Solutions to Assignment #7

For p. 95, Ex. 2:

Notice that $T$ is $\tau _ A$ for $A = \left[\begin{array}{rrr}1&1&0\\ -1&0&2\end{array}\right]$.

(a) The matrix is $A$.

(b) We must find coefficients with

$T(\alpha _ 1) = p \beta _ 1 + q \beta _ 2$

$T(\alpha _ 2) = r \beta _ 1 + s \beta _ 2$

$T(\alpha _ 3) = t \beta _ 1 + u \beta _ 2$

Details:

$T(\alpha _ 1) = A \alpha _ 1 = \left[\begin{array}{rrr}1&1&0\\ -1&0&2\end{arr... ...}{r}0\\ 1\end{array}\right] + q \left[\begin{array}{r}1\\ 0\end{array}\right]$, so $p = -3$, $q = 1$.

$T(\alpha _ 2) = A \alpha _ 2 = \left[\begin{array}{rrr}1&1&0\\ -1&0&2\end{arr... ...}{r}0\\ 1\end{array}\right] + s \left[\begin{array}{r}1\\ 0\end{array}\right]$, so $r = 1$, $s = 2$.

$T(\alpha _ 3) = A \alpha _ 3 = \left[\begin{array}{rrr}1&1&0\\ -1&0&2\end{arr... ...}{r}0\\ 1\end{array}\right] + u \left[\begin{array}{r}1\\ 0\end{array}\right]$, so $t = -1$, $u = 1$.

Therefore the matrix of $T$ relative to these bases is $\left[\begin{array}{rrr}-3&1&-1\\ 1&2&1\end{array}\right]$.

For p. 95, Ex. 3:

This says $T = \tau _ A$. The column space $W$ is the range of $T$.

For p. 95, Ex. 5.

(Outline) This says $T = \tau _ A$. Row-reduce to RREF. The columns of $A$ corresponding to the pivot columns are a basis for the range (the column space). You also know how to find a basis for the null space.