I know that there is this question (which is closed): Linear Algebra - Vector space - Help!!!! but I am specifically asking for feedback on my proof and I wasn't sure how to approach this so I uploaded a new question.
Let $A$ be a $n \times n$ matrix and let $\lambda \in R^{1}$. Define $E_{\lambda}(A) := \{v \in R^{n}: Av = \lambda v\}.$ Prove that $E_{\lambda}(A)$ is not a zero subspace if and only if $A-\lambda I_{n}$ is not invertible. I cannot use the fact that if the Kernel of a matrix is trivial then the matrix is invertible (from the Invertible Matrix Theorem).
Proof: Suppose $E_{\lambda}(A)$ is not a zero subspace. Let $\vec{x} \in E_{\lambda}(A)$ be non-zero. Then, $A\vec{x} = \lambda \vec{x}$ which implies: \begin{equation} A\vec{x} - \lambda \vec{x} = \vec{0} \end{equation} \begin{equation} (A- \lambda I_{n})\vec{x} = \vec{0} \end{equation} Note that $A- \lambda I_{n}$ is a square matrix so we can take its determinant. Since $\vec{x} \neq \vec{0}$, it follows that $A- \lambda I_{n}= \vec{0}$ which implies that $\det(A- \lambda I_{n})= \det(\vec{0}) = \vec{0}$ and we are done. Conversely, to prove that $E_{\lambda}(A)$ is not a zero subspace if $A- \lambda I_{n}$ is not invertible, assume $E_{\lambda}(A)$ is a zero subspace. (We will show that $A- \lambda I_{n}$ is invertible). Then, the only vector that solves $A\vec{x} - \lambda \vec{x}$ is $\vec{x} = \vec{0}$. In other words, $A\vec{x} = \vec{b}$ has a (unique) solution, namely $\vec{x} = \vec{0}$, where $b = \lambda \vec{x} = \vec{0}$. So, by the Invertible Matrix Theorem, A is invertible. This completes the proof. Is this proof correct? Where am I going wrong? How can I improve it?