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Eigenvalues of a 2 by 2 matrix 
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#1
Oct2505, 09:56 AM

P: 585

Hi, I'm wondering if there is some kind of shortcut for finding the eigenvalues and eigenvectors of the following matrix.
[tex] C = \left[ {\begin{array}{*{20}c} {0.8} & {0.3} \\ {0.3} & {0.7} \\ \end{array}} \right] [/tex] Solving the equation [tex]\det \left( {C  \lambda I} \right) = 0[/tex], I get [tex]\lambda = \frac{{15 \pm \sqrt {37} }}{{20}}[/tex] and then if I try to find the eigenvectors by finding the solution space of [C  (eigenvalues)I] I end up with what seems like an endless/tedious/time consuming/frustrating bunch of row operations which don't appear to lead anywhere. I'm wondering if there is an easier way to find the eigenvalues, perhaps that might reveal an easier way to find the eigenvectors. Thanks... 


#2
Oct2505, 12:17 PM

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PF Gold
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The hard part is finding the eigenvalues!
Find the corresponding eigenvectors is the same as solving [itex]Ax= \lamdax[/itex] for each eigenvalue, [itex]\lamda[/itex]. In this case solve the equation [tex] .8x+ .3y= \frac{{15 \pm \sqrt {37} }}{{20}}x[/tex] You don't need to worry about the second equation, it will be automatically satisfied by your solution to this. Of course, you can only find y in terms of x: any multiple of an eigenvector is an eigenvector. Since the entries in the matrix are given as decimals to one decimal place, I would recommend using 1.0 and 0.4 in place of [tex]\frac{{15 +\sqrt {37} }}{{20}}[/tex] and [tex]\frac{{15  \sqrt {37} }}{{20}}[/tex] respectively. 


#3
Oct2605, 03:17 AM

P: 585

Hmm...I should've seen that.
Thanks very much for your help. So the key idea is that the multiplicity of each of the eigenvalues is 1 so the eigenspace of each one must be one dimensional? If so then I can see why it would be fairly easily to extract appropriate eigenvectors. Edit: The question is from a noncalculator exam so I can't take decimal approximations. :( 


#4
Oct3105, 08:36 PM

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Eigenvalues of a 2 by 2 matrix
8A + 3B = (15 + sqrt(37))/2 A Multiply by 2 to get 16A + 6B = 15A + sqrt(37) A, or 6B = (+sqrt(37)  1) A. And the answers are the vectors: [tex]\left(\begin{array}{c} 6 \\ \pm \sqrt{37}1\end{array}\right)[/tex] The trick to remember is to put the eigenvalue equation into "kA = mB" form, and then the eigenvectors are (m,n). Carl 


#5
Oct3105, 11:40 PM

P: 585

Thanks Carl. I haven't been using a calculator to do math problems for quite a while anyway, so I'm pretty used to using the traditional ways to solve equations.
Anyway, suppose that I've been given a matrix A and I'm told to find an invertible matrix P and a diagonal matrix D such that A = PDP^1. Then the implication is that A is definitely diagonalizable right? Suppose A is a 3 by 3 matrix. A condition for A to be diagonalizable is that the algebraic multiplicity of each eignenvalue is the same as the dimension of the eigenspace corresponding to each of those eigenvalues. So say that I find an eigenvalue of A, call it lamda sub one, which has a multiplicity of 1. To find the corresponding eigenspace I need to solve the system: [tex] \left[ {A  \lambda _1 I} \right]\mathop x\limits^ \to = \mathop 0\limits^ \to [/tex] If I have: [tex] \left[ {A  \lambda _1 I} \right] = \left[ {\begin{array}{*{20}c} a & b & c \\ d & e & f \\ g & h & i \\ \end{array}} \right] [/tex] Then since I know that the multiplicity of the eigenvalue is one, and by the implications of the question(asking for A = PDP^1), can I just straight away ignore/wipe out one of the rows of the above matrix because I know the dimension of the eigenspace must be one? 


#6
Nov105, 07:05 AM

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If you've only got one eigenvector, then you can reduce the size of A by finding any two more vectors that are perpendicular to your eigenvector and perpendicular to each other. When you assemble a P inverse and P from that set of three vectors, its effect on D will be to turn it into a a matrix with a diagonal of your eigenvalue followed by a 2x2 matrix. Is that what you're getting at? Carl 


#7
Nov105, 07:10 AM

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#8
Nov105, 07:15 AM

P: 585

What I mean is, if one of my eigenvectors is lamda sub one, then to find an eigenvector corresponding to that eigenvalue, I would find the solution space of [tex]\left[ {A  \lambda _1 I} \right][/tex] which is just the eigenspace corresponding to the eigenvalue lambda sub one. From that, I can extract an eigenvector and put it into P as one of P's columns. That's the procedure I used for finding eigenvectors.
My original question rests on actually finding the eigenspace. As I mentioned before, assuming that the multiplicity of the eigenvalue lamda sub one is exactly one. Then since A is diagonalizable, the dimension of the eigenspace corresponding to that eigenvalue, must also be one. In finding the eigenspace, this comes down to finding the solution space of the matrix that I referred to in the above paragraph. Since there is one 'parameter' in the solution, then doesn't that mean that one of the rows can be ignored, since whatever solution I get out from the other two rows of the matrix, must satisify the equation represented by the 'ignored' row. I know what I've said is quite confusing but I'm pretty sure that I can do what I mentioned. I just wanted some clarification. 


#9
Nov105, 07:37 AM

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Benny, I think you understand this well, but it doesn't hurt to spend a certain amount of time solving matrix equations. Here's (a not terribly hard) one that I've been working on:
[tex]\left(\begin{array}{cccccccc} +c_pc_t & +c_ps_t & s_pc_t & s_ps_t \\ c_ps_t & +c_pc_t & +s_ps_t & s_pc_t \\ +s_pc_t & s_ps_t & +c_pc_t & +c_ps_t \\ s_ps_t & s_pc_t & c_ps_t & +c_pc_t \end{array}\right)[/tex] where [tex]c_p = \cosh(\alpha_p)[/tex], [tex]s_p = \sinh(\alpha_p)[/tex], [tex]c_t = \cos(\alpha_t)[/tex], [tex]s_t = \sin(\alpha_t)[/tex] and [tex]\alpha_p[/tex] and [tex]\alpha_t[/tex] are real numbers. The problem arises when I try to parameterize symmetry breaking of parity and time reversal symmetries by modifying the Dirac equation, more or less. I'm putting it here to show that this sort of eigenvector problem is something that you may find useful in your later studies in physics. It's also a very important part of engineering and mathematics in general. By the way, TD is right, to get P and P inverse from the eigenvectors does require that the eigenvectors be normalized and orthogonal. In your situation, since the eigenvalues have multiplicity one, the orthogonality is automatic. Carl 


#10
Nov105, 08:47 AM

P: 585

Thanks for taking the time to answer my questions today. I'll be off now, need some sleep.



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