# Meaning of the determinant of the restriction of a linear map

Suppose $$T : \mathbb{R}^n \to \mathbb{R}^n$$ is a linear map and let $$U \subset \mathbb{R}^n$$ be a $$d$$-dimensional subspace where $$0 < d < n$$ and $$\ker T = U$$. I was wondering how to make sense of the sentence

The determinant of $$T$$ restricted to $$U^\perp$$

I figure that this means that you form the map $$S: \mathbb{R}^n \to \mathbb{R}^n$$ where $$S(x) =\begin{cases} x \text{ if } x\in U\\ T(x) \text{ otherwise} \end{cases}$$ and then the determinat of $$T$$ restricted to $$U^\perp$$ is given by $$\det S$$. Is this the correct interpretation?

• Oh, I think this would only be feasible if $T(U^\perp)$ doesn't intersect $U$. I guess you have to look at it as the determinant of the linear map $T: U^\perp \to T(U^\perp)$ – Questions May 11 at 2:10
• Is the map $T$ meant to be self-adjoint with respect to the inner product? – Joppy May 11 at 7:57
• In what I'm working on $T$ is a the Hessian of a function so you can make that assumption – Questions May 11 at 15:28

If $$T: V \to V$$ is a self-adjoint operator, then $$\operatorname{Im}(T)$$ is orthogonal to $$\ker T$$, since if $$k \in \ker T$$ then $$\langle k, Tv \rangle = \langle Tk, v \rangle = 0$$ for any $$v \in V$$. Hence $$T((\ker T)^\perp) \subseteq \operatorname{Im}(T) \subseteq (\ker T)^\perp$$, so $$T$$ actually restricts to a linear endomorphism of $$(\ker T)^\perp$$.