2.6.2 So­lu­tion herm-b

Ques­tion:

A ma­trix $A$ is de­fined to con­vert any vec­tor ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x,y)$ into the vec­tor ${\skew0\vec r}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x+y,x+y)$. Ver­ify that $(\cos 45^\circ ,\sin 45^\circ)$ and $(\cos 45^\circ ,-\sin 45^\circ)$ are or­tho­nor­mal eigen­vec­tors of this ma­trix, with eigen­val­ues 2 re­spec­tively 0. Note: $\cos 45^\circ$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sin 45^\circ$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac 12\sqrt{2}$.

An­swer:

For ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\cos 45^\circ ,\sin 45^\circ)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\frac 12\sqrt{2},\frac 12\sqrt{2})$, $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac 12\sqrt{2}$ so ${\skew0\vec r}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\sqrt{2},\sqrt{2})$, and that is twice ${\skew0\vec r}$. For ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\cos 45^\circ ,-\sin 45^\circ)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\frac 12\sqrt{2},-\frac 12\sqrt{2})$, $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac 12\sqrt{2}$ so ${\skew0\vec r}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ (0,0), and that is zero times ${\skew0\vec r}$.

The square length of ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\cos 45^\circ ,\sin 45^\circ)$ is ${\skew0\vec r}\cdot{\skew0\vec r}$, which is given by the sum of the square com­po­nents: $\cos^245^\circ +\sin^245^\circ$. That is one, so the vec­tor is of length one. The same for ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(\cos 45^\circ ,-\sin 45^\circ)$. The dot prod­uct of $(\cos 45^\circ ,\sin 45^\circ)$ and $(\cos 45^\circ ,-\sin 45^\circ)$ is $\cos^245^\circ -\sin^245^\circ$. That is zero, be­cause $\cos 45^\circ$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sin 45^\circ$, so the two eigen­vec­tors are or­thog­o­nal.

In lin­ear al­ge­bra, you would write the re­la­tion­ship ${\skew0\vec r}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $A{\skew0\vec r}$ out as:

$$\left(\begin{array}{c}x_2\ y_2\end{array}\right) = \left(\b... ...ay}\right) = \left(\begin{array}{c}x+y\ x+y\end{array}\right)$$