Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!!

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.