ and $b$ over
$StimesmathbbR^k/0$. - Partition $B$ as $(B_1,B_2)$ where $B_1$ is $ntimes k_1$ and
$B_2$ is $ntimes k_2$, with $0<k_1<k$, and suppose that $AB_1=B_1L$
for some $k_1times k_1$ matrix $L$ ($mathrmcol(B_1)$ is an invariant
subspace of $A$), and that $mathrmrank(B,AB)=k+k_2.$ Then, letting
$left( b_1^prime,b_2^primeright) $ be the partition of
$b^prime$ conformable with that of $B,$ $eqref1$ becomes
$$
B_1(b_1-b_1^ast+L(ab_1^ast-a^astb_1))+B_2(b_2%
-b_2^ast)+AB_2(ab_2^ast-a^astb_2)=0.
$$
Since the columns of $(B_1,B_2,AB_2)$ are linearly independent,
$eqref1$ is satisfied if and only if $b_1-b_1^ast+L(ab_1^ast
-a^astb_1)=0$, $b_2-b_2^ast=0$, and $ab_2^ast-a^ast%
b_2=0$. Combining the second and third equalities, gives $b_2=b_2^ast
$ and $a=a^ast$, provided that $b_2neq0.$ Hence the first equality
becomes $left( I_k_1-aLright) left( b_1-b_1^astright) =0$,
which is equivalent to $b_1-b_1^ast$ because $I_k_1-aL$ is
invertible for any $ain S$. This means that $f(a,b)$ identifies $(a,b)$
provided that not all entries of $b$ associated to columns of $B$ not in
$mathrmcol(A)$ are zero (if they were we would essentially be in the
case when $mathrmcol(B)$ is an invariant subspace of $A$--in that case it is clear that
$left( a,bright) $ cannot be identified from $f(a,b)$)
Update:
Let's look at the case $k=2.$ Write $B=(B_1,B_2)$, so $B_1$ and $B_2$ are vectors. Assume
$mathrmrank(B,AB)=mathrmrank(B,AB_2)=k+1$ (the case $mathrmrank(B,AB)=2k$ is trivial, see case 1 above, and so is the case $mathrmrank(B,AB)=k$), and write $$AB_1%
=l_1B_1+l_2B_2+l_3AB_2,$$ for $l_1,l_2,l_3inmathbbR$.
From $eqref1$,
beginequation
left[ beta_1-beta_1^ast+l_1(abeta_1^ast-a^astbeta
_1)right] B_1+left[ beta_2-beta_2^ast+l_2(abeta_1^ast
-a^astbeta_1)right] B_2+left[ abeta_2^ast-a^astbeta
_2+l_3(abeta_1^ast-a^astbeta_1)right] AB_2=0.
endequation
Since $mathrmrank(B,AB_2)=k+1,$ this equality is satisfied iff
$$left{
beginarray
[c]c%
b_1-b_1^ast+l_1(ab_1^ast-a^astb_1)=0\
b_2-b_2^ast+l_2(ab_1^ast-a^astb_1)=0\
ab_2^ast-a^astb_2+l_3(ab_1^ast-a^astb_1)=0
endarray
right.$$
As a linear system in the unknowns $b_1^ast,b_2^ast,a^ast$, this
is
$$Cleft[
beginarray
[c]c%
b_1^ast\
b_2^ast\
a^ast%
endarray
right] =left[
beginarray
[c]c%
b_1\
b_2\
0
endarray
right],tag1label2$$
where
$$C=left[
beginarray
[c]ccc%
1-l_1a & 0 & l_1b_1\
-l_2a & 1 & l_2b_1\
-l_3a & a & l_3b_1+b_2%
endarray
right] .$$
If $mathrmrank(C)=3$, the only solution to the system is $left( b^ast
,a^astright) =left( b,aright) $. If $mathrmrank(C)<3$, there may
be other solutions.
