The identity matrix in, for example, $R^2$, has every vector in $R^2$ as an eigenvector, correct?

Question

Furthermore, every vector in $R^2$ corresponds to the eigenvalue of the identity matrix 1. Is this all correct?

Answer 1

This is correct: For all ${\bf v} \in \Bbb R^2$, $I {\bf v} = {\bf v} = 1 \cdot {\bf v}$, so every vector in $\Bbb R^2$ (exception $\bf 0$, which by definition is never an eigenvector) is an eigenvalue of the identity matrix $I$ of eigenvalue $1$.

Answer 2

Yes. The definition of 'eigenvalue' for a given matrix, A, is a number, $\lambda$, such that there exist a non-zero vector, v, such that $Av= \lambda v$. Since if A is the identity matrix, Av= 1*v, 1 is an eigenvalue for the identity matrix (in fact the only one). The definition of 'eigenvector', corresponding to the eigenvalue $\lambda$, is that $Av= \lambda v$. Since, for the identity matrix, that is true for every vector, every vector is an eigenvector. This is not just true for 2 by 2 matrices but for the identity transformation over any vector space.

(Some people maintain that an "eigenvector" must be non-zero but I don't like that because then you have to keep adding "and the zero-vector" to make it true, for example, that "the set of all eigenvectors corresponding to a give eigenvalue form a subspace" or this statement. I assert that the "v" in the definition of "eigenvalue" must be non-zero but an "eigenvector" does not have to be. The 0 vector is an eigenvector for every linear transformation corresponding to every eigenvalue.)