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| lauraa... |
 Posted: Fri Nov 06, 2009 9:07 am |
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A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal elements and identical off-diagonal elements (i.e., they are equal to aI+bii' for some scalars a and b, where I is the identity matrix and i the vector of all ones)
Is it true that both A and A' have i as an eigenvector? |
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| Arturo Magidin... |
| Posted: Fri Nov 06, 2009 9:14 am |
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On Nov 6, 1:07 pm, lauraa <lau... at (no spam) hotmail.com> wrote:
[quote]A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal elements and identical off-diagonal elements (i.e., they are equal to aI+bii' for some scalars a and b, where I is the identity matrix and i the vector of all ones)
Is it true that both A and A' have i as an eigenvector?
[/quote]
Take 2 by 2 matrices, A = (2,0; 0,0). A' = A. A+A' = (4,0;0,0); A'A (4,0;0,0). So in fact, A+A' = A'A. However, Ai is not a multiple of i.
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Arturo Magidin |
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| lauraa... |
| Posted: Fri Nov 06, 2009 9:26 am |
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[quote]On Nov 6, 1:07Â pm, lauraa <lau... at (no spam) hotmail.com> wrote:
A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal
elements and identical off-diagonal elements (i.e.,
they are equal to aI+bii' for some scalars a and b,
where I is the identity matrix and i the vector of
all ones)
Is it true that both A and A' have i as an
eigenvector?
Take 2 by 2 matrices, A = (2,0; 0,0). A' = A. A+A' =
(4,0;0,0); A'A =
(4,0;0,0). So in fact, A+A' = A'A. However, Ai is not
a multiple of i.
--
[/quote]
sorry if this was ambiguous. see my explanation above in parenthesis for what I meant by "both A+A' and A'A have identical diagonal elements and identical off-diagonal elements ". |
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| Arturo Magidin... |
| Posted: Fri Nov 06, 2009 9:47 am |
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On Nov 6, 1:26 pm, lauraa <lau... at (no spam) hotmail.com> wrote:
[quote]On Nov 6, 1:07 pm, lauraa <lau... at (no spam) hotmail.com> wrote:
A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal
elements and identical off-diagonal elements (i.e.,
they are equal to aI+bii' for some scalars a and b,
where I is the identity matrix and i the vector of
all ones)
Is it true that both A and A' have i as an
eigenvector?
Take 2 by 2 matrices, A = (2,0; 0,0). A' = A. A+A' > > (4,0;0,0); A'A > > (4,0;0,0). So in fact, A+A' = A'A. However, Ai is not
a multiple of i.
--
sorry if this was ambiguous. see my explanation above in parenthesis for what I meant by "both A+A' and A'A have identical diagonal elements and identical off-diagonal elements ".
[/quote]
So, every diagonal entry is equal to a+b, and every off-diagonal entry
is equal to b. And that's A and also A', so A=A'.
And you are having trouble figuring out whether Ai and A'i are
multiples of i?
Just out of curiosity: if B is *any* nxn matrix, what is the j-th
entry of Bj?
--
Arturo Magidin |
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| Arturo Magidin... |
| Posted: Fri Nov 06, 2009 9:48 am |
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On Nov 6, 1:47 pm, Arturo Magidin <magi... at (no spam) member.ams.org> wrote:
[quote]Just out of curiosity: if B is *any* nxn matrix, what is the j-th
entry of Bj?
[/quote]
Should be "of Bi", with i the column vector all of whose entries are
1.
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Arturo Magidin |
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| Arturo Magidin... |
| Posted: Fri Nov 06, 2009 10:17 am |
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On Nov 6, 1:47 pm, Arturo Magidin <magi... at (no spam) member.ams.org> wrote:
[quote]On Nov 6, 1:26 pm, lauraa <lau... at (no spam) hotmail.com> wrote:
On Nov 6, 1:07 pm, lauraa <lau... at (no spam) hotmail.com> wrote:
A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal
elements and identical off-diagonal elements (i.e.,
they are equal to aI+bii' for some scalars a and b,
where I is the identity matrix and i the vector of
all ones)
Is it true that both A and A' have i as an
eigenvector?
Take 2 by 2 matrices, A = (2,0; 0,0). A' = A. A+A' > > > (4,0;0,0); A'A > > > (4,0;0,0). So in fact, A+A' = A'A. However, Ai is not
a multiple of i.
--
sorry if this was ambiguous. see my explanation above in parenthesis for what I meant by "both A+A' and A'A have identical diagonal elements and identical off-diagonal elements ".
So, every diagonal entry is equal to a+b, and every off-diagonal entry
is equal to b. And that's A and also A', so A=A'.
