Prove that $E_{\lambda}(A)$ is not a zero subspace if and only if $A-\lambda I_{n}$ is not invertible.

Question

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?

Answer 1

Comment 1: You say "Then, $\vec{x}$ must be a non-zero vector." This isn't quite right; you could have chosen the 0 vector. What you want to say is "let $\vec{x} \in E_\lambda(A)$ be nonzero, which is possible since $E_\lambda(A)$ is not the zero subspace.

Comment 2: You say "since $\vec{x} \neq 0$ ...., $\det(A - \lambda I_n) = \det(\vec{0}) =0$." (Note that if you put a '\' before commonly used expressions such as 'det', MathJax and Latex will often typeset them properly, e.g. $\det$ vs $det$) This is not true. You can't even take the determinant of a vector, so the expression $\det(\vec{0})$ just doesn't even make sense.

Comment 3: Note you are trying to prove the converse not the contrapositive. For a statement like $A \implies B$, the converse is $B \implies A$ while the contrapositive is $\neg B \implies \neg A$ (where $\neg A$ should be read as "not $A$"). These are not the same thing.

Comment 4: I'm not quite sure what you mean by the "invertible matrix theorem", but it is true that $A$ is invertible iff $A x = b$ has a unique solution for any fixed $b$, which is what you seem to be trying to use. Unfortunately your application of it is not quite correct. In this theorem, $b$ has to fixed, i.e. independent of $x$, but you define $b:= \lambda x$, so this doesn't work.

Look at Luke Collins post if you want to see how you should prove everything.

Answer 2

The proof is correct, however you seem to be complicating it a bit, this depends how you define "invertible". Usually, $f$ is invertible if and only if it is a bijection. This is equivalent to having zero kernel for linear $f$.

Thus an easy proof is \begin{align*} E_\lambda(A)=\{0\} & \iff \{x\in\mathbb R^n:Ax=\lambda x\}=\{0\}\\ &\iff \{x\in\mathbb R^n:(A-\lambda I)x=0\} = \{0\}\\ &\iff \ker(A-\lambda I)=\{0\}. \end{align*}