Note that
$$
det(C)=left( l_1a-1right) b_2-left( l_2a+l_3right) b_1.
$$
Note that for fixed $a$, the set of $(b_1,b_2)$ such that $det(C)=0$ is a line in $mathbbR^2$.
Incidentally note that case 2. above (with $k=2$) obtains when $l_2=l_3=0.$ In that case,
$det(C)=left( l_1a-1right) b_2$, so $det(C)=0$ only if $b_2=0$
($l_1a-1=0$ is impossible as $l_1$ must be an eigenvalue of $A$, so $l_1a-1=0$ would contradict invertibility of $(I_n-aA)$). That is,
in that case, we find again the result that $b,a$ is identifiable provided
that $b_2neq0$.
linear-algebra matrices invariant-subspace
asked Jul 15 at 23:15
mark
244214
Let $A$ be a real $ntimes n$ matrix, and $B$ be a real $ntimes k$
matrix of rank $k<n$, with $mathrmcol(B)$ (the column space of $B$) not an
invariant subspace of $A$. We assume $Aneq0,I_n$. Consider $$f(a,b)=(I_n-aA)^-1Bb,$$ for real
scalar $ain S$ and $binmathbbR^k$. $S$ is the set of values of $a$ such that
$(I_n-aA)$ is invertible.
We say that $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $f(a,b)=f(a^ast
,b^ast)$ implies $(a,b)=(a^ast,b^ast)$ for any two pairs
$(a,b),(a^ast,b^ast)inOmega$.
The problem is to describe the set over which $f(a,b)$ does not identify $a$ and $b$.
What I have done so far:
Rewrite $f(a,b)=f(a^ast,b^ast)$ as
$$
B(b-b^ast)+AB(ab^ast-a^astb)=0.tag1label1
$$
Thus, $f(a,b)$ identifies $a$ and $b$ over $Omega$ if $eqref1$ implies
$(a,b)=(a^ast,b^ast)$ for all $(a,b),(a^ast,b^ast)inOmega$.
So far I understand only two particular cases:
- If $mathrmrank(B,AB)=2k$, then $eqref1$ is satisfied iff $b-b^ast=0$ and
$ab^ast-a^astb=0$, from which $(a,b)=(a^ast,b^ast)$ provided that
$bneq0$. Hence, in this case $f(a,b)$ identifies $a$ and $b$ over
$StimesmathbbR^k/0$. - Partition $B$ as $(B_1,B_2)$ where $B_1$ is $ntimes k_1$ and
$B_2$ is $ntimes k_2$, with $0<k_1<k$, and suppose that $AB_1=B_1L$
for some $k_1times k_1$ matrix $L$ ($mathrmcol(B_1)$ is an invariant
subspace of $A$), and that $mathrmrank(B,AB)=k+k_2.$ Then, letting
$left( b_1^prime,b_2^primeright) $ be the partition of
$b^prime$ conformable with that of $B,$ $eqref1$ becomes
$$
B_1(b_1-b_1^ast+L(ab_1^ast-a^astb_1))+B_2(b_2%
-b_2^ast)+AB_2(ab_2^ast-a^astb_2)=0.
$$
Since the columns of $(B_1,B_2,AB_2)$ are linearly independent,
$eqref1$ is satisfied if and only if $b_1-b_1^ast+L(ab_1^ast
-a^astb_1)=0$, $b_2-b_2^ast=0$, and $ab_2^ast-a^ast%
b_2=0$. Combining the second and third equalities, gives $b_2=b_2^ast
$ and $a=a^ast$, provided that $b_2neq0.$ Hence the first equality
becomes $left( I_k_1-aLright) left( b_1-b_1^astright) =0$,
which is equivalent to $b_1-b_1^ast$ because $I_k_1-aL$ is
invertible for any $ain S$. This means that $f(a,b)$ identifies $(a,b)$
provided that not all entries of $b$ associated to columns of $B$ not in
$mathrmcol(A)$ are zero (if they were we would essentially be in the
case when $mathrmcol(B)$ is an invariant subspace of $A$--in that case it is clear that
$left( a,bright) $ cannot be identified from $f(a,b)$)
Update:
Let's look at the case $k=2.$ Write $B=(B_1,B_2)$, so $B_1$ and $B_2$ are vectors. Assume
$mathrmrank(B,AB)=mathrmrank(B,AB_2)=k+1$ (the case $mathrmrank(B,AB)=2k$ is trivial, see case 1 above, and so is the case $mathrmrank(B,AB)=k$), and write $$AB_1%
=l_1B_1+l_2B_2+l_3AB_2,$$ for $l_1,l_2,l_3inmathbbR$.
