Matrix proof, linear algebra

Let $$R(\theta)=\begin{bmatrix} \cos\theta &-\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$

Also, $a$ and $b$ are real numbers. We suppose that $b\neq 0$ and we consider the matrix: $$A=\begin{bmatrix} a &-b \\b &a \end{bmatrix}$$

Show that it exist a unique number $\lambda>0$ and a unique number $\theta \in\ ]0,2\pi[$ such as $A=\lambda R(\theta)$.

I don't even understand the question. What is the number $\lambda$? How am I supposed to proceed here? Any help to point me in the right direction would help me a lot.

• What is $R( \theta)$ ? – Arpit Kansal Oct 15 '15 at 19:55
• Is A your second matrix? – Jerry Guern Oct 15 '15 at 20:06
• A is your matrix with a and b entries. You need to show that this is some number multiplied by R. – Paul Oct 15 '15 at 20:08
• yes, sorry about that – spexel Oct 15 '15 at 20:08

The question asks you to show this: given $b\neq 0$, and some $a$, there exist uniquely $\lambda>0$ and $\theta\in(0,2\pi)$ such that $$a=\lambda\cos\theta,\quad b=\lambda\sin\theta.$$ Squaring and summing give $$0<a^2+b^2=\lambda^2\cos^2\theta+\lambda^2\sin^2\theta=\lambda^2\implies\lambda=\sqrt{a^2+b^2}>0.$$ For $\theta$, there is either a unique $\theta_0\in(0,2\pi)$ such that $\cos(\theta_0)=a/\lambda$ (the case $a/\lambda=-1$) or there are $\theta_1<\theta_2$ also in $(0,2\pi)$ such that $\cos(\theta_1)=\cos(\theta_2)=a/\lambda$. In the first case, you are done: just set $\theta=\theta_0$. In the second case, either $b/\lambda>0$ in which case, pick $\theta=\theta_1$ or $b/\lambda<0$ in which case pick $\theta=\theta_2$. Throughout this short discussion about $\theta$, it's helpful if you draw the plots of $\sin$ and $\cos$ on $(0,2\pi)$.
• so $\lambda$ is just a variable and could be replaced by x? – spexel Oct 15 '15 at 20:20
• @spexel $\lambda$ is a number that depends on the values of $a$ and $b$, and you treat it as an unknown until you can figure out enough about the problem to understand how it depends on $a$ and $b$. – Aaron Oct 15 '15 at 20:23
• @spexel Yes, $\lambda$ is just a label for a quantity that is uniquely specified (or so the question claims) once you know $b\neq 0$ and some $a$. You can use a different label if you wish but there is very little reason to do so in this simple problem. I recommend that you stick to the labels used by the question to avoid unnecessary miscommunications. – yurnero Oct 15 '15 at 20:23
I think here, $λ^2=a^2+b^2$. The problem means it must be a scalar times a rotation matrix.