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?
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:
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 $5$.