[Math] Can real non-symmetric matrices have real eigenvalues

eigenvalues-eigenvectorslinear algebramatrices

From the spectrum theorem, we know real symmetric matrices have real eigenvalues.

But can real non-symmetric matrix have real eigenvalues?

What are the necessary and sufficient conditions for a real matrix to have real eigenvalues?

Best Answer

Yes, I asked myself the same question, and the condition:

  • real eigenvalues
  • real non symmetric matrix

can be more strict:

$\lambda_i \in\mathbb{N}, i=1, \ldots, n $ and non symmetric matrix $M \in \mathbb N^{n\times n}$

Where

$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{A}_{R} \subset \mathbb{R}$

Here is one such matrix.

$M = \left[\begin{array}{ll}2 & 3 \\ 1 & 4\end{array}\right]$

with the $\lambda_1= 1$ and $\lambda_2= 5$, $\{\lambda_1, \lambda_2\}$ $\in \mathbb N$.

Let's show there are infinitely many matrices that are non symmetric with exactly the same eigenvalues (similar).

If we decompose matrix $M$ via Schur method we get:

$\begin{array}{c} Q\to \left[ \begin{array}{cc} -0.948683 & -0.316228 \\ 0.316228 & -0.948683 \end{array} \right] \\ T\to \left[ \begin{array}{cc} 1 & 2 \\ 0 & 5 \end{array} \right] \\ M\to Q\cdot T\cdot Q^{\text{-1}} \end{array}$

So now we are having the matrix $T$ that is non symmetric, and we can alter the upper triangular value $2$, to anything other than $0$ (because of the non symmetric condition) and still the eigenvalues will be $1$ and $5$.

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