[proofplan] The strategy is purely linear-algebraic. We first show that any two outward-pointing vectors $v, w \in T_pM$ are related by $w = a v + u$ with $a > 0$ and $u \in T_p\partial M$, by working in a boundary chart where outward-pointing is detected by the sign of a single coordinate. Given any ordered basis $(e_1, \ldots, e_{k-1})$ of $T_p\partial M$, the change-of-basis matrix from $(v, e_1, \ldots, e_{k-1})$ to $(w, e_1, \ldots, e_{k-1})$ is then lower triangular with determinant $a > 0$. Hence the two bases of $T_pM$ have the same orientation, so they are simultaneously positively oriented or simultaneously negatively oriented, which is exactly the statement $\mathcal{O}_v = \mathcal{O}_w$. [/proofplan]

[step:Reduce the well-definedness statement to a comparison of ordered bases of $T_pM$]By the definition of $\mathcal{O}_v$, two outward-pointing vectors $v, w \in T_pM$ induce the same orientation on $T_p\partial M$ if and only if, for every ordered basis $(e_1, \ldots, e_{k-1})$ of $T_p\partial M$, the ordered bases \begin{align*} \mathcal{B}_v &:= (v, e_1, \ldots, e_{k-1}), \\ \mathcal{B}_w &:= (w, e_1, \ldots, e_{k-1}) \end{align*} of $T_pM$ have the same orientation with respect to $\mathcal{O}_p$. Equivalently, the change-of-basis matrix $A \in \mathbb{R}^{k \times k}$ defined by \begin{align*} \mathcal{B}_w = \mathcal{B}_v \cdot A \end{align*} (in which the $j$-th column of $A$ expresses the $j$-th vector of $\mathcal{B}_w$ in the basis $\mathcal{B}_v$) must satisfy $\det A > 0$. In the case $k = 1$, the basis $(e_1, \ldots, e_{k-1})$ is empty, $\mathcal{B}_v = (v)$, $\mathcal{B}_w = (w)$, and $A = (a)$ with $w = a v$; the same determinantal criterion applies and the argument that follows still goes through. We proceed with the general $k \geq 1$ statement.[/step]

[guided]To prove $\mathcal{O}_v = \mathcal{O}_w$, we must check that the two orientations agree on every ordered basis of $T_p\partial M$. So fix an arbitrary ordered basis $(e_1, \ldots, e_{k-1})$ of $T_p\partial M$. The orientation of $T_p\partial M$ assigned by $\mathcal{O}_v$ to this basis is determined by whether $(v, e_1, \ldots, e_{k-1})$ is positively or negatively oriented in $T_pM$; similarly for $\mathcal{O}_w$ via $(w, e_1, \ldots, e_{k-1})$. These two ordered bases of $T_pM$ have the same orientation with respect to $\mathcal{O}_p$ if and only if the change-of-basis matrix from one to the other has positive determinant — this is the defining property of an orientation on a finite-dimensional [vector space](/page/Vector%20Space). Why does it suffice to verify positivity of the determinant for **one** choice of $(e_1, \ldots, e_{k-1})$? Because once we know $\mathcal{B}_v$ and $\mathcal{B}_w$ have the same orientation in $T_pM$, they are both positively oriented or both negatively oriented, and the same conclusion automatically transfers to any other choice of $(e_1, \ldots, e_{k-1})$ via the same argument. We have therefore reduced the geometric problem to a linear-algebraic one: compute $\det A$ where $A$ is the change-of-basis matrix from $\mathcal{B}_v$ to $\mathcal{B}_w$, and show that $\det A > 0$. The $k = 1$ degenerate case (where $\partial M$ is $0$-dimensional and an orientation on $T_p \partial M$ is just a sign) collapses gracefully into the same framework: $A$ becomes a $1 \times 1$ matrix and the criterion $\det A > 0$ becomes "the scalar relating $v$ and $w$ is positive".[/guided]

