I am trying to solve this Linear algebra question but I am unsure on how to proceed and get stuck.
Define a three-dimensional ``Givens rotation'' in the 1-2 plane by $$M := \left( \begin{array}{ccc} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{array} \right). $$ Given an arbitrary $3 \times 3$ matrix $A$, find the angle $\theta$ that will produce a 0 in the $(3,1)$ entry of $M^{-1}AM$.
Solving I get the equation $a cos \theta + b sin \theta =0$ and I am unable to fix a $\theta$ after using polar coordinates for a and b. Any and all help is welcome.
Well, if
$$A := \left( \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right). $$
And combining this with the fact that we have that
$$M^{-1} := \left( \begin{array}{ccc} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{array} \right). $$
Then we only need to be concerned with the $(3,1)$ entry of $AM$, which works out to be
$$a_{31} \text{cos} (\theta) - a_{32} \text{sin}(\theta) $$
Setting this to zero, we see that: $$\theta = \text{tan}^{-1}\Big(\frac{a_{31}}{a_{32}}\Big)$$