How to prove SPACE-TMSAT is PSPACE-hard?

I understand that the language:

$$\operatorname{SPACE-TMSAT} = \{⟨M, w, 1^n⟩ : \text{DTM M accepts w in space n}\}$$

is in PSPACE since it doesn't use more than $$n$$ space. But to prove that it is PSPACE-complete, we need to do the reduction from any other PSPACE language to it. If we assume the language to run on a DTM using $$O(n^k)$$ space, how could we reduce it to SPACE-TMSAT?

Suppose $$L$$ is accepted by some machine $$M$$ using up to $$p(n)$$ space, for some polynomial $$p(n)$$. We map $$w$$ to $$\langle M, w, 1^{p(|w|)} \rangle$$.
• No, $n$ is completely arbitrary. – Yuval Filmus Apr 5 at 4:03