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I realize that when a matrix is symmetric, then it must have all real eigenvalues. However, I am doing research on matrices for my own pleasure and I cannot find a mathematical proof or explanation when a matrix will have all real eigenvalues except for when it is symmetric. I am dealing with matrices such as A below and I want to know what is it about A and its characteristic polynomial that gives it real eigenvalues (0, 0, -2)? Similarly, what is it about matrix B that gives it only one real eigenvalue (0) and the other two complex? enter image description here

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4 Answers 4

Here is one example of sufficient conditions.

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Posting just a link as an answer is rather fragile; links tend to become stale some day. At least you could try to lift from the paper the statement of the conditions you refer to. –  Marc van Leeuwen Apr 9 at 6:16

Do you know of companion matrices? See the Wikipedia link here: http://en.wikipedia.org/wiki/Companion_matrix

They are made-to-order matrices which will have the polynomial you want as its characteristic polynomial. They are far from symmetric matrices. Now start with a polynomial having your favorite real numbers as its roots, and construct the Companion matrix for that polynomial.

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I don't think the question is about how to construct matrices with real eigenvalues, but on how to recognise them. Construction is simple: just take any real triangular matrix and conjugate it by any real invertible matrix (moreover all examples can be obtained in this way). –  Marc van Leeuwen Apr 9 at 6:19
    
Ok, I see the distinction in the question which my answer does not address. –  P Vanchinathan Apr 9 at 6:35

Another approach is to construct a triangular matrix with pre-determined diagonal entries; they will be the eigenvalues, and the matrix is not symmetric.

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A necessary and sufficient condition for a matrix$~A$ to have only real eigenvalues (that is, not have any non-real complex eigenvalues) is the existence of a polynomial $P$ that splits into linear factors over the real numbers and such that $P[A]=0$. If such a polynomial exists at all, one can take the characteristic polynomial for$~A$ (but not necessarily the minimal polynomial) as$~P$. This gives a trivially valid, but fairly hard to check condition. Without using eigenvectors, it is actually not so obvious why the characteristic polynomial of a symmetric matrix should always allow such a factorisation. But I don't think one can do much better to completely characterise the case of real-only eigenvalues.

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