The below result is from Linear Algebra Done Right, 4th edition, Sheldon Axler:
5.11 linearly independent eigenvectors
Suppose $T \in \mathcal{L}(V)$. Then every list of eigenvectors of $T$
corresponding to distinct eigenvalues of $T$ is linearly independent.
Firstly, cool result. Secondly, I'm having trouble understanding one part of Axler's proof. He does a proof by contradiction. Since he assumes the result is false, he states:
… there exists a smallest positive integer $m$ such that there
exists a linearly dependent list $v_1, \ldots, v_m$ of eigenvectors
of $T$ corresponding to distinct eigenvalues $\lambda_1, \ldots,
\lambda_m$ of T (note that $m \geq 2$ because an eigenvector is, by
definition, nonzero).
My question is why must there specifically exist a smallest positive integer $m$? Is that referring to the note in the parenthesis, i.e., $m$ cannot be smaller than $2$? This part is integral to the proof, so I must know.
Best Answer
Every nonempty set of positive integers contains a smallest element.