Problems in Tutorial Note
Problem 1. Done, but might be hard, let's explain once again here.
$C[0,1]$ is a vector space, as mentioned earlier, simply because addition of continuous functions and scalar multiple of a continuous function are continuous (skipping the subtle detail which is merely messy).
$Tf$ is defined pointwise in this way: $(Tf)(x)=\int_0^xf(t)\,dt$. We may also write $Tf(x)$, with the understanding that $Tf$ is a function. Actually we seldom write $T(f)$ for linear transformations, unless the "input" is a bit complicated.
Now $T$ is linear by definition. We need to find the solution of $Tf=0$, i.e., find \[
V=\{f\in C[0,1]:Tf=0\}.
\]
To obtain set equality, we need the following usual logical step for naive set theory: \[
A\subseteq B\iff (x\in A\implies x\in B)
\] In general to show two sets $A,B$ are equal, we need to show $A\subseteq B$ and $B\subseteq A$, both can be verified by the logic above.
Now suppose $g\in V$, let's find out what can $g$ be. To explain this we need:
The Fundamental Theorem of Calculus. Let $f:[a,b]\to \R$ be continuous at $x_0\in (a,b)$, then the map $F(x)=\int_a^xf(t)\,dt$ is differentiable at $x_0$, moreover, \[
F'(x_0)=f(x_0).
\]
By definition $Tg=0$, i.e., $Tg$ is a zero function on $[0,1]$, so for each $x\in [0,1]$, \[
0=Tg(x)=\int_0^x g(t)\,dt.
\] Since $g$ is continuous on $(0,1)$, for each $x_0\in (0,1)$, by fundamental theorem of calculus one has \[
\left.\frac{d}{dx} 0 \right|_{x=x_0}=\left.\frac{d}{dx}\int_0^xg(t)\,dt\right|_{x=x_0}= g(x_0),
\] so $g(x_0)=0$, for each $x_0\in (0,1)$. Now it remains to compute $g(0)$ and $g(1)$ in order to identify the function $g$. But by continuity one has \[
g(0)=\lim_{x\to 0^+}g(x)=0,\quad g(1)=\lim_{x\to 1^-}g(x)=0,
\] so $g(x)=0$ for each $x\in [0,1]$, i.e., $g=0$, $g$ is the zero element in the vector space $C[0,1]$.
Our logic says that $g\in V\implies g\in \{0\}$, so $V\subseteq \{0\}$. But if $g\in \{0\}$, then $Tg(x)=\int_0^x g(t)\,dt=\int_0^x 0\,dt=0$, so $Tg=0$, thus $g\in V$. This logic says that \[
\{0\}\subseteq V,
\]so we have $\{0\}\subseteq V\subseteq \{0\}$, i.e., $V=\{0\}$.
Problem 2. Since \[
A=AI=A\matrixx{1&0\\0&1}=\matrixx{A\matrixx{1\\0}&A\matrixx{0\\1}}.
\] So to find $A$, it is enough to find $A\matrixx{1\\0}$ and $A\matrixx{0\\1}$.
(a) This is a standard example, $R_\theta=\matrixx{\cos\theta&-\sin \theta\\ \sin\theta&\cos\theta}$.
(b) Reflecting along $y=x$, we have $(1,0)^T\mapsto (0,1)^T$ and $(0,1)^T\mapsto (1,0)^T$, so \[
A=\matrixx{0&1\\ 1&0}.
\]
(c) The matrix is $R_{\alpha-\pi/4} \matrixx{0&1\\ 1&0}R_{\pi/4-\alpha}=\matrixx{\cos2\alpha &\sin 2\alpha\\ \sin2\alpha&-\cos2\alpha}$.
Problem 3. Done.
Problem 4. Done, might be complicated, let's explain again.
Answers. (a) B; (b) C; (c) C
Reason. (a) We prove by contradiction, suppose $T$ is 1-1. Let $A$ be the standard $m\times n$ matrix of $T$, then $T$ is 1-1 if and only if $A$ is 1-1 as a map. However, \begin{align*}
\text{$A$ is 1-1}&\iff (Ax=0\implies x=0)\\
&\iff\text{columns of $A$ are linearly independent}.
