자연계열 1번

1-1.$\displaystyle A=\begin{pmatrix}a_1&a_2\\a_3&a_4\end{pmatrix}$이면 $\displaystyle A^t=\begin{pmatrix}a_1&a_3\\a_2&a_4\end{pmatrix}$이므로 $\displaystyle AA^t=\begin{pmatrix}a_1^2+a_2^2&a_1a_3+a_2a_4\\a_1a_3+a_2a_4&a_3^2+a_4^2\end{pmatrix}$$\displaystyle \langle A,A\rangle=\operatorname{tr}(AA^t)=\operatorname{tr}\left(\begin{pmatrix}a_1^2+a_2^2&a_1a_3+a_2a_4\\a_1a_3+a_2a_4&a_3^2+a_4^2\end{pmatrix}\right)=a_1^2+a_2^2+a_3^2+a_4^2$이다.

1-2.$\displaystyle \langle A,B\rangle=\operatorname{tr}(AB^t)=\operatorname{tr}\left(\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\right)=\operatorname{tr}\left(\begin{pmatrix}-1&-1\\-1&-1\end{pmatrix}\right)=-2$이고,$\displaystyle \|A\|=\sqrt4=2$, $\displaystyle \|B\|=\sqrt2$이므로두 행렬 $\displaystyle A$와 $\displaystyle B$의 사잇각을 $\displaystyle \theta$라고 하면 $\displaystyle \cos\theta=\frac{\langle A,B\rangle}{\|A\|\|B\|}=\frac{-2}{2\sqrt2}=-\frac1{\sqrt2}$이다.

1-3.$\displaystyle \|A\|=2$이면 $\displaystyle \|A\|^2=a_1^2+a_2^2+a_3^2+a_4^2=4$이다.이를 만족하는 정수 $\displaystyle a_1,a_2,a_3,a_4$를 찾아보면,$\displaystyle a_1^2=a_2^2=a_3^2=a_4^2=1$즉, $\displaystyle a_1=\pm1,a_2=\pm1,a_3=\pm1,a_4=\pm1$인 경우의 수가 $\displaystyle 2^4=16$가지그리고$\displaystyle a_1^2=4$ ($\displaystyle a_2^2=a_3^2=a_4^2=0$), $\displaystyle a_2^2=4$ ($\displaystyle a_1^2=a_3^2=a_4^2=0$),$\displaystyle a_3^2=4$ ($\displaystyle a_1^2=a_2^2=a_4^2=0$), $\displaystyle a_4^2=4$ ($\displaystyle a_1^2=a_2^2=a_3^2=0$)인 경우,즉$\displaystyle (a_1=\pm2,a_2=a_3=a_4=0)$, $\displaystyle (a_2=\pm2,a_1=a_3=a_4=0)$,$\displaystyle (a_3=\pm2,a_1=a_2=a_4=0)$, $\displaystyle (a_4=\pm2,a_1=a_2=a_3=0)$,인 경우 8가지로모두 $\displaystyle 16+8=24$가지 있다.