2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.