# Show that if all sub vectorspace are A-invariant, A must be scaled identity matrix

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.

• Very clear, thank you. Commented Dec 3, 2023 at 19:34