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melanie
science forum beginner

Joined: 15 Jul 2005
Posts: 13

Posted: Mon Jun 26, 2006 7:33 pm    Post subject: Matrix problem; only brute force?

I have a question which I can not figure out
nice way of doing it except the brute force calculation.
I would be appreciated if someone could help me.

Let A be 2x2 matrix such that

[ 4 , 1 ]
[ -4 , 0 ]

find all alpha(real number) and
2x2 real synmetric matrix B such that

A* B* A_t = (alpha)* B

where A_t denotes transposition of A
Chip Eastham
science forum Guru

Joined: 01 May 2005
Posts: 412

Posted: Mon Jun 26, 2006 7:58 pm    Post subject: Re: Matrix problem; only brute force?

melanie wrote:
 Quote: I have a question which I can not figure out nice way of doing it except the brute force calculation. I would be appreciated if someone could help me. Let A be 2x2 matrix such that [ 4 , 1 ] [ -4 , 0 ] find all alpha(real number) and 2x2 real synmetric matrix B such that A* B* A_t = (alpha)* B where A_t denotes transposition of A

Since A is a given matrix, it might be better
to rewrite the problem in this fashion:

Find all (real symmetric) B such that for
some (real scalar) alpha:

B^-1 * A * B = (alpha) (A_t)^-1

I'm not clear on your reason for restricting
B to be symmetric, but even for a general
matrix B to satisfy the condition above,
it must be that the spectrum (eigenvalues)
of A is the same as alpha times the set
of reciprocal eigenvalues of A (A and A_t,
its transpose, have the same spectrum).

In any case it narrows the possible values
of alpha considerably! Suppose A has
distinct eigenvalues r,s. Then either:

alpha = r^2 = s^2

or alpha = r*s.

regards, chip
Chip Eastham
science forum Guru

Joined: 01 May 2005
Posts: 412

Posted: Mon Jun 26, 2006 8:04 pm    Post subject: Re: Matrix problem; only brute force?

Chip Eastham wrote:
 Quote: melanie wrote: I have a question which I can not figure out nice way of doing it except the brute force calculation. I would be appreciated if someone could help me. Let A be 2x2 matrix such that [ 4 , 1 ] [ -4 , 0 ] find all alpha(real number) and 2x2 real synmetric matrix B such that A* B* A_t = (alpha)* B where A_t denotes transposition of A Since A is a given matrix, it might be better to rewrite the problem in this fashion: Find all (real symmetric) B such that for some (real scalar) alpha: B^-1 * A * B = (alpha) (A_t)^-1 I'm not clear on your reason for restricting B to be symmetric, but even for a general matrix B to satisfy the condition above, it must be that the spectrum (eigenvalues) of A is the same as alpha times the set of reciprocal eigenvalues of A (A and A_t, its transpose, have the same spectrum). In any case it narrows the possible values of alpha considerably! Suppose A has distinct eigenvalues r,s. Then either: alpha = r^2 = s^2 or alpha = r*s.

Oops, I just noticed your particular A has
a double eigenvalue r = s = 2. So for the
specific matrix A =

[ 4 , 1 ]
[ -4 , 0 ]

for the oversight.

regards, chip
Ignacio Larrosa Cañestro1
science forum Guru Wannabe

Joined: 02 May 2005
Posts: 112

Posted: Mon Jun 26, 2006 8:07 pm    Post subject: Re: Matrix problem; only brute force?

En el
mensaje:33245839.1151350438870.JavaMail.jakarta@nitrogen.mathforum.org,
melanie <melanie@yahoo.com> escribió:
 Quote: I have a question which I can not figure out nice way of doing it except the brute force calculation. I would be appreciated if someone could help me. Let A be 2x2 matrix such that [ 4 , 1 ] [ -4 , 0 ] find all alpha(real number) and 2x2 real synmetric matrix B such that A* B* A_t = (alpha)* B where A_t denotes transposition of A

A = [4, 1; -4, 0], A_t = [4, -4; 1, 0], B = [x, y, z, t] ==>

A* B* A_t = [16x + 4y + 4z + t, - 16x -4z; - 16x - 4y, 16x] = [kx, ky; kz,
kt]