\end{align*} Let $A=\matrixx{a_1&\cdots &a_n}$, where $a_i\in \R^m$. In tutorial note 2 we have shown that every set of $m+1$ vectors in $\R^m$ must be linearly dependent. From that, it is also not hard to show that every set of $k\ge m+1$ vectors must be linearly dependent in $\R^m$. Hence, $\{a_1,\dots,a_n\}\subseteq \R^m$ being linearly independent must have cardinality $n\leq m$. But $m<n$ by assumption of (a), we have \[
n\leq m<n,
\] this is absurd. What's wrong with that? We arrive this absurd by assuming that $T$ is 1-1, that means this assumption is wrong, namely, $T$ is not 1-1.
Remark. If we have the concept: dimension, then we can provide a much simpler proof to that $T:\R^n\to \R^m$ with $n>m$ can never be injective. We revisit this example later.
Remark. We can imitate the proof in tutorial note 2 to show that every set of $k\ge m+1$ vectors, say $\{v_1,v_2,\dots,v_k\}$, in $\R^m$ is linearly dependent, i.e., the system $\matrixx{v_1&v_2&\cdots&v_k}x=0$ has nontrivial solution, by showing that there must be at least one free variables.
(b) Consider $T:\R^n\to \R^n$ defined by $Tv=v$ for all $v\in \R^n$, this is injective. But $S:\R^n\to \R^n$ defined by $Sv=0$ for all $v$ is not injective, so no conclusion can be drawn.
(c) The map $T(x_1,\dots,x_n)^T:=(x_1,\dots,x_n,0,\dots,0)^T$ is injective, but $T=0$ is not injective.
Problem 5. Done.
Conceptual Detail
In this tutorial we need the concepts from the study of functions, possibly taught in calculus course already. Let's specialize to study those we need in linear algebra.
Definition. (i) A function $f:X\to Y$ is said to be injective or one-one or 1-1 if \[Remark. For $f:X\to Y$, $X$ is called domain and $Y$ is called codomain. The set $f(X):=\{f(x):x\in X\}$ is called the range of $f$.
x\neq y\implies f(x)\neq f(y).
\] Equivalently, in terms of logic the above definition is the same as \[
f(x)=f(y)\implies x=y.
\] The second formulation is usually (well, possibly always!) the one we use in mathematics.
(ii) A function $f:X\to Y$ is said to be surjective or onto if for every $y\in Y$, there is $x\in X$ such that $f(x)=y$.
Proposition. Given an $m\times n$ real matrix $A$, the following are equivalent.Proof. To prove three statements are equivalent, we use the usual strategy that \[
- $A$ is injective as a map.
- $Ax=0\implies x=0$
- $\nul A:=\{x\in \R^n:Ax=0\}=\{0\}$.
1.\implies 2.\implies 3.\implies 1.
\] ($1.\Rightarrow 2.$) Suppose $A$ is injective as a map. Let $Ax=0$, we need to show $x=0$. But $A0=0$, so $Ax=0=A0$, by injectivity, $x=0$, so we are done.
($2.\Rightarrow 3.$) Suppose $Ax=0\implies x=0$, we try to show $\nul A=\{0\}$. As explained in problem 1 we need to show two set inclusion. Suppose $u\in \nul A$, then $Au=0$ by definition, thus $u=0$ by the hypothesis 2., so $u\in \{0\}$, i.e., \[
\nul A\subseteq \{0\}.
\] Conversely, suppose $u\in \{0\}$, i.e., $u=0$, then $Au=0$, so $u\in \nul A$, i.e., $\{0\}\subseteq \nul A$, so $\nul A=\{0\}$.
($3. \Rightarrow 1.$) Suppose $\nul A=\{0\}$. To show $A$ is injective as a map, let $x,y\in \R^n$ be such that $Ax=Ay$, we need to show $x=y$. To do that, since \[
A(x-y)=0,
\] so $x-y\in \nul A=\{0\}$, i.e., $x=y$, as was to be shown.
Q.E.D.
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