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

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: $$$$A\vec{x} - \lambda \vec{x} = \vec{0}$$$$ $$$$(A- \lambda I_{n})\vec{x} = \vec{0}$$$$ Note that $$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?

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.

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*}

• I would suggest that the given proof is not correct and the OP asked not just for a proof but if their proof was correct. Note the list of issues pointed out in my answer. Apr 17, 2020 at 18:30