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?
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.
> 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 ".
> > 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'.
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?
> 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.
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)
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).
> > 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.
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 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.
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)
> > 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.
> > > 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 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.
> On Nov 9, 8:02 am, lauraa <lau...@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), 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)
> > On Nov 9, 8:02 am, lauraa <lau...@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),
This wasn't clear to me (though I didn't try long); how did you derive this is the only solution?
> > > > > 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? > > > 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
> > > 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. I have not found counterexamples under the condition the A+A' and A'A do not have zero entries.
On Nov 12, 11:47 am, lauraa <lau...@hotmail.com> wrote:
> > > > 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
> > > > > 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) > > 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
> 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!
