determining elements of a matrix from the determinant's value?

Discussion:

(too old to reply)

daverc31

2006-10-02 15:13:08 UTC

Hi,
I am a theoretical chemist, which could explain my low mathematical
skills.
Here is my question. Let' say I have a 3x3 symetrical matrix like this
:

a11 b*a12 b*a13
MAT = b*a12 a22 b*a23
b*a13 b*a23 a33

I know the value of det(MAT) and the values of the aii and aij. I am
looking for the value of b.
It's easy for a 3x3 problem but I need a general way to do that for
larger matrices.
Furthermore, such matrices seem to have one and only one negative value
for b.
Can I prove that? I used the Descarte's sign rule for simpler cases
(aij=aii=1) but I get stuck for general cases which I submit to you.
I hope I am clear and the solution is not obvious,

any help, advice, idea is most welcome

Hans Mittelmann

2006-10-02 23:40:29 UTC

Permalink

Post by daverc31
Hi,
I am a theoretical chemist, which could explain my low mathematical
skills.
Here is my question. Let' say I have a 3x3 symetrical matrix like this
a11 b*a12 b*a13
MAT = b*a12 a22 b*a23
b*a13 b*a23 a33
I know the value of det(MAT) and the values of the aii and aij. I am
looking for the value of b.
It's easy for a 3x3 problem but I need a general way to do that for
larger matrices.
Furthermore, such matrices seem to have one and only one negative value
for b.
Can I prove that? I used the Descarte's sign rule for simpler cases
(aij=aii=1) but I get stuck for general cases which I submit to you.
I hope I am clear and the solution is not obvious,
any help, advice, idea is most welcome

Hi,
expanding the determinant along the diagonal one sees that
det(MAT)= b^(n-1)*det(A)

Dave Dodson

2006-10-03 02:27:25 UTC

Permalink

Post by Hans Mittelmann
expanding the determinant along the diagonal one sees that
det(MAT)= b^(n-1)*det(A)

Er. No. let A = the 3x3 matrix of all 1's and b=2. Then det(A)=0 and
Det(MAT)=5.

You can't expand a determinant along the diagonal, but only along any
single row or column.

Dave

Hans Mittelmann

2006-10-03 04:29:55 UTC

Permalink

Post by Dave Dodson

Post by Hans Mittelmann
expanding the determinant along the diagonal one sees that
det(MAT)= b^(n-1)*det(A)

Er. No. let A = the 3x3 matrix of all 1's and b=2. Then det(A)=0 and
Det(MAT)=5.
You can't expand a determinant along the diagonal, but only along any
single row or column.
Dave

You are right. The expansion theorem is more general I believe, but
along the diagonal does not work.
Let some linalg specialists comment.

Ray Koopman

2006-10-03 06:55:07 UTC

Permalink

[a11 0 0 ] [ 1 b*c12 b*c13]
MAT = [ 0 a22 0 ] * [b*c21 1 b*c23], where cij = aij/aii.
[ 0 0 a33] [b*c31 b*c32 1 ]

The right-hand factor is b*(C - I) + I.
Its determinant is Product[b*(L_i - 1) + 1, i = 1...n],
where the L_i are the eigenvalues of C.
So det[MAT] is a polynomial in b.

daverc31

2006-10-03 13:28:18 UTC

Permalink

Thank you very much. This is preciseliy what I was looking for and
after all I could have found it with a bit more effort. However, I
should amit I do not know how obvious it is that the determinant of

b*(C - I) + I
is
Product[b*(L_i - 1) + 1, i = 1...n]

it can very well be trivial but my linear algebra is rusted. Any help?

Post by Ray Koopman

[a11 0 0 ] [ 1 b*c12 b*c13]
MAT = [ 0 a22 0 ] * [b*c21 1 b*c23], where cij = aij/aii.
[ 0 0 a33] [b*c31 b*c32 1 ]
The right-hand factor is b*(C - I) + I.
Its determinant is Product[b*(L_i - 1) + 1, i = 1...n],
where the L_i are the eigenvalues of C.
So det[MAT] is a polynomial in b.

Robert Israel

2006-10-03 16:30:53 UTC

Permalink

Post by daverc31
Thank you very much. This is preciseliy what I was looking for and
after all I could have found it with a bit more effort. However, I
should amit I do not know how obvious it is that the determinant of
b*(C - I) + I
is
Product[b*(L_i - 1) + 1, i = 1...n]
it can very well be trivial but my linear algebra is rusted. Any help?

