+0  
 
0
689
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avatar+2353 

Hi there!

I'm stuck trying to find the eigenvalues for this matrix

$$\begin{pmatrix}
3 & -1 & 0\\
-1 & 2 & -1 \\
0 & -1 & 3
\end{pmatrix}$$

I end up having the characteristic equation

$$- \lambda ^3 + 8 \lambda ^2 -20 \lambda + 12 = 0$$

I guess there must be something I missed, since solving this by hand can be quite cumbersome.

So is there something which I should've found by observation or is there an easy way to solve this which I missed.

 

Reinout 

 Jun 2, 2014

Best Answer 

 #5
avatar+893 
+5

Subtracting lambda from each of the diagonal elements and expanding the determinant along the top row gets you

$$(3-\lambda)\{(2-\lambda)(3-\lambda)-1\}-(3-\lambda)=0.$$

(3 - lambda) comes out naturally as a common factor.

If you choose to expand the whole lot though you get -19lambda rather than -20lambda.

 Jun 2, 2014
 #1
avatar+893 
0

Should be -19 rather than -20, further, (lambda-3) comes out naturally as a common factor when you expand the determinant.

 Jun 2, 2014
 #2
avatar+2353 
0

edit: Aaah I found it 

 Jun 2, 2014
 #3
avatar+33603 
+5

I get  $$-\lambda^3+8\lambda^2-19\lambda+12=0$$

Multiply through by -1 (not essential - I just find it easier that way!)

$$\lambda^3-8\lambda^2+19\lambda-12=0$$

If we assume that, because you are doing it by hand, the solutions are integers, then consider

$$(\lambda-a)(\lambda-b)(\lambda-c)$$

The three constants must multiply together such that $$abc=12$$

If they are each integers then we have either 1, 2 and 6 or 1, 3 and 4 (ok a couple of them might have negative signs attached!).

The sum of the three must also be 8, and since 1+2+6 is not 8, and 1+3+4 is, there is a good chance that the three eigenvalues are 1, 3 and 4.

Plugging each of these into the LHS of the first equation above does indeed give zero.

 Jun 2, 2014
 #4
avatar+2353 
0

Thank you Alan, that was really helpfull :)

Now that I come to think of it,

for the eigenvalues it is always true that

$$\lambda_1 + \lambda_2 + \lambda_3 = trace (A)$$

and

$$\lambda_1*\lambda_2*\lambda_3 = det(A)$$

 

In this case

$$det(A) = 12\\
trace(A) = 8\\$$

 Jun 2, 2014
 #5
avatar+893 
+5
Best Answer

Subtracting lambda from each of the diagonal elements and expanding the determinant along the top row gets you

$$(3-\lambda)\{(2-\lambda)(3-\lambda)-1\}-(3-\lambda)=0.$$

(3 - lambda) comes out naturally as a common factor.

If you choose to expand the whole lot though you get -19lambda rather than -20lambda.

Bertie Jun 2, 2014

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