I'm trying to show the following:
Let $A$ be an $n \times n$ matrix over a field $F$ such that every sub vectorspace of $F^n$ is invariant with respect to $A$, then $A$ must be of the form $\lambda I_n$ with $\lambda \in F$.
Intuitively, for any $v \in F^n$, since $\operatorname{span}(v)$ is invariant with respect to $A$, it must be an eigenvector of $A$. And since this is true for all such $v$, then $A$ can only have the above form. But how do I show that last step formally?
Let $(v_i)_{i\in[n]}$ be a basis for $F^n$, you have already established that for each basis vector we must have
$$Av_i =\lambda_{i}v_i $$
Our goal is simply to show that for any $i,j\in[n]$, we have $$\lambda_i=\lambda_j$$
So let $i,j\in[n]$, then consider the vector $w=v_i+v_j$, and its associated eigenvalue $\lambda_w$, note that by linearity we then have
$$Aw = \lambda_iv_i + \lambda_jv_j = \lambda_w w = \lambda_wv_i + \lambda_wv_j $$
But since each vector is uniquely expressable as a linear combination of basis vectors, this must mean that
$$\lambda_i=\lambda_j=\lambda_w$$
And since $A$ scales each basis vector $v_i$ the same, it must be a scalar multiple of the identity (this part is not tricky to show formally).
Thought process to see how one might have come to consider the vector $w$, it helps to first think as if you are doing a proof by contradiction:
If all the basis vectors weren't scaled equally, this means there are two which are scaled differently
Now you've identified two basis vectors, and you need to arrive at a contradiction, at this point you need a bit of inspiration, but its not too difficult to think it might be a good idea to add the two problematic vectors, and enforce the requirement that this vector should also scale.
