![]() ![]() As it is also non-empty, by connectedness of $V$ we have that im $(f)$ is the whole of $V$. Then by continuity, $f(X)=Y$, so im $(f)$ is closed.īy invariance of domain we also know that im $(f)$ is open. Then we have a sequence of vectors $X_i\in V$ with $f(X_i)\to Y$ and $f(X_i)$ Cauchy, so $X_i$ is Cauchy. ![]() Let $Y$ be an accumulation point of im $(f)$. Then $f$ is clearly injective and continuous, and im $(f)$ is non-empty. Suppose that $f$ is a rigid motion of $V$. Let $V$ be a finite dimensional real vector space with Euclidean norm. The cases $\lambda\leq0$ and $\lambda\in (0,1)$ follow analogously. An affine transformation is any transformation that preserves collinearity (i.e., all points lying on a line initially still lie on a line after transformation) and ratios of distances (e.g., the midpoint of a line segment remains the midpoint after transformation). Then $Y=\lambda g(X)$ is the unique vector satisfying $||Y||=\lambda||X||$ and $g(X)$ lies on the line segment $OY$.Īs $X$ lies on the line segment $O(\lambda X)$, we know $g(X)$ lies on the line segment $O(g(\lambda X))$. Notice that these segments are parallel, since they are perpendicular to the same line. The reflection line, m, is the perpendicular bisector of the segments joining each point to its image. The vectors $Z$ in the line segment $XY$ are characterized by the condition $||X-Y|| =||X-Z|| ||Z-Y||$, so $g$ preserves line segments. A reflection is called a rigid transformation or isometry because the image is the same size and shape as the pre-image. Bijectivity of an arbitrary rigid motion follows. Suppose that $g$ is a rigid motion of $V$, fixing the origin $O$. Solution $2$ (Geometric): Let $V$ be a finite dimensional real vector space with Euclidean norm. This video describes the difference between rigid motion and non-rigid motion, a concept you learn about in a high school Geometry course. In my view the one by is the best of these 3, as it is the most succinct and self-contained, but the others are interesting in their own right. Variants of this question have been asked so many times I expect someone has also posted the geometric solution too, at some point. Credit for the topological solution to for the comment here. However it uses invariance of domain instead, which is actually less trivial to prove than linearity. The topological one is noteworthy as it takes a different route to most solutions, by not showing that any maps are linear. ![]() Here are two alternative arguments: one geometric and one topological. If $\ g(x)=f(x)-f(0)\ $ then $\ g\ $ must be an orthogonal linear transformation. ![]()
0 Comments
Leave a Reply. |