Exercise 5.4

As in exercise 5.3, we will write the \(i\)-th component of a vector \(\boldsymbol{x}\) as \(x_i\).

For the first identity:

\[\begin{split}\boldsymbol{x} \cdot (\boldsymbol{y} + \boldsymbol{z}) &= \begin{bmatrix}x_1\\x_2\end{bmatrix} \cdot (\begin{bmatrix}y_1\\y_2\end{bmatrix} + \begin{bmatrix}z_1\\z_2\end{bmatrix}) \\ &= \begin{bmatrix}x_1\\x_2\end{bmatrix} \cdot \begin{bmatrix}y_1+z_1\\y_2+z_2\end{bmatrix} \\ &= x_1(y_1 + z_1) + x_2(y_2 + z_2) \\ &= x_1y_1 + x_1z_1 + x_2y_2 + x_2z_2 \\ &= (x_1y_1 + x_2y_2) + (x_1z_1 + x_2z_2) \\ &= (\begin{bmatrix}x_1\\x_2\end{bmatrix} \cdot \begin{bmatrix}y_1\\y_2\end{bmatrix}) + (\begin{bmatrix}x_1\\x_2\end{bmatrix} \cdot \begin{bmatrix}z_1\\z_2\end{bmatrix}) \\ &= \boldsymbol{x} \cdot \boldsymbol{y} + \boldsymbol{x} \cdot \boldsymbol{z}\end{split}\]

For the second:

\[\begin{split}(k_1 \boldsymbol{x}) \cdot (k_2 \boldsymbol{y}) &= \begin{bmatrix}k_1x_1\\k_1x_2\end{bmatrix} \cdot \begin{bmatrix}k_2y_1\\k_2y_2\end{bmatrix} \\ &= (k_1x_1)(k_2y_1) + (k_1x_2)(k_2y_2) \\ &= k_1 k_2 (x_1y_1 + x_2y_2) \\ &= k_1 k_2 (\boldsymbol{x} \cdot \boldsymbol{y})\end{split}\]