91Ó°ÊÓ

A rotating matrix rotates an object about a single axis of links, any ambiguous vector. The rotation matrix is much more complex than the measurement and translation matrix as the entire top left 3x3 matrix is required to produce complex rotations.

If A is a rotation-scaling matrix where the rotation is by the angle$\mathrm{φ}$ and the scaling is by a,

Then $A=\left[\begin{array}{cc}a\cos \mathrm{φ}& -\sin \mathrm{φ}\\ \sin \mathrm{φ}& a\cos \mathrm{φ}\end{array}\right]$.

To find eigenvalues of A, compute $\det (A-\mathrm{λ}I)=0$.

$\begin{array}{rcl}\left[\begin{array}{cc}a\cos \mathrm{φ}-\mathrm{λ}& -\sin \mathrm{φ}\\ \sin \mathrm{φ}& a\cos \mathrm{φ}-\mathrm{λ}\end{array}\right]& =& 0\\ (a\cos \mathrm{φ}-\mathrm{λ}{)}^2+{\sin }^2\mathrm{φ}& =& 0\end{array}$

Solve further as follows:

${\mathrm{λ}}^2-2a\mathrm{λ}\cos \mathrm{φ}+\left(a^2{\cos }^2\mathrm{φ}+{\sin }^2\mathrm{φ}\right)=0$

Now, solve the obtained quadratic equation as follows:

$\begin{array}{rcl}{\mathrm{λ}}_{1,2}& =& \frac{2a\cos \mathrm{φ}Â\pm \sqrt{4a^2{\cos }^2\mathrm{φ}-4\left(a^2{\cos }^2\mathrm{φ}+{\sin }^2\mathrm{φ}\right)}}{2}\\ & =& a\cos \mathrm{φ}Â\pm i\cos \mathrm{φ}\end{array}$

The two distinct complex eigenvalues on a 2x2 matrix so A is diagonalizable over $\mathrm{â„‚}$. Hence, the given statement is true.