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Matrix problem; only brute force?
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Chip Eastham
science forum Guru


Joined: 01 May 2005
Posts: 412

PostPosted: Mon Jun 26, 2006 11:35 pm    Post subject: Re: Matrix problem; only brute force? Reply with quote

Ignacio Larrosa Cañestro wrote:
Quote:
En el mensaje:4gat4oF1mfg3kU1@individual.net,
Ignacio Larrosa Cañestro <ilarrosaQUITARMAYUSCULAS@mundo-r.com> escribió:
En el
mensaje:33245839.1151350438870.JavaMail.jakarta@nitrogen.mathforum.org,
melanie <melanie@yahoo.com> 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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Ignacio Larrosa Cañestro1
science forum Guru Wannabe


Joined: 02 May 2005
Posts: 112

PostPosted: Mon Jun 26, 2006 8:17 pm    Post subject: Re: Matrix problem; only brute force? Reply with quote

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 <melanie@yahoo.com> 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.
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Ignacio Larrosa Cañestro1
science forum Guru Wannabe


Joined: 02 May 2005
Posts: 112

PostPosted: Mon Jun 26, 2006 8:07 pm    Post subject: Re: Matrix problem; only brute force? Reply with quote

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
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Chip Eastham
science forum Guru


Joined: 01 May 2005
Posts: 412

PostPosted: Mon Jun 26, 2006 8:04 pm    Post subject: Re: Matrix problem; only brute force? Reply with quote

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 ]

you ask about, alpha must be 4. Sorry
for the oversight.

regards, chip
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Chip Eastham
science forum Guru


Joined: 01 May 2005
Posts: 412

PostPosted: Mon Jun 26, 2006 7:58 pm    Post subject: Re: Matrix problem; only brute force? Reply with quote

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
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melanie
science forum beginner


Joined: 15 Jul 2005
Posts: 13

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