[/quote]
Oops; misread the problem. Rather, this holds for A+A' (and for AA');
but presumably, the actual values in A+A' and in AA' are not
necessarily the same.
Sorry about that.
Arturo Magidin |
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| Arturo Magidin... |
| Posted: Fri Nov 06, 2009 12:13 pm |
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On Nov 6, 1:07 pm, lauraa <lau... at (no spam) hotmail.com> wrote:
[quote]A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal elements and identical off-diagonal elements (i.e., they are equal to aI+bii' for some scalars a and b, where I is the identity matrix and i the vector of all ones)
[/quote]
You have that every diagonal entry of A+A' is equal to some value a;
and every off-diagonal value of A+A' is equal to some value b. Also,
every diagonal entry of A'A is equal to some value c; and every off-
diagonal entry of A'A is equal to some value d.
[quote]Is it true that both A and A' have i as an eigenvector?
[/quote]
Take A = (1,1;-1,1), A'=(1,-1;1,1). hen A+A' = (2,0;0,2), and A'A (2,0;0,2). But (1,1)' is not an eigenvalue of A, since A(1,1)' (2,0).
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Artur0 Magidin |
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| lauraa... |
| Posted: Sun Nov 08, 2009 8:38 am |
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[quote]A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal
elements and identical off-diagonal elements (i.e.,
they are equal to aI+bii' for some scalars a and b,
where I is the identity matrix and i the vector of
all ones)
Is it true that both A and A' have i as an
eigenvector?
Take A = (1,1;-1,1), A'=(1,-1;1,1). hen A+A' =
(2,0;0,2), and A'A =
(2,0;0,2). But (1,1)' is not an eigenvalue of A,
since A(1,1)' =
(2,0).
[/quote]
Thanks for that. I should have specified that A+A' and A'A have nonzero entries, so let me restate the problem.
A is a real matrix, ' denotes transposition, and i is a vector of all ones. Suppose that A+A'=aI+bii' and that A'A=cI+dii' for some nonzero scalars a,b,c and d. Show that i is an eigenvector of A. |
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| Gerry Myerson... |
| Posted: Sun Nov 08, 2009 8:53 pm |
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In article
<181851611.36139.1257705562544.JavaMail.root at (no spam) gallium.mathforum.org>,
lauraa <lauraa at (no spam) hotmail.com> wrote:
[quote]A is a real matrix and ' denotes transposition.
Suppose both A+A' and A'A have identical diagonal
elements and identical off-diagonal elements (i.e.,
they are equal to aI+bii' for some scalars a and b,
where I is the identity matrix and i the vector of
all ones)
Is it true that both A and A' have i as an
eigenvector?
Take A = (1,1;-1,1), A'=(1,-1;1,1). hen A+A' =
(2,0;0,2), and A'A =
(2,0;0,2). But (1,1)' is not an eigenvalue of A,
since A(1,1)' =
(2,0).
Thanks for that. I should have specified that A+A' and A'A have nonzero
entries, so let me restate the problem.
A is a real matrix, ' denotes transposition, and i is a vector of all ones.
Suppose that A+A'=aI+bii' and that A'A=cI+dii' for some nonzero scalars a,b,c
and d. Show that i is an eigenvector of A.
[/quote]
If there's a counterexample with the matrices having some zeros,
then I'd suspect there'd be a counterexample with the matrices
having non-zero entries, so I'd ask: do you have some reason
to think it's true? Have you done some examples?
Alternatively, do you get any mileage out of going
A' = a I + b J - A (where I'm writing J for ii'),
so c I + d J = A' A = a A + b J A - A^2 ?
--
Gerry Myerson (gerry at (no spam) maths.mq.edi.ai) (i -> u for email) |
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| lauraa... |
| Posted: Mon Nov 09, 2009 4:02 am |
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[quote]A is a real matrix, ' denotes transposition, and i
is a vector of all ones.
Suppose that A+A'=aI+bii' and that A'A=cI+dii' for
some nonzero scalars a,b,c
and d. Show that i is an eigenvector of A.
If there's a counterexample with the matrices having
some zeros,
then I'd suspect there'd be a counterexample with the
matrices
having non-zero entries, so I'd ask: do you have some
reason
to think it's true? Have you done some examples?
[/quote]
Yep have tried pretty hard to find counterexamples, but no luck.
[quote]Alternatively, do you get any mileage out of going
A' = a I + b J - A (where I'm writing J for ii'),
so c I + d J = A' A = a A + b J A - A^2 ?
[/quote]
Thanks for the tip, I appreciate it.
