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Is it possible to design this?

by S_David
Tags: design
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S_David
#1
Aug21-13, 06:07 PM
P: 597
Hello,

I wish to design the matrix P such that it satisfies the following:

[tex]P^TQP=Q^T[/tex]

where all matrices are N-by-N. Is it possible?

Thanks
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mfb
#2
Aug22-13, 08:52 AM
Mentor
P: 11,631
For all Q?
Did you try some simple cases, e.g. Q=[[0,1],[0,0]]?

I doubt this is possible in general, so searching for counterexamples looks like a good idea.

Edit: And that is a good idea for a counterexample if you combine it with another matrix.
S_David
#3
Aug22-13, 11:03 AM
P: 597
Quote Quote by mfb View Post
For all Q?
Did you try some simple cases, e.g. Q=[[0,1],[0,0]]?

I doubt this is possible in general, so searching for counterexamples looks like a good idea.

Edit: And that is a good idea for a counterexample if you combine it with another matrix.
Actually, yes I wanted it in general, but if there are some special cases of Q I would like to know what are they, if possible. I know for example if Q is circulant, then any permutation matrix will satisfies the above equality. Unfortunately, in my case Q is not circulant!!

Thanks

mfb
#4
Aug22-13, 05:21 PM
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P: 11,631
Is it possible to design this?

Quote Quote by S_David View Post
Actually, yes I wanted it in general
P=[[a,c][b,d]] and Q=[[0,1],[0,0]] lead to the conditions a=d=0 and bc=1, and this leads to a contradiction if you try to apply this to Q=[[1,0],[0,0]].
So in general, this is not possible.
S_David
#5
Aug28-13, 08:39 AM
P: 597
Quote Quote by mfb View Post
P=[[a,c][b,d]] and Q=[[0,1],[0,0]] lead to the conditions a=d=0 and bc=1, and this leads to a contradiction if you try to apply this to Q=[[1,0],[0,0]].
So in general, this is not possible.
OK, what if the matrix is Hermitian?
mfb
#6
Aug28-13, 02:41 PM
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P: 11,631
Did you test some simple cases, as I suggested?
I would not expect that being Hermitian helps, but I don't know.
S_David
#7
Aug28-13, 02:53 PM
P: 597
Quote Quote by mfb View Post
Did you test some simple cases, as I suggested?
I would not expect that being Hermitian helps, but I don't know.
After trying, I don't think it helps neither.
Office_Shredder
#8
Aug28-13, 03:12 PM
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PF Gold
P: 4,500
The right hand side is a linear operation on the space of nxn matrices (which is a vector space of dimension n2), and the left hand side is a linear operation on the space of nxn matrices picked from a vector space of dimension n2. But the set of all linear transformations on the space of nxn matrices has dimension (n2)2 = n4. So in general most linear transformations on a set of matrices cannot be represented as conjugation by a matrix like the left hand side of your equation.

This doesn't answer the specific question of whether transposing is a special case, but it does heavily suggest that there will not exist such a matrix P, and as mfb said you should look for counterexamples in small dimensions.

There is a general disproof I believe that works for when Q is Hermitian. Plugging in Q the identity matrix we get that Pt = P-1. Plugging in Q an arbitrary real symmetric matrix, we want to find P that commutes with every Q which is symmetric. Checking Q an arbitrary elementary symmetric matrix (has only a 1 in the (i,j) and (j,i) spots and 0 everywhere else) we find P has to be a scalar multiple of the identity matrix, which because its transpose is its inverse means it has to be the identity matrix or negative the identity matrix. Checking Q basically any non real Hermitian matrix shows that neither of those work.

If testing simple cases didn't help, then you should show us what cases you tested and we can show you where you went wrong in missing the counterexample
S_David
#9
Aug28-13, 03:59 PM
P: 597
Do you mean it is possible for Hermitian matrices? I tried a matrix [a b; conj(b) a] which is Hermitian and P is a real matrix. But I was wondering since for a Hermitian matrix we have the following eigenvalue decomposition:

[tex]\mathbf{Q}=\mathbf{U}\Sigma\mathbf{U}^H[/tex]

where the eigenvalues are real, which implies that:

[tex]\mathbf{U}^H\mathbf{Q}\mathbf{U}=\Sigma=\Sigma^T=\mathbf{U}^T\mathbf{Q} ^T\mathbf{U}^*[/tex]

if we can use that in a way.
Office_Shredder
#10
Aug28-13, 04:40 PM
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P: 4,500
Quote Quote by S_David View Post
Do you mean it is possible for Hermitian matrices?
No, I gave a proof that it isn't possible for Hermitian matrices


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