Why is the following statement true?
If all entries in a Matrix $K=A^TA=AA^T$ are greater than or equal to 0, then the highest eigenvalue has a corresponding eigenvector whose entries are also greater than or equal to 0.
2026-03-25 11:53:00.1774439580
Proving Matrix properties in conjunction with eigenvalues
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This is part of what is known as Perron-Frobenius theorem. The (real and nonnegative) eigenvalue that is equal to the spectral radius of $K$ is called the Perron eigenvalue, and any nonnegative eigenvector (usually normalised so that the sum of all elements, rather than the Euclidean norm, is equal to $1$) corresponding to this eigenvalue is called a Perron vector.
Showing that the Perron eigenvalue exists is relatively easy (e.g. see the above Wikipedia entry). The difficult part is to prove the existence of a Perron vector. When $K$ is doubly stochastic, the Perron eigenvalue is $1$ and the existence of a Perron vector is trivial: just take the all-one vector. If $K$ is entrywise positive and you only want to prove that there exists an entrywise positive eigenvector corresponding to some positive eigenvalue, there is also a simple proof (see Calvin Lin's answer, for instance). Yet, for the general case, I am not aware of any simple proof of the existence of Perron vector.