From $eqref1$,
beginequation
left[ beta_1-beta_1^ast+l_1(abeta_1^ast-a^astbeta
_1)right] B_1+left[ beta_2-beta_2^ast+l_2(abeta_1^ast
-a^astbeta_1)right] B_2+left[ abeta_2^ast-a^astbeta
_2+l_3(abeta_1^ast-a^astbeta_1)right] AB_2=0.
endequation
Since $mathrmrank(B,AB_2)=k+1,$ this equality is satisfied iff
$$left{
beginarray
[c]c%
b_1-b_1^ast+l_1(ab_1^ast-a^astb_1)=0\
b_2-b_2^ast+l_2(ab_1^ast-a^astb_1)=0\
ab_2^ast-a^astb_2+l_3(ab_1^ast-a^astb_1)=0
endarray
right.$$
As a linear system in the unknowns $b_1^ast,b_2^ast,a^ast$, this
is
$$Cleft[
beginarray
[c]c%
b_1^ast\
b_2^ast\
a^ast%
endarray
right] =left[
beginarray
[c]c%
b_1\
b_2\
0
endarray
right],tag1label2$$
where
$$C=left[
beginarray
[c]ccc%
1-l_1a & 0 & l_1b_1\
-l_2a & 1 & l_2b_1\
-l_3a & a & l_3b_1+b_2%
endarray
right] .$$
If $mathrmrank(C)=3$, the only solution to the system is $left( b^ast
,a^astright) =left( b,aright) $. If $mathrmrank(C)<3$, there may
be other solutions.
Note that
$$
det(C)=left( l_1a-1right) b_2-left( l_2a+l_3right) b_1.
$$
Note that for fixed $a$, the set of $(b_1,b_2)$ such that $det(C)=0$ is a line in $mathbbR^2$.
Incidentally note that case 2. above (with $k=2$) obtains when $l_2=l_3=0.$ In that case,
$det(C)=left( l_1a-1right) b_2$, so $det(C)=0$ only if $b_2=0$
($l_1a-1=0$ is impossible as $l_1$ must be an eigenvalue of $A$, so $l_1a-1=0$ would contradict invertibility of $(I_n-aA)$). That is,
in that case, we find again the result that $b,a$ is identifiable provided
that $b_2neq0$.
linear-algebra matrices invariant-subspace
asked Jul 15 at 23:15
mark
244214
edited Jul 24 at 10:02
asked Jul 15 at 23:15
mark
244214
asked Jul 15 at 23:15
mark
244214
asked Jul 15 at 23:15
mark
244214
244214
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Here are some ideas.
We consider the relation $(I_n-aA)^-1Bb=v$ where $A,B$ and the vector $v$ are known and $a,b=[b_i]$ are unknown. Note that $ngeq k,operatornamerank(B)=k$ and $1/anotin operatornamespectrum(A)$.
Then $Bb=(I-aA)v=v-aAv$, that is $Bb+aAv=v$. Let $B=[B_1,cdots,B_k]$.
Assume that $A,B,v$ are randomly chosen. Then the $(B_i)_i,Av=w,v$ are random vectors in $mathbbR^n$.
