Let $A$ be an $n\times n$ matrix with real coefficients. Prove or disprove:
(a) if $\lambda\in\mathbb{R}$ satisfies $Ax=\lambda x$ for some vector $x$, then $\lambda$ is an eigenvalue of $A$;
(b) if $x\in\mathbb{R}^n$ satisfies $Ax=\lambda x$ for some $\lambda\in\mathbb{R}$, then $x$ is an eigenvector of $A$.
Make clear what definitions you are using for eigenvalue and eigenvector.
This is a problem from a past graduate school entrance exam, and I'm quite confused about it. These are $\textit{almost}$ the $\textbf{definition}$ of eigenvalue and eigenvector, but they didn't specify that $x$ is not the zero vector. So I'm not sure what there is to disprove even, as it's direct from definition. My question, then, is if there's another way to define eigenvalue and eigenvector that lends to an actual proof for this problem.
Best Answer
Let's fix a field $k=\mathbb{R}$ or $\mathbb{C}$ and $A$ is an $n\times n$ matrix with entries in $k$.
By definition, an eigenvalue of $A$ is $\lambda\in k$ such that $$Ax=\lambda x,\quad x\in k^n\setminus\{0\}.$$ Thus, we cannot expect assertion (a) to be true. Say $A=0$. The only eigenvalue of $A$ is $0$, but every $\lambda$ satisfies $Ax=\lambda x$ if $x=0$.
However, the definition of an eigenvector may depend on authors. I think the standard definition requires an eigenvector to be nonzero: an eigenvector is $x\in k^n\setminus\{0\}$ such that $$Ax=\lambda x, \quad \lambda\in k.$$ If you use this definition then (b) is not true for $x=0$ is never an eigenvector.
A less common definition of eigenvector is as follows. If $\lambda$ is an eigenvalue of $A$, then any $x\in k^n$ that solves $Ax=\lambda x$ is called an eigenvector of $A$, associated with $\lambda$. If this convention is used then (b) becomes true! Some discussions can be found here.
Because there are two different definitions of eigenvectors, you need to specify which definition you are using.