Solutions to Assignment #4

Reminder: Midterm is in Haines A18.

Errata: In Handout J, J-8(b), the card numbers are A, 2, 3, 4, not A, 1, 2, 3. The reference should be to G-8, not G-9.

p. I 9, solution to E-5: line 3 should say $t _ 1 = t _ 2$.

p. L 1, solution to p. 15, Ex. 2: line 4 should say $1-i/((1+i)(1-i)) = (1-i)/2 = \frac 12 - \frac 12 i$.

Solutions:

First, a comment on a situation that often arises.

Recall that an implication $P \Rightarrow Q$ is equivalent to the ``contrapositive'' form not$\;Q \Rightarrow$   not$\;P$, in which $P$ and $Q$ are negated and switched. The contrapositive of the contrapositive is the original implication. Notice that the contrapositive is equivalent to the original; it is not the same as the opposite implication $Q \Rightarrow P$ (the ``converse'').

Often a statement to be proved will have the form $a \neq b \Rightarrow c \neq d$. It is often simpler and clearer to prove the contrapositive: $c = d \Rightarrow a = b$.

For example, if you are asked to prove that a function $f$ is one-to-one, which says $x _ 1 \neq x _ 2 \Rightarrow f(x _ 1) \neq f(x _ 2)$, it is often better to prove the contrapositive form $f(x _ 1) = f(x _ 2) \Rightarrow x _ 1 = x _ 2$.

Other examples occur in the solutions below to problems about Latin squares.

Using the contrapositive is so frequent that in simple situations you can use it without calling attention to it.

The contrapositive can also be viewed as a version of proof by contradiction.

p. 66, Ex. 6. The row-reduced echelon form is $\left[\begin{array}{rrrrr}1&7&0&3&0\\ 0&0&1&5&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{array}\right]$.

(a) A basis for $V$ consists of the nonzero rows of the row-reduced echelon form. As the text does, let's call the three nonzero rows $\rho _ 1, \rho _ 2, \rho _ 3$ (rho's).

(b) The elements of $V$ are vectors of the form $c _ 1 \rho _ 1 + c _ 2 \rho _ 2 + c _ 3 \rho _ 3$.

(c) For $(x _ 1,\ldots, x _ 5) \in V$, using the fact that we can see a pattern like the identity matrix in the first, third, and fifth entries of $\rho _ 1, \rho _ 2, \rho _ 3$ when compared, we see that the coefficients needed are $c _ 1 = x _ 1, c _ 2 = x _ 3, c _ 3 = x _ 5$.

Ex. 7. Let's write $[A\vert Y]$ for the augmented matrix. We know that row-reducing the augmented matrix doesn't change the solutions, so we can work with its row-reduced echelon form $[E\vert Z]$. There are only two possibilities: (i) The last row of $[E\vert Z]$ looks like $(0,\ldots, 0,1)$. In that case there is an equation $0 = 1$ and the system (row-reduced or not) cannot be solved. (ii) The rows of $[E\vert Z]$ all look like $(0,\ldots, 0,1,\ldots, z _ i)$ (perhaps with no 0's to start but at least with the 1). In this case we can solve for the pivot variables in terms of non-pivot variables and constants, and there are solutions.

p. 73, Ex. 1. (b) and (e) are linear transformations, since each output coordinate is a linear combination of input coordinate values.

(a), (c), and (e) all fail the condition that $T(rv) = rT(v)$, for example if $r = 3$ and $x _ 1$ is any nonzero value.

Ex's 2 and 3 are not on Midterm #1, although you do need to know the similar concepts for matrices. (For matrices we didn't talk about the range, but rather the column space, which turns out to be the same thing.) These problems are based on reading, to prepare for the discussion in class.

Ex. 2. Let $n =$   dim$V$. For the zero transformation, the range is $\{$0$\}$, the rank is 0, the null space is $V$, and the nullity is $n$. For the identity transformation, the range is $V$, the rank is $n$, the null space is $\{$0$\}$, and the nullity is 0.

Ex. 3. For Example 2: Let's stick to $F = \mathbb{R}$; I think the problem is harder than intended otherwise. Notice that $V$ is the space of all polynomial functions, not just polynomials up to some degree. The range is $V$ and the null space is the subspace of constant functions.

For Example 5: Notice that $T(f)$, or $Tf$, has the property that $(Tf)(0) = 0$ and also $Tf$ has a continuous derivative, as mentioned in the text. The range consists of all functions with these two properties. To see this, let $g$ be such a function and consider $h = T(g')$, which is given by $h(x) = \int _ 0 ^ x g'(t) dt = g(x) - g(0) = g(x)$, by a version of the fundamental theorem of calculus. Thus $g = T(g')$ and is in the range. The null space consists of all functions $f$ with $Tf = 0$, which says $\int _ 0 ^ x f(t) dt = 0$ for all $x$. Taking the derivative of both sides we get $f(x) = 0$, again by a version of the fundamental theorem of calculus. Therefore the null space consists only of the 0 function.