The above equation can be rewritten $sum_ileq kb_iB_i+aw=v$, that is, we want to decompose the vector $v$ on the $k+1$ vectors $(B_i)_i,w$; moreover, the decomposition must be unique.
It is possible with probability $1$ when $n=k+1$; otherwise, it is possible with probability $0$.
When $A$ is random, it's invertible with probability $1$. Here, $A$ is not assumed to be invertible and the question is: how to choose $v$ so that the decomposition exists and is unique? In particular, the vectors $vinker(A)$ are not convenient. I think that there are many such subcases and I stand prudently on the sidelines.
edited Jul 20 at 19:10
Botond
3,8782632
answered Jul 18 at 13:10
loup blanc
20.4k21549
thanks for ideas. perhaps I should modify my aim, please see update. It is probably true that $a$ and $b$ are identified except on sets of measure zero. maybe there's a simple way to see that
– mark
Jul 20 at 17:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here are some ideas.
We consider the relation $(I_n-aA)^-1Bb=v$ where $A,B$ and the vector $v$ are known and $a,b=[b_i]$ are unknown. Note that $ngeq k,operatornamerank(B)=k$ and $1/anotin operatornamespectrum(A)$.
Then $Bb=(I-aA)v=v-aAv$, that is $Bb+aAv=v$. Let $B=[B_1,cdots,B_k]$.
Assume that $A,B,v$ are randomly chosen. Then the $(B_i)_i,Av=w,v$ are random vectors in $mathbbR^n$.
The above equation can be rewritten $sum_ileq kb_iB_i+aw=v$, that is, we want to decompose the vector $v$ on the $k+1$ vectors $(B_i)_i,w$; moreover, the decomposition must be unique.
It is possible with probability $1$ when $n=k+1$; otherwise, it is possible with probability $0$.
When $A$ is random, it's invertible with probability $1$. Here, $A$ is not assumed to be invertible and the question is: how to choose $v$ so that the decomposition exists and is unique? In particular, the vectors $vinker(A)$ are not convenient. I think that there are many such subcases and I stand prudently on the sidelines.
edited Jul 20 at 19:10
Botond
3,8782632
answered Jul 18 at 13:10
loup blanc
20.4k21549
thanks for ideas. perhaps I should modify my aim, please see update. It is probably true that $a$ and $b$ are identified except on sets of measure zero. maybe there's a simple way to see that
– mark
Jul 20 at 17:51
add a comment |Â
up vote
1
down vote
Here are some ideas.
We consider the relation $(I_n-aA)^-1Bb=v$ where $A,B$ and the vector $v$ are known and $a,b=[b_i]$ are unknown. Note that $ngeq k,operatornamerank(B)=k$ and $1/anotin operatornamespectrum(A)$.
Then $Bb=(I-aA)v=v-aAv$, that is $Bb+aAv=v$. Let $B=[B_1,cdots,B_k]$.
Assume that $A,B,v$ are randomly chosen. Then the $(B_i)_i,Av=w,v$ are random vectors in $mathbbR^n$.
The above equation can be rewritten $sum_ileq kb_iB_i+aw=v$, that is, we want to decompose the vector $v$ on the $k+1$ vectors $(B_i)_i,w$; moreover, the decomposition must be unique.
It is possible with probability $1$ when $n=k+1$; otherwise, it is possible with probability $0$.
When $A$ is random, it's invertible with probability $1$. Here, $A$ is not assumed to be invertible and the question is: how to choose $v$ so that the decomposition exists and is unique? In particular, the vectors $vinker(A)$ are not convenient. I think that there are many such subcases and I stand prudently on the sidelines.
edited Jul 20 at 19:10
Botond
3,8782632
answered Jul 18 at 13:10
loup blanc
20.4k21549
thanks for ideas. perhaps I should modify my aim, please see update. It is probably true that $a$ and $b$ are identified except on sets of measure zero. maybe there's a simple way to see that
– mark
Jul 20 at 17:51
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here are some ideas.