I wasn't able to find a proof (or a counterexample) though. |
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| Arturo Magidin... |
| Posted: Mon Nov 09, 2009 6:17 am |
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Guest
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On Nov 9, 8:02 am, lauraa <lau... at (no spam) hotmail.com> wrote:
[quote]A is a real matrix, ' denotes transposition, and i
is a vector of all ones.
Suppose that A+A'=aI+bii' and that A'A=cI+dii' for
some nonzero scalars a,b,c
and d. Show that i is an eigenvector of A.
If there's a counterexample with the matrices having
some zeros,
then I'd suspect there'd be a counterexample with the
matrices
having non-zero entries, so I'd ask: do you have some
reason
to think it's true? Have you done some examples?
Yep have tried pretty hard to find counterexamples, but no luck.
Alternatively, do you get any mileage out of going
A' = a I + b J - A (where I'm writing J for ii'),
so c I + d J = A' A = a A + b J A - A^2 ?
Thanks for the tip, I appreciate it.
I wasn't able to find a proof (or a counterexample) though.
[/quote]
Your matrix A must have all diagonal entries the same.
If you require A+A' and A'A to have no zero entries, there are no 2x2
examples: for it comes down to finding numbers r and s such that
r^2=s^2 (the (1,2) and (2,1) entries of A). If r=s, then (1,1) is an
eigenvector of both; if r=-s, then you get the off-diagonal entries
equal to 0.
In the 3x3 case, it comes down to finding numbers b, c, d, e, f, g
(the (1,2), (1,3), (2,3), (2,1) (3,1), and (3,2) entries,
respectively), such that
b+e = c+f = d+g (so the off-diagonal entries of A+A' are all equal);
e^2+f^2 = b^2+g^2 = c^2+d^2
(so the diagonal entries of A'A are all equal); and
fg = ed = bc (so the off-diagonal entries of A'A are all equal).
This does not take into account the non-zero entries issue.
--
Arturo Magidin |
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| lauraa... |
| Posted: Mon Nov 09, 2009 8:49 am |
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[quote]On Nov 9, 8:02Â am, lauraa <lau... at (no spam) hotmail.com> wrote:
A is a real matrix, ' denotes transposition,
and i
is a vector of all ones.
Suppose that A+A'=aI+bii' and that A'A=cI+dii'
for
some nonzero scalars a,b,c
and d. Show that i is an eigenvector of A.
If there's a counterexample with the matrices
having
some zeros,
then I'd suspect there'd be a counterexample with
the
matrices
having non-zero entries, so I'd ask: do you have
some
reason
to think it's true? Have you done some examples?
Yep have tried pretty hard to find counterexamples,
but no luck.
Alternatively, do you get any mileage out of
going
A' = a I + b J - A (where I'm writing J for ii'),
so c I + d J = A' A = a A + b J A - A^2 ?
Thanks for the tip, I appreciate it.
I wasn't able to find a proof (or a counterexample)
though.
Your matrix A must have all diagonal entries the
same.
If you require A+A' and A'A to have no zero entries,
there are no 2x2
examples: for it comes down to finding numbers r and
s such that
r^2=s^2 (the (1,2) and (2,1) entries of A). If r=s,
then (1,1) is an
eigenvector of both; if r=-s, then you get the
off-diagonal entries
equal to 0.
In the 3x3 case, it comes down to finding numbers b,
c, d, e, f, g
(the (1,2), (1,3), (2,3), (2,1) (3,1), and (3,2)
entries,
respectively), such that
b+e = c+f = d+g (so the off-diagonal entries of A+A'
are all equal);
e^2+f^2 = b^2+g^2 = c^2+d^2
(so the diagonal entries of A'A are all
A are all equal); and
fg = ed = bc (so the off-diagonal entries of A'A
A are all equal).
This does not take into account the non-zero entries
issue.
[/quote]
Thanks. Since you're system of 6 equations in 6 variables yields (b = f, g = c, e = c, d = f), then A is equal to (a, b, c ; c, a, b ; b, c, a) and thus has i as an eigenvector. So you have proved that my statement is correct for n=2 and n=3. (The statement is: If, for a real nxn matrix A, A+A'=aI+bii' and A'A=cI+dii' for some nonzero scalars a,b,c and d, then i is an eigenvector of A)
But what about the case of general n? |
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| Arturo Magidin... |
| Posted: Mon Nov 09, 2009 9:48 am |
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On Nov 9, 12:49 pm, lauraa <lau... at (no spam) hotmail.com> wrote:
[quote]On Nov 9, 8:02 am, lauraa <lau... at (no spam) hotmail.com> wrote:
A is a real matrix, ' denotes transposition,
and i
is a vector of all ones.
