Proving that $Aw\in \langle w\rangle \implies A$ is of the form $\lambda I_n$

If $$w=\begin{bmatrix} w_1\\ \vdots\\w_n \end{bmatrix}$$ is a vector in $$K^n$$ for a field $$K$$ and $$A= \begin{bmatrix} \lambda & & \\ & \ddots & a_{ij}\\ & & \lambda \end{bmatrix}$$ is a $$n\times n$$ matrix with entries on $$K$$ and $$\lambda\in K^{*}$$ then $$Aw=\lambda w+e_i a_{ij} w_j\notin\langle w\rangle$$ if $$i\ne j$$. And so $$\langle A w\rangle\ne \langle w\rangle$$ where $$\langle w\rangle$$ is the $$1$$ dimensional vector space generated by $$w$$.

But if $$A$$ is a general matrix in $$GL_n(K)$$ that is not of the form $$\lambda I_n$$ then $$Aw=\sum\limits_{i=1}^n\sum\limits_{j=1}^n a_{ij}w_je_i$$.

How to show that $$Aw\notin \langle w\rangle$$?

Can we decompose $$A$$ into $$\sum A_{ij}$$ where each $$A_{ij}$$ has a simple form like above (so $$A_{ij}w\notin \langle w\rangle$$) and say that $$Aw=\underbrace{\sum A_{ij}w}_{\notin \langle w\rangle} \notin \langle w\rangle$$?

For context the question asks to prove that $$AW=W~~\forall W\in\{\text{sub-vector-spaces of } K^n \text{ with dimension 1}\} \iff A=\lambda I_n$$ for some $$\lambda\in K^{*}$$

The condition $$Aw\in\langle w\rangle$$, applied to elements of the standard basis of $$K^n$$, implies that $$A$$ is a diagonal matrix. Now, for elements $$w_1\neq w_2$$ of the basis, being linearly independent, $$Aw_1=\lambda_1 w_1$$ and $$Aw_2=\lambda_2 w_2$$ together with $$A(w_1+w_2)=\lambda(w_1+w_2)$$ imply $$\lambda_1=\lambda_2(=\lambda)$$.