Prove that the largest eigenvalue (in terms of magnitude) of a non-symmetric matrix is real.

eigenvalues-eigenvectorsmatricesmatrix decomposition

Let $A\in \mathbb{R} ^{n\times n}$ be a non-symmetric matrix with
eigenvalues $\lambda _i$, $i=1,…,n$ satisfying $$\left | \lambda_1\right | >\left | \lambda _2\right |\geqslant \left | \lambda _3\right |\geqslant …\geqslant \left | \lambda _n\right |.$$

Prove that $\lambda _1$ is real.

Since A is a non-symmetric matrix and I wanted to tackle this problem using power iteration, I decomposed A as $$A=X\Lambda X^{-1}.$$

And I tried to solve it using some power iteration properties. But I have no idea how to proceed.

Edit: I made a mistake at the beginning. There should be a gap between the largest eigenvalue and other eigenvalues.

Best Answer

The complex eigenvalues of a real matrix come in conjugate pairs, as they are roots of the matrix's characteristic polynomial, which is real. If $\lambda_1$ is not real, its conjugate $\lambda_1^*$ is also an eigenvalue and has the same magnitude, meaning $\lambda_1$ does not have a strictly largest magnitude – contradiction. Hence $\lambda_1$ must be real.