We consider the relation $(I_n-aA)^-1Bb=v$ where $A,B$ and the vector $v$ are known and $a,b=[b_i]$ are unknown. Note that $ngeq k,operatornamerank(B)=k$ and $1/anotin operatornamespectrum(A)$.
Then $Bb=(I-aA)v=v-aAv$, that is $Bb+aAv=v$. Let $B=[B_1,cdots,B_k]$.
Assume that $A,B,v$ are randomly chosen. Then the $(B_i)_i,Av=w,v$ are random vectors in $mathbbR^n$.
The above equation can be rewritten $sum_ileq kb_iB_i+aw=v$, that is, we want to decompose the vector $v$ on the $k+1$ vectors $(B_i)_i,w$; moreover, the decomposition must be unique.
It is possible with probability $1$ when $n=k+1$; otherwise, it is possible with probability $0$.
When $A$ is random, it's invertible with probability $1$. Here, $A$ is not assumed to be invertible and the question is: how to choose $v$ so that the decomposition exists and is unique? In particular, the vectors $vinker(A)$ are not convenient. I think that there are many such subcases and I stand prudently on the sidelines.
edited Jul 20 at 19:10
Botond
3,8782632
answered Jul 18 at 13:10
loup blanc
20.4k21549
Here are some ideas.
We consider the relation $(I_n-aA)^-1Bb=v$ where $A,B$ and the vector $v$ are known and $a,b=[b_i]$ are unknown. Note that $ngeq k,operatornamerank(B)=k$ and $1/anotin operatornamespectrum(A)$.
Then $Bb=(I-aA)v=v-aAv$, that is $Bb+aAv=v$. Let $B=[B_1,cdots,B_k]$.
Assume that $A,B,v$ are randomly chosen. Then the $(B_i)_i,Av=w,v$ are random vectors in $mathbbR^n$.
The above equation can be rewritten $sum_ileq kb_iB_i+aw=v$, that is, we want to decompose the vector $v$ on the $k+1$ vectors $(B_i)_i,w$; moreover, the decomposition must be unique.
It is possible with probability $1$ when $n=k+1$; otherwise, it is possible with probability $0$.
When $A$ is random, it's invertible with probability $1$. Here, $A$ is not assumed to be invertible and the question is: how to choose $v$ so that the decomposition exists and is unique? In particular, the vectors $vinker(A)$ are not convenient. I think that there are many such subcases and I stand prudently on the sidelines.
edited Jul 20 at 19:10
Botond
3,8782632
answered Jul 18 at 13:10
loup blanc
20.4k21549
edited Jul 20 at 19:10
Botond
3,8782632
edited Jul 20 at 19:10
Botond
3,8782632
edited Jul 20 at 19:10
Botond
3,8782632
3,8782632
answered Jul 18 at 13:10
loup blanc
20.4k21549
answered Jul 18 at 13:10
loup blanc
20.4k21549
answered Jul 18 at 13:10
loup blanc
20.4k21549
20.4k21549
thanks for ideas. perhaps I should modify my aim, please see update. It is probably true that $a$ and $b$ are identified except on sets of measure zero. maybe there's a simple way to see that
– mark
Jul 20 at 17:51
add a comment |Â
thanks for ideas. perhaps I should modify my aim, please see update. It is probably true that $a$ and $b$ are identified except on sets of measure zero. maybe there's a simple way to see that
– mark
Jul 20 at 17:51
thanks for ideas. perhaps I should modify my aim, please see update. It is probably true that $a$ and $b$ are identified except on sets of measure zero. maybe there's a simple way to see that
– mark
Jul 20 at 17:51
thanks for ideas. perhaps I should modify my aim, please see update. It is probably true that $a$ and $b$ are identified except on sets of measure zero. maybe there's a simple way to see that
– mark
Jul 20 at 17:51
add a comment |Â
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