[step:Express $w$ as $a v + u$ with $a > 0$ and $u \in T_p \partial M$, using a boundary chart]Choose a boundary chart $(U, \varphi)$ at $p$, i.e., $U \subseteq M$ is open, $p \in U$, and $\varphi: U \to \varphi(U) \subseteq \mathbb{H}^k$ is a diffeomorphism onto a relatively open subset of $\mathbb{H}^k := \{x \in \mathbb{R}^k : x_k \geq 0\}$, mapping $p$ to a point with $x_k = 0$. The differential \begin{align*} d\varphi_p: T_pM \to \mathbb{R}^k \end{align*} is a linear isomorphism. Let $\rho: \mathbb{R}^k \to \mathbb{R}$ be the linear coordinate projection $\rho(y) = y_k$. Since $\partial M \cap U = \varphi^{-1}(\{y \in \mathbb{R}^k : y_k = 0\})$, a vector $\xi \in T_pM$ lies in $T_p\partial M$ exactly when $(\rho \circ d\varphi_p)(\xi) = 0$. Hence \begin{align*} d\varphi_p(T_p\partial M) &= \{y \in \mathbb{R}^k : y_k = 0\}. \end{align*} For any $\xi \in T_pM$, write $\pi_k(\xi) := (d\varphi_p(\xi))_k$ for the $k$-th coordinate of $d\varphi_p(\xi)$. Then $\pi_k: T_pM \to \mathbb{R}$ is a linear functional with kernel exactly $T_p\partial M$. By the definition of "outward-pointing" stated in the theorem, $\pi_k(v) < 0$ and $\pi_k(w) < 0$. Since $v \notin T_p\partial M$ and $T_p\partial M$ has codimension $1$ in $T_pM$, the subspace $\mathbb{R} v \oplus T_p\partial M = T_pM$. Hence there exist unique $a \in \mathbb{R}$ and $u \in T_p\partial M$ with \begin{align*} w &= a v + u. \end{align*} Applying $\pi_k$ and using $\pi_k(u) = 0$ gives $\pi_k(w) = a \, \pi_k(v)$. Since $\pi_k(v) < 0$ and $\pi_k(w) < 0$, dividing yields \begin{align*} a &= \frac{\pi_k(w)}{\pi_k(v)} > 0. \end{align*}[/step]

[guided]We need a concrete decomposition $w = a v + u$ in which we can detect the sign of $a$. The defining property of an outward-pointing vector is local and chart-dependent, so we work in a boundary chart $(U, \varphi)$ at $p$. By definition, $\varphi$ is a diffeomorphism onto an open subset of the closed half-space $\mathbb{H}^k = \{x_k \geq 0\}$, sending $p$ to a point with $x_k = 0$. The differential $d\varphi_p: T_pM \to \mathbb{R}^k$ is then a linear isomorphism. Let $\rho: \mathbb{R}^k \to \mathbb{R}$ be the coordinate projection $\rho(y) = y_k$. Since $\partial M \cap U = \varphi^{-1}(\{y \in \mathbb{R}^k : y_k = 0\})$, a vector $\xi \in T_pM$ lies in $T_p\partial M$ exactly when $(\rho \circ d\varphi_p)(\xi) = 0$. Therefore $d\varphi_p$ sends $T_p\partial M$ onto the hyperplane $\{y \in \mathbb{R}^k : y_k = 0\}$. To detect outward-pointing-ness we extract the $k$-th coordinate: define $\pi_k(\xi) := (d\varphi_p(\xi))_k$. This is a linear functional $T_pM \to \mathbb{R}$ whose kernel is precisely $T_p\partial M$, since $d\varphi_p$ sends $T_p \partial M$ onto $\{y_k = 0\}$ bijectively. Outward-pointing exactly means $\pi_k < 0$, by definition. Now we use the fact that $v \notin T_p\partial M$. Since $\ker \pi_k = T_p\partial M$ has codimension $1$, the one-dimensional subspace $\mathbb{R} v$ is complementary to $T_p\partial M$ in $T_pM$; that is, $T_pM = \mathbb{R} v \oplus T_p\partial M$. This direct-sum decomposition gives **unique** scalars $a \in \mathbb{R}$ and $u \in T_p\partial M$ with $w = a v + u$. The key sign check: apply $\pi_k$. Linearity and $u \in \ker \pi_k$ yield \begin{align*} \pi_k(w) = a\, \pi_k(v) + \pi_k(u) = a\, \pi_k(v). \end{align*} Both $\pi_k(v)$ and $\pi_k(w)$ are negative [real numbers](/page/Real%20Numbers), so their ratio is positive, giving $a > 0$. A subtle point: the choice of boundary chart $(U, \varphi)$ might appear to affect the value of $a$. It does not — $a$ is intrinsic, defined by the direct-sum decomposition $T_pM = \mathbb{R} v \oplus T_p\partial M$ — and the sign of $\pi_k(v)$ is also intrinsic, since the equivalence of "for some chart" and "for any chart" in the definition of outward-pointing is a standard consequence of the fact that boundary-preserving diffeomorphisms of $\mathbb{H}^k$ have differentials at boundary points whose $(k,k)$-entry is strictly positive. The chart is merely a convenient witness.[/guided]

