Prove matrix identity involving null spaces

Question

Let $A \in \mathsf{M}_n$ and suppose that $ \lambda \in \mathbb{F}$. Prove that $ E_\lambda = \mathsf{N}(\lambda I_n - A)$.

$N$ stands for the null space, $I_n$ stands for the n x n identity matrix, and $E_\lambda$ is the eigenspace of $\lambda$

Answer 1

I suspect by $E_\lambda$ you mean the eigenspace of $\lambda$, which can be described by $$ E_\lambda :=\{x \mid x=0\ \text{or}\ x\ \text{is an eigenvector of $A$ with eigenvalue $\lambda$}\} = \{x \mid Ax=\lambda x\}. $$

Then $x\in E_\lambda$ iff $Ax=\lambda x$ iff $$ 0=\lambda x-Ax=\lambda I_nx-Ax = (\lambda I_n-A)x $$ iff $x$ is in the null space of $\lambda I_n-A$. Therefore $E_\lambda = \mathsf{N}(\lambda I_n-A)$.

Answer 2

Note that \begin{align*} E_\lambda &= \{x : Ax=\lambda x\} \\ &= \{x : \lambda x- Ax=\mathbf{0}\} \\ &= \{x : (\lambda\cdot I-A)x=\mathbf{0}\} \\ &= \mathsf{N}(\lambda\cdot I-A) \end{align*}