# Polar co-ordinates, Jordan form, Axler textbook

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: $$\theta = \text{tan}^{-1}\Big(\frac{a_{31}}{a_{32}}\Big)$$