ProblemPage 59 # 2:
Give an example of a function \(f:\mathbf{R}^2\to\mathbf{R}\) such that
\[
f(a v) = af(v)
\]
for all \(a\in\mathbf{R}\) and all \(v\in\mathbf{R}^2\), but \(f\) is not
linear.
Solution:
An interesting observation is that if \(f\) has a Taylor series
expansion around \((0,0)\), then \(f\) has to be linear. (You might
try proving this.) Anyway, it shows that we need to look for a
function whose derivative fails to exist.
Someone suggested the function \(f(x,y)=\sqrt{xy}\). This almost works,
but it has a couple of problems. First, it's value isn't in \(\mathbf{R}\)
if \(xy<0\). Second, it satisfies
\[
f(ax,ay) = \sqrt{a^2xy} = |a|f(x,y).
\]
However, this example suggests an approach. The problem with taking square
roots is that \(\sqrt{x}\) is only defined if \(x\ge0\) and the values
all satisfy \(\sqrt{x}\ge0\). But if we take cube roots, then \(\sqrt[3]{x}\)
is defined for all real values of \(x\), and the cube root of \(x^3\)
is always equal to \(x\). So here are some functions that
have the desired property:
\[
f_1(x,y) = \sqrt[3]{x^3+y^3},\qquad
f_2(x,y) = \sqrt[3]{xy^2},\qquad
f_3(x,y) = \sqrt[7]{x^7 + xy^6}.
\]
And you can make up many others of a similar
nature. Let's check that \(f_1\) is not linear. We have
\[
f_1\bigl((1,0)+(0,1)\bigr) = f(1,1) = \sqrt[3]{2}
\quad\text{and}\quad
f_1(1,0)+f_1(0,1) = \sqrt[3]{1}+\sqrt[3]{1}=2.
\]
One can also create weirder sorts of non-linear functions that have the
desired property. Here's an example:
\[
f(x,y) = \begin{cases}
x &\text{if \(y=0\) or if \(x/y\) is a rational number,} \\
0 &\text{if \(y\ne0\) and \(x/y\) is an irrational number,} \\
\end{cases}
\]
I'll leave you to check that \(f(ax,ay)=f(x,y)\). To see that \(f\) is not
linear, we use the fact that \(\pi=3.14159\ldots\) is not rational. Then
\[
f(\pi,1) + f(1-\pi,1) = 0 + 0 = 0
\quad\text{and}\quad
f\bigl((\pi,1)+(1-\pi,1)\bigr) = f(1,1) = 1.
\]
Pretty weird!
Problem Page 59 # 4:
Suppose that \(T\) is a linear map from \(V\) to \(\mathbf{F}\). Prove
that if \(u\in V\) is not in null\((T)\), then
\[
V = \text{null}(T) \oplus \{au : a\in\mathbf{F}\}.
\]
Solution:
The fact that \(T\) is a map from \(V\) to \(\mathbf{F}\) means
that the values of \(T\) are in \(\mathbf{F}\), i.e., they are scalars.
So \(T(u)\in\mathbf{F}\) is a scalar, and since we are given
that \(u\) is not in null\((T)\), that scalar is not \(0\). To emphasize
that it is a scalar, we'll call it \(b\). In other words,
\[
b = T(u) \in \mathbf{F}\quad\text{and}\quad b \ne 0.
\]
Now let \(v\in V\) be any vector. We want to show that \(v\) is the sum
of a vector in null\((T)\) and a vector in \(\{au:a\in\mathbf{F}\}\). The
idea is to try to subtract a multiple of \(u\) from \(v\) to get something
that's in null\((T)\). So we look for a scalar \(t\) so that
\[
T(v-tu) = 0.
\]
By linearity, this is
\[
T(v) - tT(u) = 0.
\]
But remember that \(T(u)=b\) is a non-zero scalar, i.e., \(b\) is a non-zero
number, so it has a multiplicative inverse \(b^{-1}\). So we should
set \(t=b^{-1}T(v)\), where remember that \(T(v)\) is also a scalar.
To emphasize that \(T(v)\) is a scalar, let's call it \(c\), so \(c=T(v)\).
With this notation, we take \(t=b^{-1}c\).
This means that if we write \(v\) as
\[
v = (v - b^{-1}cu) + b^{-1}cu,
\]
then the vector in parentheses is in null\((T)\), since
\[
T(v-b^{-1}cu) = T(v) - b^{-1}cT(u) = c - b^{-1}cb = 0,
\]
while the vector \(b^{-1}cu\) is clearly in \(\{au:a\in\mathbf{F}\}\), since
it is a scalar multiple of \(u\). This proves that every vector
in \(V\) is a sum of a vector in null\((T)\) and a vector in
\(\{au:a\in\mathbf{F}\}\), so this proves that
\[
V = \text{null}(T) + \{au:a\in\mathbf{F}\}.
\]
Finally, to show that it is a direct sum, we need to show that
\[
\text{null}(T) \cap \{au:a\in\mathbf{F}\} = \{0\}.
\]
So let \(v\) be a vector in the intersection. Then \(v=au\) for some
scalar \(a\). But also \(v\) is in null\((T)\), so
\[
0 = T(v) = T(au) = aT(u).
\]
We are given that \(T(u)\ne0\), so we conclude that \(a=0\), and hence
that \(v=au=0\).
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