(16-k)x + 4y + 4z + t = 0

- 16x - ky - 4z = 0

- 16x - 4y - kz = 0

16x - kt = 0

But

|C| = |[16-k, 4, 4, 1; -16, -k, -4, 0; -16, -4, -k, 0; 16, 0, 0, - k]| =
(k - 4)^4

If k =/= 4, that system is homogeneus and determined ===> x = y = z 0 t = 0
is the only solution

If k = 4, the rank of C is 2, and you can set two values of {x, y, z, t} and
solve the others for it. By example,

x = t/4

y = - z - t

--
Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
Ignacio Larrosa Cañestro1
science forum Guru Wannabe

Joined: 02 May 2005
Posts: 112

Posted: Mon Jun 26, 2006 8:17 pm    Post subject: Re: Matrix problem; only brute force?

En el mensaje:4gat4oF1mfg3kU1@individual.net,
Ignacio Larrosa Cañestro <ilarrosaQUITARMAYUSCULAS@mundo-r.com> escribió:
 Quote: En el mensaje:33245839.1151350438870.JavaMail.jakarta@nitrogen.mathforum.org, melanie escribió: I have a question which I can not figure out nice way of doing it except the brute force calculation. I would be appreciated if someone could help me. Let A be 2x2 matrix such that [ 4 , 1 ] [ -4 , 0 ] find all alpha(real number) and 2x2 real synmetric matrix B such that A* B* A_t = (alpha)* B where A_t denotes transposition of A A = [4, 1; -4, 0], A_t = [4, -4; 1, 0], B = [x, y, z, t] == A* B* A_t = [16x + 4y + 4z + t, - 16x -4z; - 16x - 4y, 16x] = [kx, ky; kz, kt] (16-k)x + 4y + 4z + t = 0 - 16x - ky - 4z = 0 - 16x - 4y - kz = 0 16x - kt = 0 But |C| = |[16-k, 4, 4, 1; -16, -k, -4, 0; -16, -4, -k, 0; 16, 0, 0, - k]| = (k - 4)^4 If k =/= 4, that system is homogeneus and determined ===> x = y = z 0 t = 0 is the only solution If k = 4, the rank of C is 2, and you can set two values of {x, y, z, t} and solve the others for it. By example, x = t/4 y = - z - t

Sorry
--
Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com, I don't see the word 'symmetric'.
Then, simply do z = y.
Chip Eastham
science forum Guru

Joined: 01 May 2005
Posts: 412

Posted: Mon Jun 26, 2006 11:35 pm    Post subject: Re: Matrix problem; only brute force?

Ignacio Larrosa Cañestro wrote:
 Quote: En el mensaje:4gat4oF1mfg3kU1@individual.net, Ignacio Larrosa Cañestro escribió: En el mensaje:33245839.1151350438870.JavaMail.jakarta@nitrogen.mathforum.org, melanie escribió: I have a question which I can not figure out nice way of doing it except the brute force calculation. I would be appreciated if someone could help me. Let A be 2x2 matrix such that [ 4 , 1 ] [ -4 , 0 ] find all alpha(real number) and 2x2 real synmetric matrix B such that A* B* A_t = (alpha)* B where A_t denotes transposition of A A = [4, 1; -4, 0], A_t = [4, -4; 1, 0], B = [x, y, z, t] == A* B* A_t = [16x + 4y + 4z + t, - 16x -4z; - 16x - 4y, 16x] = [kx, ky; kz, kt] (16-k)x + 4y + 4z + t = 0 - 16x - ky - 4z = 0 - 16x - 4y - kz = 0 16x - kt = 0 But |C| = |[16-k, 4, 4, 1; -16, -k, -4, 0; -16, -4, -k, 0; 16, 0, 0, - k]| = (k - 4)^4 If k =/= 4, that system is homogeneus and determined ===> x = y = z 0 t = 0 is the only solution If k = 4, the rank of C is 2, and you can set two values of {x, y, z, t} and solve the others for it. By example, x = t/4 y = - z - t Sorry... I don't see the word 'symmetric'. Then, simply do z = y.

Based on Igancio's analysis we can formulate a
recipe.

Let v be a right (column) eigenvector of A: Av = k*v

Let B = v v', where v' is the (row) transpose of v.

If v is a real vector, then B is real symmetric.

Now AB = kB, and by symmetry A B A_t = (k^2) B.

regards, chip

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