C being a real symmetric matrix, it can be diagonalized as
C = U D U^(-1) where D is a diagonal matrix with the eigenvalues
on the diagonal. Then (b (C-I) + I) = U (b (D-I) + I) U^(-1)
so det(b (C-I) + I) = det(b (D-I) + I)
= product_i (b (L_i - 1) + 1)

Actually it works for any matrix C, using the Jordan canonical
form rather than diagonalization, with the eigenvalues
counted by algebraic multiplicity.

Robert Israel ***@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Paul Abbott

2006-10-04 07:23:45 UTC

Permalink

Post by Robert Israel

C being a real symmetric matrix, it can be diagonalized as
C = U D U^(-1) where D is a diagonal matrix with the eigenvalues
on the diagonal.

However, C is _not_ symmetric in general because cij = aij/aii != cji.

Cheers,
Paul

_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul

daverc31

2006-10-04 08:29:12 UTC

Permalink

Post by Robert Israel
C being a real symmetric matrix, it can be diagonalized as
C = U D U^(-1) where D is a diagonal matrix with the eigenvalues
on the diagonal. Then (b (C-I) + I) = U (b (D-I) + I) U^(-1)
so det(b (C-I) + I) = det(b (D-I) + I)
= product_i (b (L_i - 1) + 1)
Actually it works for any matrix C, using the Jordan canonical
form rather than diagonalization, with the eigenvalues
counted by algebraic multiplicity.

Thank you. I had withdrawn this post after I worked on it and realized
I could prove it.
However, thanks for the answer

daverc31

2006-10-03 14:39:53 UTC

Permalink

Thank you very much. that's preceisely what I was looking for.

Post by Ray Koopman

[a11 0 0 ] [ 1 b*c12 b*c13]
MAT = [ 0 a22 0 ] * [b*c21 1 b*c23], where cij = aij/aii.
[ 0 0 a33] [b*c31 b*c32 1 ]
The right-hand factor is b*(C - I) + I.
Its determinant is Product[b*(L_i - 1) + 1, i = 1...n],
where the L_i are the eigenvalues of C.
So det[MAT] is a polynomial in b.

andy2O

2006-10-03 23:20:03 UTC

Permalink

Hi All,

I'm just interested by the post, so out of pure curiosity: does anyone
know if the type of equation given by the OP, i.e.

det(M-alpha*I)=beta

come up frequently with non-zero beta? What sort of problem led you to
the equation you posted, daverc31? I can't immediately think of an
obvious interpretation in the usual terms of geometrical
transformations for non-zero beta - can anyone else? Does it have a
name, or general application?

Any thoughts welcomed...
Best wishes,
andy

(I thought of pseudospectra at first - but that appears to be the norm
of (M-alpha*I), not the determinant...)

Post by Ray Koopman

[a11 0 0 ] [ 1 b*c12 b*c13]
MAT = [ 0 a22 0 ] * [b*c21 1 b*c23], where cij = aij/aii.
[ 0 0 a33] [b*c31 b*c32 1 ]
The right-hand factor is b*(C - I) + I.
Its determinant is Product[b*(L_i - 1) + 1, i = 1...n],
where the L_i are the eigenvalues of C.
So det[MAT] is a polynomial in b.

daverc31

2006-10-04 08:27:01 UTC

Permalink

Post by andy2O
What sort of problem led you to
the equation you posted, daverc31? I can't immediately think of an
obvious interpretation in the usual terms of geometrical
transformations for non-zero beta - can anyone else? Does it have a
name, or general application?

Hi,
I mentioned I am a theoretical chemist. I don't know how much people in
this list are interested in such problems so I make it short.
We write the electronic wave function F in a basis which we can
diagonalize easily. We obtain eigenvalues and eigenvectors which are
the molecular orbitals and their respective energies. We can then get
the total energy of the system.
We want to compute now the wave function in terms of configuration
interaction (IC method). We write then
F=c1*f1 + c2*f3 + c3*f3
The fi are configurations. We know f1, f2 and f3 and we know they
diagonalize the operator H which gives to us the total energy (which we
computed previously) but we don't know H, only its structure.
Then come the problem of knowing the determinant of a matrix, its shape
but not its values. The values of H can then lead us to the ci
coefficients, i.e. to the interpretation of the wave function thus to
the properties of the molecules we apply this method for.

Thanks for your interest