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I need to prove existence and uniqueness of two factors for matrix operations but I don't know at all what I should do, here is the exercise:

$$ R(\theta)=\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} $$ I have already proved that $R(\theta)^n=R(n\theta)$. Now I need to prove that

For $a,b\in\mathbb{R}$ with $b\ne0$ and $$A=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$ there exist only one $\lambda>0$ and only one $\theta\in\mathopen{]}0,2\pi\mathclose{[}$ such that $$ A=\lambda R(\theta).$$

How can I prove it?

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    FYI, *uniqueness, not unicity – 2017-01-01
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    @parsiad Thanks I corrected it! – 2017-01-01
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    @parsiad that's a cursive n – 2017-01-01
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    Hint: matrices of this form are isomorphic to complex numbers. – 2017-01-01
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    @parsaid in fact, it seems that [both are technically correct](http://english.stackexchange.com/a/225098/46438) (or at least, both are in the OED). However, "uniqueness" is much more common among native English speakers and is more likely to be understood. – 2017-01-02

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Suppose $\lambda$ and $\theta$ exist. Then $A=\lambda R(\theta)$, so $\det A=\det(\lambda R(\theta))$ and so $$ a^2+b^2=\lambda^2 $$ Thus $\lambda=\sqrt{a^2+b^2}$. Now your task is to see that there exists a unique $\theta\in (0,2\pi)$ with $$ \cos\theta=\frac{a}{\sqrt{a^2+b^2}},\qquad\sin\theta=\frac{b}{\sqrt{a^2+b^2}} $$ Hint: The point with coordinates $\biggl(\dfrac{a}{\sqrt{a^2+b^2}},\dfrac{b}{\sqrt{a^2+b^2}}\biggr)$ belongs to the circle with radius $1$ and center the origin.

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    Thank you so much for your answer, I can't understand why lambda² = a² + b² ? – 2017-01-01
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    @TrevörAnneDenise calculate the determinant of each of $A$ and $\lambda R(\theta)$ – 2017-01-01
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    @John11 oh thank you, I understand for this part now but I can't understand the rest, I have difficulties to see how proving that will prove the original problem ? I don't know how to prove it either – 2017-01-01
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    @TrevörAnneDenise $\mathrm{\LaTeX}$ specialist here. `;-)` I added a further hint. – 2017-01-01
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    Ah ah thank you !! I am working on it – 2017-01-01
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    We can discard $\theta=\pi$ because $b=\lambda \sin \pi=0$. – 2017-01-01
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    Since it's on that circle and that we are using sin and cos and cos and thêta is between 0 and 2Pi, I do understand that once we will have proven that it exists, it will be unique, but Don't know how to prove that I exists or why we are trying to prove that in the first place... – 2017-01-01
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    @TrevörAnneDenise Connect the point on the circle with the origin: $\theta$ is the angle between this line and the positive $x$-semiaxis. Definition of cosine and sine. – 2017-01-01
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    @egreg I am starting to understand it a little bit better, especially why we are proving it this way (as always it looks soo obvious once I realize it…), so actually, now that we have proven that lambda is equal to sqrt(a² + b²) we only care about proving the existence and uniqueness of thêta right ? – 2017-01-01
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    Oh wait, belong to circle means that it is on the "circle line" and not inside of the circle right ? – 2017-01-01
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    @TrevörAnneDenise Right so! – 2017-01-01
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    @TrevörAnneDenise You might be interested in knowing that this is basically the expression of a complex number in trigonometric form. `;-)` – 2017-01-02