An identity matrix has only one eigenvalue, 1, but an infinite amount of eigenvectors corresponding to that eigenvalue, correct?

Question

Futhermore, if the this is the identity transformation for $R^2$, then the $span([1,0],[0,1])$ is the basis for the eigenspace of the transformation. Is that all correct or have I a misunderstanding somewhere?

Edit: thank you everyone!

Answer 1

This is correct.

In general, a scalar $\lambda$ is an eigenvalue of $A$ if $\lambda\cdot I-A$ is singular. The eigenspace associated to an eigenvalue $\lambda$ of $A$ is defined as $E_\lambda=\operatorname{Null}(\lambda\cdot I-A)$. Each vector in $E_\lambda$ is called an eigenvector of $A$.

Note that $\dim E_\lambda\geq 1$, so each eigenvalue has infinitely many eigenvectors associated to it.

Answer 2

It's correct, but not "full" - suppose that $Av =\lambda v$. Then for any nonzero real $c$, $cv$ is an eigenvector of $A$, so the fact that it's the identity matrix is not relevant for the fact that it has infinitely many eigenvectors.

Answer 3

Whenever you are dealing with vector spaces over an infinite field $k$, any eigenvalue gives rise to an infinite number of eigenvectors.

Let $V$ denote any vector space over $k$. If $T\colon V\rightarrow V$ is a linear transformation with an eigenvalue $\lambda$, then the eigenvectors associated with $T$ form a nontrivial linear subspace $0\neq V_\lambda \subset V$. (That is, $V_\lambda := \mathrm{ker}(T-\lambda I)$ where $I$ is the identity matrix.) There are many ways to see that a nontrivial vector space over $k$ must have cardinality greater than or equal to $k$. For any nonzero vector $v\in V_\lambda$, we have an injection $\alpha\mapsto \alpha v$ from $k$ into $V_\lambda$. Why is this an injection? A left inverse is given by $v\mapsto \alpha^{-1}v$ (i.e. use that $k$ is a field and $\alpha$ is nonzero). @enedil describes something to this effect in their answer.

Examples of infinite fields: $\mathbb R$, $\mathbb C$, etc.

Examples of finite fields: $\mathbb Z /p$ for $p$ a prime and more generally $\mathbb F_{p^n}$ (finite fields of characteristic $p$ for $p$ a prime).