I am studying some topics and I faced to the following statement:

[...] all tangent vectors are of this form, form the 1-dimensional vector space of real multiples of the matrix $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $

It may be a naive question, but: how can I interpret/visualize this matrix as a 1-dimension vector space (a line)?

Edit 1: I am working with $SO(2) \in \mathbb{S}^1$.

Edit 2: This matrix is supposed to represent the tangent line of the unit circle at $x=1$.

  • $\begingroup$ Identify $r\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ with $r$. Or $-r$ if you like. $\endgroup$ – Elliot G Jun 10 at 19:35
  • $\begingroup$ Hi @ElliotG , could you elaborate a bit more on your comment? I did not understand your point. Thanks. $\endgroup$ – Paulo Araujo Jun 10 at 19:56
  • $\begingroup$ It depends what your question really is. If the goal is to interpret multiples of the matrix $A=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ as a $1$-dimensional vector space, the answer is just that the vectors are of the form $rA=\begin{bmatrix} 0 & -r\\ r & 0 \end{bmatrix}$ for any $r\in\Bbb R$. You can check $rA+sA=(r+s)A$ and $r(sA)=(rs)A$, for example. That's all it really means to be a vector space. $\endgroup$ – Elliot G Jun 10 at 20:33


Consider the three dimensional space $\mathbb R^3$ and the vector $v=(1,0,-1)^T$. This vector defines a line containing all the vectors of coordinates $(t,0,-t)$ for $t \in \mathbb R$.

Back to your case

The given matrix (let’s call it $A$) can be used to define a line in a four dimensional space. The line contains all the matrices $tA$ for $t \in \mathbb R$.

  • $\begingroup$ thanks for the comment. The analogy I got, a vector from the origin to the mentioned point forms a line in the xz-plane. Taking the multiples of t will make me stay in that line. However, I am not understanding how I can use this analogy with the matrix in a 2D space. $\endgroup$ – Paulo Araujo Jun 10 at 19:54
  • 1
    $\begingroup$ The $2 \times 2$ real matrices are a four dimensional linear space. And in any dimensional linear space a non zero vector (here a matrix) spans a one dimensional linear subspace, i.e. a « line ». $\endgroup$ – mathcounterexamples.net Jun 10 at 19:58
  • $\begingroup$ That is still abstract to me. For example, this matrix is supposed to be the tangent line of the unit circle at x=1. The problem is that I couldn't visualize that. $\endgroup$ – Paulo Araujo Jun 10 at 20:25
  • $\begingroup$ If you have the unit circle $S^1$ in $\Bbb R^2$, the line tangent at $(1,0)$ can be visualized as the line $x=1$. Otherwise your question doesn't really make sense because a $2\times 2$ matrix is not an element of $\Bbb R^2$. Formally, though, the tangent space is just an abstract $1$-dimensional vector space, and nothing more. $\endgroup$ – Elliot G Jun 10 at 20:30
  • $\begingroup$ You are right. I must edit the question. I am working with $SO(2) \in \mathbb{S}^1$. If I calculate the tangent space at $t=0$, I end up with something similar to what you commented in the original post. I was trying to visualize how that matrix ends up being a straight vertical line. That was the main question, if there is a way, at least in 2D, to visualize how the matrix becomes the tangent line. $\endgroup$ – Paulo Araujo Jun 10 at 20:42

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