Failure in finging eigenvectors

[ELCA]

When we find eigenvectors of a symmetric metrix containing off-diagonal 0 components, Some failure occurs in values or ordering its eigenvectors.
e.g., for a matix
0.9700 0 0
0 0.9600 0.0200
0 0.0200 0.8800
Labview answers
1 1 0
0.98 -0.95 -0.28
0.27 -0.26 1
for eigenvalues
.97 .96 .88.

Beside the normalizing factor,
correct eigenvalues will be

1.0 0 0
0 -0.9732 0.2298
0 -0.2298 -0.9732.

check it and give me the solution for this trouble please

[Zvezdana S.]

Actually, this is the correct result that LabVIEW is returning. Here is some math:
The way to check if the eigenvalues results are correct, is to solve the equation det(A-L*I)=0, where A is the original matrix, L is the eigenvector, and I is a diagonal matrix.
The way to calculate this equation, is to use only one result for L, lets user L3=0.88. Then ,the matrix (A-L*I) is
0.09 0 0
0 0.08 0
1 0.24 0
Determinant of this matrix is 0, so L3 is a correct eigenvalue. You would get that det(A-L*I)=0 if you calculate it for any of the L's LabVIEW returned.
Now, to calculate these Eigenvalues by hand, you would the use the same equation det(A-L*I)=0. So we would look for the determinant of the following matrix to be 0:
0.97-L 0 0
0 0.96-L
0
0 0 0.88-L
The det of this matrix is (0.97-L)(0.96-L)(0.88-L)=0. From here, you would get that the eigenvalues are L1=0.97, L2=0.96, L3=0.88, the same as the LabVIEW results.

Zvezdana S.
National Instruments

[ELCA]

Thank you for your kind answer.
I checked it out that the eigenvectors satisfied the characteristic equation. But, I wonder what makes the different sets of eigenvectors with the same eigenvalues in that problem. As you know, the second set of the eigenvectors also satisfy the equation with the same eigenvalues. Is it general that an eigenvalue problem has more than two eigenvector directions? Is there any dependance on algorithms?