Suppose that A+A'=aI+bii' and that A'A=cI+dii'
for
some nonzero scalars a,b,c
and d. Show that i is an eigenvector of A.
If there's a counterexample with the matrices
having
some zeros,
then I'd suspect there'd be a counterexample with
the
matrices
having non-zero entries, so I'd ask: do you have
some
reason
to think it's true? Have you done some examples?
Yep have tried pretty hard to find counterexamples,
but no luck.
Alternatively, do you get any mileage out of
going
A' = a I + b J - A (where I'm writing J for ii'),
so c I + d J = A' A = a A + b J A - A^2 ?
Thanks for the tip, I appreciate it.
I wasn't able to find a proof (or a counterexample)
though.
Your matrix A must have all diagonal entries the
same.
If you require A+A' and A'A to have no zero entries,
there are no 2x2
examples: for it comes down to finding numbers r and
s such that
r^2=s^2 (the (1,2) and (2,1) entries of A). If r=s,
then (1,1) is an
eigenvector of both; if r=-s, then you get the
off-diagonal entries
equal to 0.
In the 3x3 case, it comes down to finding numbers b,
c, d, e, f, g
(the (1,2), (1,3), (2,3), (2,1) (3,1), and (3,2)
entries,
respectively), such that
b+e = c+f = d+g (so the off-diagonal entries of A+A'
are all equal);
e^2+f^2 = b^2+g^2 = c^2+d^2
(so the diagonal entries of A'A are all
A are all equal); and
fg = ed = bc (so the off-diagonal entries of A'A
A are all equal).
This does not take into account the non-zero entries
issue.
Thanks. Since you're system of 6 equations in 6 variables yields (b = f, g = c, e = c, d = f),
[/quote]
This wasn't clear to me (though I didn't try long); how did you derive
this is the only solution?
--
Arturo Magidin |
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| lauraa... |
| Posted: Tue Nov 10, 2009 8:52 am |
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Guest
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[quote]A is a real matrix, ' denotes transposition,
and i is a vector of all ones.
Suppose that A+A'=aI+bii' and that A'A=cI+dii'
for some nonzero scalars a,b,c
and d. Show that i is an eigenvector of A.
If there's a counterexample with the matrices
having some zeros, then I'd suspect there'd be a
counterexample wit the matrices
having non-zero entries, so I'd ask: do you
have some reason
to think it's true? Have you done some examples?
[/quote]
[quote]In the 3x3 case, it comes down to finding numbers
b, c, d, e, f, g
(the (1,2), (1,3), (2,3), (2,1) (3,1), and (3,2)
entries,
respectively), such that
b+e = c+f = d+g  (so the off-diagonal entries of
A+A' are all equal);
 e^2+f^2 = b^2+g^2 = c^2+d^2
(so the diagonal entries of A'A are all
equal); and
fg = Â ed = Â bc (so the off-diagonal entries of
A'A are all equal).
This does not take into account the non-zero
entries issue.
Thanks. Since you're system of 6 equations in 6
variables yields (b = f, g = c, e = c, d = f), then A
is equal to (a, b, c ; c, a, b ; b, c, a) and thus
has i as an eigenvector. So you have proved that my
statement is correct for n=2 and n=3. (The statement
is: If, for a real nxn matrix A, A+A'=aI+bii' and
A'A=cI+dii' for some nonzero scalars a,b,c and d,
then is an eigenvector of A)
But what about the case of general n?
This wasn't clear to me (though I didn't try long);
how did you derive
this is the only solution?
[/quote]
I've just used Maple without doing any further checks
Have not found any counterexamples or a proof for my statement (for general n) unfortunately |
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| lauraa... |
| Posted: Thu Nov 12, 2009 7:47 am |
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Guest
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[quote]The statement is:
If, for a real nxn matrix A, A+A'=aI+bii' and
A'A=cI+dii' for some nonzero scalars a,b,c and d,
then is an eigenvector of A)
[/quote]
[quote]Have not found any counterexamples or a proof for my
statement (for general n) unfortunately
[/quote]
Hope it's OK if I ask for help/advice about this once more. I'm still stuck but perhaps someone can give me some further hints on how to procede?
When n=4, an example that the nonzero condition on the entries of A+A' and A'A is needed is the matrix
A=(1, 0, 1, 0 ; 0, 1, 0, -1 ; -1, 0, 1, 0; 0, 1, 0, 1), which gives A+A'=A'A=2I and has not the vector i of all ones as an eigenvector. I have not found counterexamples under the condition the A+A' and A'A do not have zero entries.
Thanks in advance |
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