Finding eigenvalues and eigenvectors

  • Thread starter fargoth
  • Start date
  • #1
315
6
is there any trick for finding the eigenvalues and vectors for this kind of matrix?
[tex]
\left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\
1 & 0 & \sqrt{\frac{3}{2} & 0 & 0 \\
0 & \sqrt{\frac{3}{2} & 0 & \sqrt{\frac{3}{2} & 0 \\
0 & 0 & \sqrt{\frac{3}{2} & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
\end{array}
\right)
[/tex]

i mean, i can tell the eigenvalues are 2,1,0,-1,-2... and i can tell the eigenvectors would have a=e and b=d... but thats because i know what this matrix is... but if i'll see some matrix with different values then this roaming around.... i dont know what i'll do, i dont think trying to solve the standard polynom of it is a good idea... and after knowing the eigenvalues one has to solve the set of equations to find the eigenvectors -yuck!-
 
Last edited:

Answers and Replies

  • #2
TD
Homework Helper
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Well it's a symmetrical matrix so it's surely diagonalizable.
I don't see any special 'trick' although the classical way (add [itex]-\lambda[/itex] on the main diagonal and compute the determinant) shouldn't be too hard thanks to the many 0's... (e.g. expand the determinant to the first row or column).
 
  • #3
315
6
yeah, im just REALLY lazy :tongue2:
i thought there's a trick that would make me see in a sec the right solution...
i only know the trick for block diagonal, and its not useful here...
 
  • #4
TD
Homework Helper
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Perhaps there is, but I then I don't know it :smile:

If you know the (normal) method and you're only interested in the solution, why not use a computer program?
 
  • #5
315
6
well, the advantage of knowing helpful shortcuts is that you develop some intuition, which is pretty useful...

for example, if i got a block diagonal matrix i know the different blocks are orthogonal, which means vectors with parts that belong only to a certain block would stay in that block... and thats why i can find eigenvectors seperately for each block.
 
Last edited:
  • #6
TD
Homework Helper
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Of course, it was just a suggestion if you were only looking for the answer.
It's by far a better choice to do it yourself, if you wish to develop your mathematical intuition :smile:
 

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