The discussion clarifies that when finding eigenvectors, there is a free choice of variables due to the scalar multiplication property of eigenvectors. For any non-zero scalar \( c \), the vector \( \begin{pmatrix} x \\ y \end{pmatrix} = c \begin{pmatrix} -5 \\ 1 \end{pmatrix} \) represents an eigenvector corresponding to a specific eigenvalue. This property allows for infinite eigenvector representations, as any linear combination of eigenvectors also results in another eigenvector, confirming that the set of all eigenvectors corresponding to a given eigenvalue forms a subspace.
PREREQUISITESStudents and professionals in mathematics, particularly those studying linear algebra, as well as anyone working with systems that involve eigenvalues and eigenvectors in applied fields such as engineering and physics.
synkk said:
Hi, I don't quite understand when finding eigenvectors there is usually a free choice of variables to pick, and you can sub in any number to find an eigenvector. Could anyone please explain how this works (and why you can sub in any number), as this usually comes up in vector problems with 3 variables also.