2.6.1 So­lu­tion herm-a


A ma­trix $A$ is de­fined to con­vert any vec­tor ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x{\hat\imath}+y{\hat\jmath}$ into ${\skew0\vec r}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2x{\hat\imath}+4y{\hat\jmath}$. Ver­ify that ${\hat\imath}$ and ${\hat\jmath}$ are or­tho­nor­mal eigen­vec­tors of this ma­trix, with eigen­val­ues 2, re­spec­tively 4.


Take $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 to get that ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hat\imath}$ trans­forms into ${\skew0\vec r}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2{\hat\imath}$. There­fore ${\hat\imath}$ is an eigen­vec­tor, and the eigen­value is 2. The same way, take $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 to get that ${\hat\jmath}$ trans­forms into $4{\hat\jmath}$, so ${\hat\jmath}$ is an eigen­vec­tor with eigen­value 4. The vec­tors ${\hat\imath}$ and ${\hat\jmath}$ are also or­thog­o­nal and of length 1, so they are or­tho­nor­mal.

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...
...rray}\right) = \left(\begin{array}{c}2x\ 4y\end{array}\right)

In short, vec­tors are rep­re­sented by columns of num­bers and ma­tri­ces by square ta­bles of num­bers.