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JEMebius
science forum Guru Wannabe

Joined: 24 Mar 2005
Posts: 209

Posted: Mon Jul 17, 2006 1:46 pm    Post subject: Re: symmetry in eigenvectors-

Vector y=(-x2 ,x1 ,-x4 ,x3) is obtained from vector x=(x1, x2, x3, x4)
by a double 4D rotation over 90 deg in the 1-2 and the 3-4 planes.

Whereas it is true that the other three eigenvectors of A are orthogonal
to x, there is no a priori reason to expect that y =must= be one of
these three.

However, by a suitable 4D coordinate transformation one can arrange
things such in the new coordinates x'1, x'2, x'3, x'4 the vector x is
unchanged: x = (x'1, x'2, x'3, x'4), and one of the remaining three
eigenvectors becomes (-x'2, x'1, -x'4, x'3).
It is a worthwhile exercise to find out which 4x4 symmetric matrices
have (x1, x2, x3, x4) and (-x2, x1, -x4, x3) as two of their eigenvectors.

Ciao: Johan E. Mebius

Al wrote:

 Quote: Hi! Maybe someone could help me with the following. It would be very helpful! Given a real symmetric 4x4 matrix A (it's a covariance matrix), and one real eigenvector x=(x1,x2,x3,x4). Is it now possible to show that y=(-x2,x1,-x4,x3) is an eigenvector of A, too? Thanks, Al
Chip Eastham
science forum Guru

Joined: 01 May 2005
Posts: 412

Posted: Mon Jul 17, 2006 1:32 pm    Post subject: Re: symmetry in eigenvectors

Al wrote:
 Quote: Hi! Maybe someone could help me with the following. It would be very helpful! Given a real symmetric 4x4 matrix A (it's a covariance matrix), and one real eigenvector x=(x1,x2,x3,x4). Is it now possible to show that y=(-x2,x1,-x4,x3) is an eigenvector of A, too?

No, it isn't. While y is orthogonal to x, it is not true that
just because A is real symmetric, and x is an eigenvector
that y must also be an eigenvector.

It is true that eigenvectors of real symmetric A for distinct
eigenvalues must be orthogonal.

Here's a way to construct counterexamples to your
underlying claim. Take x (normalized) and extend
it to an orthonormal basis {x, u, v, w} for R^4 such
that y is not a multiple of any single basis element.

Define A = x'x + 2 u'u + 3 v'v + 4 w'w.

Clearly A is real symmetric. Due to orthonormality
of the basis, the eigenvalues are shown to be 1,2,3,4
corresponding to resp. basis x,u,v,w of eigenvectors.

Since y is not a multiple of any of these, y is not an
eigenvector of A.

regards, chip
A N Niel
science forum Guru

Joined: 28 Apr 2005
Posts: 475

Posted: Mon Jul 17, 2006 1:16 pm    Post subject: Re: symmetry in eigenvectors

<alalcoolj@gmx.de> wrote:

 Quote: Hi! Maybe someone could help me with the following. It would be very helpful! Given a real symmetric 4x4 matrix A (it's a covariance matrix), and one real eigenvector x=(x1,x2,x3,x4). Is it now possible to show that y=(-x2,x1,-x4,x3) is an eigenvector of A, too?

No.
Say A is diagonal matrix 1,1,1,0. Then (1,0,1,0) is an eigenvector,
but (0,1,0,1) is not.
Al1131
science forum beginner

Joined: 01 Mar 2006
Posts: 4

 Posted: Mon Jul 17, 2006 1:09 pm    Post subject: symmetry in eigenvectors Hi! Maybe someone could help me with the following. It would be very helpful! Given a real symmetric 4x4 matrix A (it's a covariance matrix), and one real eigenvector x=(x1,x2,x3,x4). Is it now possible to show that y=(-x2,x1,-x4,x3) is an eigenvector of A, too? Thanks, Al

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