Tuesday, 28 June 2016

matrices - Find x0 when a 3x3 symmetric matrix has equal eigenvalues



The question goes like this: There is a symmetric matrix:A=[3000x202x]



Find the value(s) of x for which A has at most two distinct eigenvalues. (Eigenvalues like 3,2,2)



In my attempts to solve this problem, I got the characteristic equation as:

λ3(2x+3)λ2+(x2+3x2)λ3(x24)=0
I am unable to proceed any further than this. Should I try to solve for λ by putting appropriate values in the equation, then find x?



Is there any property that I seem to be missing?


Answer



Observe that
A[100]=3[100].
Thus λ=3 is an eigenvalue of this matrix.



Also observe

A[011]=(x+2)[011].
Thus λ=x+2 is an eigenvalue of this matrix as well.



Now the sum of the eigenvalues is the trace of the matrix. Let the other eigenvalue be λ3, then
3+(x+2)+λ3=2x+3λ3=x2.



So the three eigenvalues are 3,x+2 and x2. We want at most two distinct eigenvalues. Observe that when x=1,5 then two of them are equal, hence only two distinct eigenvalues.



When x=1, the eigenvalues are 3,3,1.




When x=5, the eigenvalues are 3,7,3.



When x1,5, the eigenvalues are all distinct.



For no value of x can all the eigenvalues be the same.






Further addition to the solution:




In case you are not aware of the trace result, you can still get the third eigenvalue by observing that
A[011]=(x2)[011].
Thus λ=x2 is an eigenvalue of this matrix as well.


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