Lecture 21
Definition 11.1.1. Let $X$ be a set together with the operations of addition, $+\, \colon X \times X \to X,$ and multiplication, $\pd \,\colon \R \times X \to X,$ (we usually write $ax$ instead of $a \pd x$). $X$ is called a vector space (or a real vector space) if the following conditions are satisfied:
Definition 11.1.1. Let $X$ be a set together with the operations of addition, $+\, \colon X \times X \to X,$ and multiplication, $\cdot \,\colon \R \times X \to X,$ (we usually write $ax$ instead of $a \cdot x$). $X$ is called a vector space (or a real vector space) if the following conditions are satisfied:
An example vector space is $\R^n.$ Addition and multiplication by a scalar is done componentwise.
If $v = (v_1,v_2,\ldots,v_n),$ $w = (w_1,w_2,\ldots,w_n) \in \R^n,$ and $a \in \R$, then
$\ds v+w \coloneqq (v_1,v_2,\ldots,v_n) + (w_1,w_2,\ldots,w_n) \qquad $
$\ds \,= (v_1+w_1,v_2+w_2,\ldots,v_n+w_n)$
$\ds \;\; a v \coloneqq a (v_1,v_2,\ldots,v_n)$ $ = (a v_1, a v_2,\ldots, a v_n) .$
The set $X \coloneqq \{ 0 \}$ is a vector space. 🤯
The operations are defined as:
$0 + 0 \coloneqq 0\;$ and $\;a0 \coloneqq 0\;$ ($a\in \R$).
$X$ is the smallest possible vector space.
The space $C([0,1],\R)$ of continuous functions on the interval $[0,1]$ is a vector space. For two functions $f,g\in C([0,1],\R)$ and $a \in \R,$ we make the obvious definitions of $f+g$ and $af$:
$ (f+g)(x) \coloneqq f(x) + g(x), \; $ $ \,(af) (x) \coloneqq a\bigl(f(x)\bigr) . $
The 0 is the function that is identically zero. That is, $f(x)=0$ for all $x\in [0,1].$
Remark: If $X$ is a vector space, to check that a subset $S \subset X$ is a vector subspace, we only need to show:
The following concepts should be
familiar to you from
MATH1051.
Definition 11.2.1. Suppose $X$ is a vector space, $x_1, x_2, \ldots, x_k \in X$ are vectors, and $a_1, a_2, \ldots, a_k \in \R$ are scalars. Then \begin{equation*} a_1 x_1 + a_2 x_2 + \cdots + a_k x_k \end{equation*} is called a linear combination of the vectors $x_1, x_2, \ldots, x_k$.
If $Y \subset X$ is a set, then the span of $Y,$ or in notation $\text{span}(Y),$ is the set of all linear combinations of all finite subsets of $Y.$ By convention, define $\text{span}(\emptyset) \coloneqq \{ 0 \}$.
Definition 11.2.2. A set of vectors $\{ x_1, x_2, \ldots, x_k \} \subset X$ is linearly independent if the equation \begin{equation} a_1 x_1 + a_2 x_2 + \cdots + a_k x_k = 0 \end{equation} has only the trivial solution $a_1 = a_2 = \cdots = a_k = 0.$
A set that is not linearly independent is linearly dependent.
A linearly independent set of vectors $B$ such that $\text{span}(B) = X$ is called a basis of $X.$
Remarks about dimension
If a vector space $X$ contains a linearly independent set of $d$ vectors, but no linearly independent set of $d+1$ vectors, then we say the dimension of $X$ is $d$, and we write $\dim \, X \coloneqq d.$
If for all $d \in \N$ the vector space $X$ contains a set of $d$ linearly independent vectors, we say $X$ is infinite-dimensional and write $\dim \, X \coloneqq \infty.$
Remarks about dimension
$\dim \, X = d$ if and only if $X$ has a basis of $d$ vectors (and so every basis has $d$ vectors).
If $\dim \, X = d$ and a set $Y$ of $d$ vectors spans $X,$ then $Y$ is linearly independent.
