# Is this set of matrices closed under matrix multiplication?

2x2 matrices of the form: \begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix} x is a real number

multiplying the matrix by itself produces \begin{bmatrix}\cos^2x-\sin^2x&-2\cos x\sin x\\2\sin x\cos x&\cos^2x-\sin^2x\end{bmatrix}

Is this product matrix still in the set? It seems not to be in my mind, but I'm not sure.

• Observe that $-2\cos x\sin x = -\sin(2x)$. (I expect that you already know this, right?) Is that the hint you need to go on by yourself? – MJD Sep 8 '17 at 20:19
• First, what trig identities do you know? Second, showing something is closed under multiplication involves multiplying two different things together, not multiplying something by itself. – Jason DeVito Sep 8 '17 at 20:19
• Notice that the matrix you got is $\begin{bmatrix}\cos(x+x)&-\sin(x+x)\\ \sin(x+x)&\cos(x+x)\end{bmatrix}$. A similar argument with two distinct matrices of the set will give the analog for $x+y$ and you can then conclude. – Prasun Biswas Sep 8 '17 at 20:20
• @JasonDeVito But if OP suspects that the claim might be false, then trying some simpler special cases to see if they work is a reasonable way to proceed. – MJD Sep 8 '17 at 20:20
• @MJD: Agreed! To the op: No, because $x$ denotes a fixed real number. For example, that reasoning can be used to prove he following (false) theorem: Any two numbers of the form $x^2$ are equal. – Jason DeVito Sep 8 '17 at 20:26

## 2 Answers

To check if $S:=\left\{\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}:x\in\Bbb R\right\}$ is closed under multiplication you would need to multiply $$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{bmatrix}$$ not $$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}$$ because the two matrices need not be the same. So we need to show $$\begin{bmatrix}\cos(x)&-\sin(x)\\\sin(x)&\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)&-\sin(y)\\\sin(y)&\cos(y)\end{bmatrix}=\begin{bmatrix}\cos(x)\cos(y)-\sin(x)\sin(y)&-\cos(x)\sin(y)-\sin(x)\cos(y)\\\sin(x)\cos(y)+\cos(x)\sin(y)&-\sin(x)\sin(y)+\cos(x)\cos(y)\end{bmatrix}$$ is in $S$. Recall the identities: $$(1)\qquad\sin(x\pm y)=\sin(x)\cos(y)\pm\sin(y)\cos(x)\\(2)\qquad \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)$$

• You answer the question of the OP, no doubt, but I am a little surprised that you do not mention the geometrical background (rotation matrices) which gives an immediate understanding of the question (see my answer). – Jean Marie Sep 8 '17 at 21:46
• @JeanMarie that is a fine answer (+1). I was just trying to address the OP's line of reasoning for showing the set is closed under multiplication (i.e. by directly showing $A,B\in S\implies AB\in S$) by showing him/her the mistake they made and how to correct it. – Dave Sep 8 '17 at 22:22
• Thanks, and yes you did it in a paedogical manner by spotting at once the error. I just reacted in this way because I see that geometric intuition is maybe the largest lack in contemporary mathematical education... – Jean Marie Sep 8 '17 at 22:27

There is a geometrical interpretation that sheds light on all this :

$$R_x=\begin{bmatrix}\cos x&-\sin x\\ \sin x& \ \ \cos x \end{bmatrix}$$

is a rotation matrix (https://en.wikipedia.org/wiki/Rotation_matrix).

Thus, what has been done in the answer given by @Dave is plainly

$$R_x \times R_y = R_{x+y}$$

Using sentences : Rotation with angle $y$ followed by rotation by angle $x$ is rotation by angle $x+y$.

This is why you have a closed set.

One can say more : the set of rotation matrices is a (commutative) group for multiplication.