0. Diagonal matrices are easy

As you have seen in the past, diagonal matrices are especially easy. They represent problems in which the variables work separately. For example, let $D = \matp{rr}{5&0\\ 0&3}$. Here are some ways in which $D$ is easy:

• $D$ is the coefficient matrix of linear equations such as $\left \{ \begin{array}{llll} 5x && = & 8\\ &3y& =&7 \end{array} \right .$, easily solved separately.

• For x$= \matp{c}{x\\ y}$, we have $\tau _ D($x$) = \matp{rr}{5&0\\ 0&3}\matp{c}{x\\ y}= \matp{c}{5x\\ 3y}$, a nonuniform scaling in which $x$ and $y$ are multiplied by separate constants.

• $D ^ 2 = \matp{rr}{25&0\\ 0&9}$, obtained by treating each diagonal entry separately.

Figure shows what $\tau _ D$ does to a unit square and vectors along the $x$- and $y$-axes.

Figure: A diagonal transformation

udir/diagonal.eps

Problem U-1. What, specifically, does $\tau _ D$ do to vectors along the $x$-axis? Along the $y$-axis? How about a vector pointing left along the $x$-axis?