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Let matrix $A$ be $$\begin{bmatrix} -5& 1& 0& 0\\ a &2& 1 &0\\ 0& 1 &1 &1\\ 0 &0&1& 0 \end{bmatrix}$$ where $a$ is a constant between 1 and 3.

Show that the dominant eigenvalue is real.

Thanks a lot!!

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What did you do? –  Sigur Jan 3 '13 at 19:55
1  
calculating explicitly is not possible Gershgorin circle theorem gives info for the absolute value?? –  Salih Ucan Jan 3 '13 at 20:23

1 Answer 1

up vote 5 down vote accepted

$A$ is a real tridiagonal matrix. One property of real tridiagonal matrices is this: if the signs of the entries in the upper and lower diagonals are symmetric (i.e. the $(i,\,i+1)$-th and $(i+1,\,i)$-th entries have the same sign for every $i$), then the matrix is similar to a real symmetric matrix and hence all of its eigenvalues are real. Now this is your case here.

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How to show that @user1551 ? –  Salih Ucan Jan 3 '13 at 22:41
    
See these slides, for instance. –  user1551 Jan 4 '13 at 9:31

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