# Polar Decomposition of 2x2 Matrix

I have the following homework problem and I just don't know how to go about starting it. Is it asking me to find a unique value of ϕ? I just can't see any other solution apart from when ϕ = θ. So my attempt to solve was to start by looking at A of the first type. I graphed out the standard cosine and sine functions to see if there existed two values such that Sinθ = Sinϕ and Cosθ = Cosϕ. However, graphically no such angles existed, thus I figured that the first type of A can only be true when ϕ = θ. And similarly for the type 2 case, it looks to be only true when ϕ = -θ

Let $$A$$ be a $$2\times2$$ matrix with real entries having one of the following forms: $$A = \left[ \begin{array}{cc} r \cos{\theta} & -r\sin{\theta}\\ r\sin{\theta} & r\cos{\theta} \end{array} \right] \qquad \text{or} \qquad A = \left[ \begin{array}{cc} r \cos{\theta} & r\sin{\theta}\\ r\sin{\theta} & -r\cos{\theta} \end{array} \right]$$ for some $$r>0$$ and $$0 \leq \theta < 2 \pi$$. Prove that for some $$0 \leq \varphi < 2\pi$$ we can write $$A = r R_{\varphi} \qquad \text{or} \qquad A = r R R_{\varphi}$$ where $$R = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right]$$ is a reflection and $$R_{\varphi} = \left[ \begin{array}{cc} \cos{\varphi} & -\sin{\varphi} \\ \sin{\varphi} & \cos{\varphi} \end{array} \right]$$ is a rotation. Note it is not necessarily the case that $$\varphi = \theta$$.

• It seems to be just asking you to prove such a $\varphi$ exists, not necessarily to determine what it is. When you write "I just can't see any other solution apart from ...", what sort of work have you done to check this? Please provide it here, including anything you are stuck on or unsure of. Thanks. – John Omielan Mar 12 at 7:41
• So my attempt to solve was to start by looking at A of the first type. I graphed out the standard cosine and sine functions to see if there existed two values such that Sinθ = Sinϕ and Cosθ = Cosϕ. However, graphically no such angles existed, thus I figured that the first type of A can only be true when ϕ = θ. And similarly for the type 2 case, it looks to be only true when ϕ = -θ. – GilNorth Mar 12 at 8:20
• Thanks for the info. As many people don't read the comments, or at least not all of them, I suggest you add what you wrote above to your question text. Thanks. – John Omielan Mar 12 at 8:24
• Have you tried something simple like solving $A=rRR_\phi$ for $R_\phi$ ($R$ is obviously invertible)? Alternatively, think about what these formulas mean geometrically: $rR_\phi$ is a scaled rotation, while $rRR_\phi$ is a composition of scaling, reflection and rotation. – amd Mar 12 at 19:14

If $$A$$ is of the first type you can just choose $$\phi=\theta$$. In the second case, just take $$\phi=-\theta$$.