Specifically, suppose $\lambda_1$ is an eigenvalue of the matrix $A$, and that it has multiplicity of 2.
Also suppose that the eigenspace corresponding to $\lambda_1$ consists solely of vectors of the form
$$
\begin{pmatrix}
{\alpha/2+\beta\\\alpha\\\beta}
\end{pmatrix}
=
\alpha\begin{pmatrix}
{\frac12\\1\\0}
\end{pmatrix}
+\beta\begin{pmatrix}
{1\\0\\1}
\end{pmatrix}
$$
In this case I believe we would call $\begin{pmatrix}
{1\\0\\1}
\end{pmatrix}$ and $\begin{pmatrix}
{\frac12\\1\\0}
\end{pmatrix}$
eigenvectors corresponding to the eigenvalue $\lambda_1$
Since any vector, $v$, of the form $v=\begin{pmatrix}
{\alpha/2+\beta\\\alpha\\\beta}
\end{pmatrix}$ satisfies $Av=\lambda_1 v$ how come the basis vectors are the eigenvectors?More specifically, I am wondering if can I just say that "the eigenvectors corresponding to eigenvalue $\lambda_1$ are all vectors of the form $v=\begin{pmatrix}
{\alpha/2+\beta\\\alpha\\\beta}
\end{pmatrix}$?
Is it that both approaches are okay, but using vectors that form a basis is more informative (for example giving info about dimension of the eigenspace)?
Best Answer
All nonzero vectors $x$ satisfying $A x = \lambda x$ are eigenvectors. Not just the basis vectors. It's just that a convenient way to specify a vector space is to list a basis for it.