3

$$F(x)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix}$$

Is it very obvious (in the sense that without any calculations) that $[F(x)]^{-1}=F(-x)$?

My book directly writes this without any explanation. How is this evident without calculation?

Soham
  • 10,120
  • 2
    If I am not mistaken this matrix describes a rotation around a certain angle $x$. The inversion of this would be indeed the rotation around the angle $-x$ which overall makes it pretty clear that $F^{-1}(x)=F(-x)$. – mrtaurho Feb 13 '19 at 19:38
  • @mrtaurho Actually its not in rotation context (Maybe it implies rotation, I don't know). Also, I am a beginner. Don't know much about rotation matrices ;-/ – Soham Feb 13 '19 at 19:40
  • 2
    see here maybe? https://math.stackexchange.com/q/716578/515527 and here for some intuition + explanation https://math.stackexchange.com/q/363652/515527 – Zacky Feb 13 '19 at 19:41
  • 1
    One way to assert this is by noticing that $F(x)^TF(x)=I$ for any $x$, so the inverse of $F$ is the transpose of $F$ (meaning that it's orthogonal). The matrix $F$ satisfies $F(x)^T=F(-x)$, using the symmetries of sine and cosine. This proves the claim. – Josh B. Feb 13 '19 at 19:42
  • 1
    @JoshB. Interesting approach. How could one notice that $F^T(x)F(x)=I$ without computing the product? Is it sufficient to deduce this relation yet alone from the fact that $|F(x)|=|F^T(x)|=1$$($which I would claim is directly viewable from the matrix itself$)$? – mrtaurho Feb 13 '19 at 19:47
  • 1
    I highly recommend watching this youtube playlist: https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab , chapters 3, 4, and 5 are of high relevance to this question. The presentation of the geometric intuition behind central concepts in linear algebra is excellent for someone new to the subject, in my opinion. –  Feb 13 '19 at 19:57
  • 1
    @mrtaurho To be honest, I'm not really sure how to notice this in general, since I mainly knew to do this from experience. Part of it is the fact that each column has a magnitude of $1$, which may ask the question "is this matrix orthogonal?" As you said, since the matrix norm for both $F$ and $F^T$ are both $1$, this also implies that the inverse could be the transpose, but doesn't guarantee it. – Josh B. Feb 14 '19 at 20:04

2 Answers2

3

Yes, it is obvious if you know multiplication by blocks and the matrix of a rotation (of angle $x$) in the plane:

$$R(x)=\begin{bmatrix}\cos x&-\sin x\\\sin x & \cos x\end{bmatrix}, \qquad R(x)^{-1}=R(-x)=\begin{bmatrix}\cos x&\sin x\\-\sin x & \cos x\end{bmatrix}.$$ Now, consider your matrix as a block matrix: $$F(x)=\begin{bmatrix}R(x)&\mathbf 0 \\ \mathbf 0 & 1\end{bmatrix},$$ where $\mathbf 0$ denotes the null $2{\times} 1$ column vector in the first row and the null $1{\times}2$ row vector in the second row . Using block multiplication, we see at once that this matrix is invertible and its inverse is $$F(x)^{-1}=\begin{bmatrix}R(x)^{-1}&\mathbf 0 \\ \mathbf 0 & 1\end{bmatrix}=\begin{bmatrix}R(-x)&\mathbf 0 \\ \mathbf 0 & 1\end{bmatrix}. $$

Bernard
  • 179,256
2

This matrix $F(\theta)$ represents a linear transformation that fixes the $z$ axis and rotates the $xy$-plane by an angle of $\theta$ around the origin. The inverse mapping is of course therefore fixing $z$ and rotating the $xy$-plane by an angle of $-\theta$, which is precisely $F(-\theta)$!

YiFan Tey
  • 17,740
  • 4
  • 30
  • 73