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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?

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  • $\begingroup$ 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 $T: U^\perp \to T(U^\perp)$ $\endgroup$
    – J.B.R
    May 11, 2019 at 2:10
  • $\begingroup$ Is the map $T$ meant to be self-adjoint with respect to the inner product? $\endgroup$
    – Joppy
    May 11, 2019 at 7:57
  • $\begingroup$ In what I'm working on $T$ is a the Hessian of a function so you can make that assumption $\endgroup$
    – J.B.R
    May 11, 2019 at 15:28

1 Answer 1

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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 $(\ker T)^\perp$.

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