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Finding eigenvectors of 3x3 matrix

  • #1
Hi,

I have an issue with zeroing the 3x3 matrix to find the eigenvector.

I have found the characteristic equation for the 3 eigenvalues.

the matrix is

1 1/2 1/3
1/2 1/3 1/4
1/3 1/4 1/5

The equation i got is -A^3 + (23/15)A^2 - (127/720)A + (1/2160) which correspond to A = 0.1223, 0.0027 & 1.41.

I have tried row swapping, division and multiplication but cannot get the correct eigenvector.

Please help. Thank you.
 

Answers and Replies

  • #2
ehild
Homework Helper
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What did you get?

ehild
 
  • #3
Ray Vickson
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Hi,

I have an issue with zeroing the 3x3 matrix to find the eigenvector.

I have found the characteristic equation for the 3 eigenvalues.

the matrix is

1 1/2 1/3
1/2 1/3 1/4
1/3 1/4 1/5

The equation i got is -A^3 + (23/15)A^2 - (127/720)A + (1/2160) which correspond to A = 0.1223, 0.0027 & 1.41.

I have tried row swapping, division and multiplication but cannot get the correct eigenvector.

Please help. Thank you.
No: the characteristic polynomial of A is ##x^3 - (23/15) x^2 + (127/720) x -(1/1260)## (although you wrote its negative, which is also true---but the usual definition of characteristic polynomial is that its coefficient is +1 on the highest power). You need to distinguish between ##x## (or ##\lambda##) and ##A##; if fact, the defintion of the characteristic polynomial is
[tex] \text{char. polynomial} \;\; P(x) \equiv \det(xI - A)[/tex]

And of course, A is a matrix so your statement that A = 0.1223, 0.0027 and 1.41 makes no sense; it would make sense to say that the eigenvalues of A are 1.408318927, .268734034e-2, and 0.1223270659.

I realize that your 'errors' are just errors of expression, but it is a good idea to express yourself properly, because doing what you did on an exam is a surefire way to lose marks. Anyway, math is about precision, at least in part.

As to your main question: you do not show your work, so I cannot be sure, but I suspect your inability to get eigenvectors is due to excessive roundoff before solving the linear equations. If I use your eigenvalues ##\lambda## exactly as you wrote them and then ask Maple to solve the system ##(\lambda I - A)x = 0## it chokes; it reports all three linear systems as being inconsistent. However, if I increase the accuracy to 10 digits and use the values I wrote above, Maple gets three solutions.

The fact is that linear algebra computations are notoriously sensitive to roundoff errors and to numerical instability, so need to be done carefully and to more presicision than you may think. If you do not have access to Maple or Mathematica, you can use a speadsheet to deal with the problem; for example, EXCEL has a Solver tool you can use.
 
  • #4
No: the characteristic polynomial of A is ##x^3 - (23/15) x^2 + (127/720) x -(1/1260)## (although you wrote its negative, which is also true---but the usual definition of characteristic polynomial is that its coefficient is +1 on the highest power). You need to distinguish between ##x## (or ##\lambda##) and ##A##; if fact, the defintion of the characteristic polynomial is
[tex] \text{char. polynomial} \;\; P(x) \equiv \det(xI - A)[/tex]

And of course, A is a matrix so your statement that A = 0.1223, 0.0027 and 1.41 makes no sense; it would make sense to say that the eigenvalues of A are 1.408318927, .268734034e-2, and 0.1223270659.

I realize that your 'errors' are just errors of expression, but it is a good idea to express yourself properly, because doing what you did on an exam is a surefire way to lose marks. Anyway, math is about precision, at least in part.

As to your main question: you do not show your work, so I cannot be sure, but I suspect your inability to get eigenvectors is due to excessive roundoff before solving the linear equations. If I use your eigenvalues ##\lambda## exactly as you wrote them and then ask Maple to solve the system ##(\lambda I - A)x = 0## it chokes; it reports all three linear systems as being inconsistent. However, if I increase the accuracy to 10 digits and use the values I wrote above, Maple gets three solutions.

The fact is that linear algebra computations are notoriously sensitive to roundoff errors and to numerical instability, so need to be done carefully and to more presicision than you may think. If you do not have access to Maple or Mathematica, you can use a speadsheet to deal with the problem; for example, EXCEL has a Solver tool you can use.

Thanks, it really was the rounding off issue. I am solving it by hand without programs. I will be careful with the precision and notations. Thanks for the advice.
 

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