View Full Version : Eigen question

mammothrob

05-23-2007, 07:33 PM

For which real numbers c and d does the matrix have real eigenvalues and three orthogonal eignvectors?

120

2dc

053

I have some thoughts on this but have not gotten too far.

The matrix in question should be self adjoint right?

I tried solving for the adj. and the numbers i get dont work...

Couldnt I just use gram schmidt with arbtirary numbers pluged in c and d and make the collumns orthonormal?

hmmmm

mammothrob

05-24-2007, 06:03 AM

does the real spectral theorem have anything to do with this?

Opalg

05-25-2007, 02:39 PM

Call the matrix A. If A has three real eigenvectors and three orthogonal eigenvectors, let D be the diagonal matrix whose diagonal entries are the eigenvalues and let P be the orthogonal matrix whose columns are the corresponding eigenvectors. Then A=PDP^T,which is selfadjoint. So yes, you're right to say that A is selfadjoint, and therefore we must have c=5.

Conversely, if A is selfadjoint then it is diagonalisable. In other words, provided that the eigenvalues are real it will have three orthogonal eigenvectors. So everything depends on knowing whether all three eigenvalues are real. But the eigenvalue equation \det(A-\lambda I)=0 is a nasty cubic equation with d as a parameter. And that is where I get stuck: I can't see any way of deciding when this equation has three real roots.

JakeD

05-25-2007, 04:37 PM

Call the matrix A. If A has three real eigenvectors and three orthogonal eigenvectors, let D be the diagonal matrix whose diagonal entries are the eigenvalues and let P be the orthogonal matrix whose columns are the corresponding eigenvectors. Then A=PDP^T,which is selfadjoint. So yes, you're right to say that A is selfadjoint, and therefore we must have c=5.

Conversely, if A is selfadjoint then it is diagonalisable. In other words, provided that the eigenvalues are real it will have three orthogonal eigenvectors. So everything depends on knowing whether all three eigenvalues are real. But the eigenvalue equation \det(A-\lambda I)=0 is a nasty cubic equation with d as a parameter. And that is where I get stuck: I can't see any way of deciding when this equation has three real roots.

The eigenvalues of a selfadjoint matrix are always real.

http://planetmath.org/encyclopedia/Eige ... eReal.html (http://planetmath.org/encyclopedia/EigenvaluesOfAHermitianMatrixAreReal.html)

A polynomial has all real roots if and only if it splits over the real numbers (fairly obvious). You know well there is a "formula" for the roots of quadratics, but there is also a solution by radicals to cubics (although very involved). This is probably rediculously overcomplicated for this problem, but you should be able to find your answer this way if you are determined enough. I wouldn't reccomend it though.

Opalg

05-26-2007, 03:51 AM

The eigenvalues of a selfadjoint matrix are always real.

Good point (sometimes one overlooks the blindingly obvious). So d can have any real value.

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