[step:Compute the change-of-basis matrix and its determinant]Expand $u \in T_p \partial M$ in the basis $(e_1, \ldots, e_{k-1})$ as \begin{align*} u &= \sum_{i=1}^{k-1} c_i \, e_i, \qquad c_1, \ldots, c_{k-1} \in \mathbb{R}. \end{align*} Combined with the result of the previous step, \begin{align*} w &= a v + c_1 e_1 + c_2 e_2 + \cdots + c_{k-1} e_{k-1}. \end{align*} For $i = 1, \ldots, k-1$ the coordinate expansion $e_i = 0 \cdot v + \sum_{j=1}^{k-1} \delta_{ij} e_j$ holds. Reading off the coefficients gives the change-of-basis matrix $A \in \mathbb{R}^{k \times k}$ from $\mathcal{B}_v$ to $\mathcal{B}_w$: \begin{align*} A &= \begin{pmatrix} a & 0 & 0 & \cdots & 0 \\ c_1 & 1 & 0 & \cdots & 0 \\ c_2 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ c_{k-1} & 0 & 0 & \cdots & 1 \end{pmatrix}. \end{align*} This matrix is lower triangular. The [determinant of a triangular matrix](/theorems/3299) equals the product of its diagonal entries, so \begin{align*} \det A &= a \cdot \underbrace{1 \cdot 1 \cdots 1}_{k-1 \text{ factors}} = a > 0, \end{align*} using $a > 0$ from the previous step.[/step]

[guided]We now translate the decomposition $w = a v + u$ into a matrix and compute its determinant. The first column of $A$ encodes how to express the first vector of $\mathcal{B}_w$, namely $w$, in terms of $\mathcal{B}_v = (v, e_1, \ldots, e_{k-1})$. From $w = a v + c_1 e_1 + \cdots + c_{k-1} e_{k-1}$ the first column is $(a, c_1, c_2, \ldots, c_{k-1})^\top$. For $j = 2, \ldots, k$, the $j$-th vector of $\mathcal{B}_w$ is $e_{j-1}$, which equals $0 \cdot v + e_{j-1}$ in $\mathcal{B}_v$. Hence the $j$-th column of $A$ is $(0, 0, \ldots, 1, \ldots, 0)^\top$ with the $1$ in position $j$. Assembling these columns gives the explicit lower-triangular form displayed above. Why is the form lower-triangular and not upper-triangular? The convention $\mathcal{B}_w = \mathcal{B}_v \cdot A$ places the coefficients of each new basis vector into the **columns** of $A$. The new first vector $w$ has a generally non-zero expansion, contributing the entire first column; each new $e_i$ for $i \geq 1$ has only one non-zero coefficient (in row $i+1$, corresponding to itself in $\mathcal{B}_v$), so columns $2, \ldots, k$ are standard basis vectors of $\mathbb{R}^k$. The result is the matrix shown. The determinant of a lower-triangular matrix is the product of its diagonal entries, so \begin{align*} \det A = a \cdot 1 \cdot 1 \cdots 1 = a, \end{align*} and we have already shown $a > 0$. Note that the off-diagonal entries $c_1, \ldots, c_{k-1}$ play no role in the determinant computation — only the sign of $a$ matters, and it is precisely the sign extracted from the outward-pointing condition.[/guided]

[step:Conclude that $\mathcal{O}_v = \mathcal{O}_w$] Fix any ordered basis $(e_1, \ldots, e_{k-1})$ of $T_p\partial M$. By the previous step, the change-of-basis matrix $A$ from $\mathcal{B}_v = (v, e_1, \ldots, e_{k-1})$ to $\mathcal{B}_w = (w, e_1, \ldots, e_{k-1})$ satisfies $\det A > 0$. By the defining property of an orientation on a finite-dimensional real [vector space](/page/Vector%20Space), two ordered bases of $T_pM$ have the same $\mathcal{O}_p$-orientation if and only if the change-of-basis matrix between them has positive determinant. Therefore $\mathcal{B}_v$ and $\mathcal{B}_w$ are simultaneously positively oriented or simultaneously negatively oriented with respect to $\mathcal{O}_p$. By the definition of $\mathcal{O}_v$ and $\mathcal{O}_w$ recalled in Step 1, this means: $(e_1, \ldots, e_{k-1})$ is $\mathcal{O}_v$-positively oriented if and only if it is $\mathcal{O}_w$-positively oriented. As $(e_1, \ldots, e_{k-1})$ was an arbitrary ordered basis of $T_p\partial M$, the orientations $\mathcal{O}_v$ and $\mathcal{O}_w$ agree on every ordered basis, hence are equal as orientations of $T_p\partial M$. This completes the proof that the induced boundary orientation at $p$ is independent of the choice of outward-pointing transverse vector. [/step]