Eigenvector
lauraa
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?
Arturo Magidin
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.
--
Arturo Magidin
lauraa
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 ".
Arturo Magidin
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
Arturo Magidin
> Just out of curiosity: if B is *any* nxn matrix, what is the j-th
> entry of Bj?
Should be "of Bi", with i the column vector all of whose entries are
1.
--
Arturo Magidin
Arturo Magidin
> On Nov 6, 1:26 pm, lauraa <lau...@hotmail.com> wrote:
>
>
>
> > > On Nov 6, 1:07 pm, lauraa <lau...@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'.
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
Arturo Magidin
>
> 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)
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.
> 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).
--
Artur0 Magidin
lauraa
> >
> > 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)
>
> 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.
Gerry Myerson
<181851611.36139.12577...@gallium.mathforum.org>,
lauraa <lau...@hotmail.com> wrote:
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 (ge...@maths.mq.edi.ai) (i -> u for email)
lauraa
> > 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.
Arturo Magidin
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
lauraa
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?
Arturo Magidin
This wasn't clear to me (though I didn't try long); how did you derive
this is the only solution?
--
Arturo Magidin
lauraa
> > > > > 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?
> > > 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.
> > 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?
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
lauraa
> > > 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)
> Have not found any counterexamples or a proof for my
> statement (for general n) unfortunately
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
Arturo Magidin
> > > > 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)
> > Have not found any counterexamples or a proof for my
> > statement (for general n) unfortunately
>
> 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.
You already had a 2x2 example with A+A' and A'A having off-diagonal
zero entries (and neither A nor A' had zero entries). It should be
pretty easy to come up with examples, certainly at every even
dimension.
> I have not found counterexamples under the condition the A+A' and A'A do not have zero entries.
>
> Thanks in advance
Did you find the corresponding system of equations and feed it into
Mathematica? If there is a counterexample, I would expect it to be in
even dimension, and possibly at n=4 (at n=2, the lack of "room" may
prevent one).
If you let A = (a, b, c, d; r, a, e, f; s, t, a, g; u, v, w, a), then
you get
b+r = c+s = d+u = e+t = f+v = g+w
r^2 + s^2 + u^2 = b^2 + t^2 + v^2 = c^2 + e^2 + w^2 = d^2+f^2+g^2
st+vu = re +uw = fr + gs = bc+wv = db+gr = cd+ef
The condition of nonzero entries is achieved by requiring b+r=/=0, a=/
=0, and a(b+r)+st+vu=/=0.
--
Arturo Magidin
lauraa
> > > > > 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)
> > 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.
> Did you find the corresponding system of equations
> and feed it into
> Mathematica? If there is a counterexample, I would
> expect it to be in
> even dimension, and possibly at n=4 (at n=2, the lack
> of "room" may
> prevent one).
>
>
> If you let A = (a, b, c, d; r, a, e, f; s, t, a, g;
> u, v, w, a), then
> you get
>
> b+r = c+s = d+u = e+t = f+v = g+w
>
> r^2 + s^2 + u^2 = b^2 + t^2 + v^2 = c^2 + e^2 + w^2 =
> d^2+f^2+g^2
>
> st+vu = re +uw = fr + gs = bc+wv = db+gr = cd+ef
>
> The condition of nonzero entries is achieved by
> requiring b+r=/=0, a=/
> =0, and a(b+r)+st+vu=/=0.
thanks for the tip, it might have been useful to try and find a counterexample. I've tried to use "solve" in maple for that system of equations, but did not get a solution in the few hours I let it run.
Meanwhile I have written a little program in matlab that generates random matrices (with equal entries on the main diagonal) and then checks if they represent counterexamples to my statement above. I have done this up to n=4, let the program run for a long time, but have not found any counterexamples
Any further suggestion on how to go about (dis)proving the statement at the top of this post would be very gratefully received!