For $\R^n$ we define the standard basis of $\R^n$ as
$e_1 \coloneqq (1,0,0,\ldots,0) ,$
$e_2 \coloneqq (0,1,0,\ldots,0) , $
$\vdots $
$e_n \coloneqq (0,0,0,\ldots,1). $
The dimension of $\R^n$ is $n.$
i.e., $\dim (\R^n)=n.$
Definition 11.3.1. A mapping $A \colon X \to Y$ of vector spaces $X$ and $Y$ is linear (we also say $A$ is a linear transformation or a linear operator) if for all $a \in \R$ and all $x,y \in X,$ \begin{equation*} A(a x) = a A(x), \;\text{and} \; A(x+y) = A(x)+A(y) . \end{equation*}
We usually write $Ax$ instead of $A(x)$ if $A$ is linear. If $A$ is one-to-one and onto, then we say $A$ is invertible, and we denote the inverse by $A^{-1}.$ If $A \colon X \to X$ is linear, then we say $A$ is a linear operator on $X.$
The vector space $L(X,Y)$
We write $L(X,Y)$ for the set of all linear transformations from $X$ to $Y,$ and just $L(X)$ for the set of linear operators on $X.$
If $a \in \R$ and $A,B \in L(X,Y),$ define the operations $A+B$ and $aA$ by \begin{equation*} (A+B)(x) := Ax + Bx, \;\;\, (aA)(x) := aAx . \end{equation*}
The vector space $L(X,Y)$
Furthermore, if $A \in L(Y,Z)$ and $B \in L(X,Y),$ define the operation $AB$ as the composition $A \circ B,$ that is, \begin{equation*} ABx := A(Bx) . \end{equation*}
Finally, denote by $I \in L(X)$ the identity: the linear operator such that $Ix = x$ for all $x.$
In particular, $L(X,Y)$ is a vector space. 😃
Theorem 11.3.1. If $A \in L(X,Y)$ is invertible, then $A^{-1}$ is linear.
Proof. Let $a \in \R$ and $y \in Y.$ Since $A$ is invertible, then is onto and one-to-one, that is
Theorem 11.3.1. If $A \in L(X,Y)$ is invertible, then $A^{-1}$ is linear.
Proof.
Thus
$A^{-1}(ay)$ $= A^{-1}(aAx)$ $= A^{-1}\bigl(A(ax)\bigr)$
$= ax $ $= aA^{-1}(y). \;\;\; $
Theorem 11.3.1. If $A \in L(X,Y)$ is invertible, then $A^{-1}$ is linear.
Proof.
Now let $y_1,y_2 \in Y,$ and $x_1, x_2 \in X$ such that $Ax_1 = y_1$ and $Ax_2 = y_2,$ then
$A^{-1}(y_1+y_2)$ $= A^{-1}(Ax_1+Ax_2)\qquad \qquad\qquad$
$ \qquad \;\;\;\;= A^{-1}\bigl(A(x_1+x_2)\bigr)$ $= x_1+x_2 $
$\quad = A^{-1}(y_1) + A^{-1}(y_2). \;\bs $
Theorem 11.3.2. If $X$ is a finite-dimensional vector space and $A \in L(X),$ then $A$ is one-to-one if and only if it is onto.
Let $\{ x_1,x_2,\ldots,x_n \}$ be a basis for $X$ and $A\in L(X).$
$\nec$ First suppose $A$ is one-to-one. Let $c_1,c_2,\ldots,c_n$ be such that
$0 =\ds \sum_{k=1}^n c_k \, Ax_k$ $=\ds A\sum_{k=1}^n c_k \, x_k .$
Since $A$ is one-to-one, the only vector that is taken to 0 is 0 itself.
Then $\;0 = \ds \sum_{k=1}^n c_k \, x_k\,$ and $\,c_k = 0\,$ for all $\,k$.
Let $\{ x_1,x_2,\ldots,x_n \}$ be a basis for $X$ and $A\in L(X).$
$\nec$ This means that $\{ Ax_1, Ax_2, \ldots, Ax_n \}$ is a linearly independent set. Since the dimension is $n$, we can deduce that $\{ Ax_1, Ax_2, \ldots, Ax_n \}$ spans $X.$ Thus any point $x \in X$ can be written as
$x =\ds \sum_{k=1}^n a_k \, Ax_k $ $= \ds A\sum_{k=1}^n a_k \, x_k ,$
so $A$ is onto.
Let $\{ x_1,x_2,\ldots,x_n \}$ be a basis for $X$.
$\suf$ Now suppose $A$ is onto. As $A$ is determined by the action on the basis, every element of $X$ is in the span of $\{ Ax_1, Ax_2, \ldots, Ax_n \}.$ Suppose that for some $c_1,c_2,\ldots,c_n,$
$0 =\ds A\sum_{k=1}^n c_k \, x_k $ $=\ds \sum_{k=1}^n c_k \, Ax_k .$
As $\{ Ax_1, Ax_2, \ldots, Ax_n \}$ span $X,$ the set is linearly independent, and hence $c_k = 0$ for all $k.$ In other words, if $Ax = 0$, then $x=0.$ If $Ax = Ay,$ then $A(x-y) = 0$ and so $x=y.$ This means that $A$ is one-to-one. $\;\bs$
The last fact to keep in mind about the vector space $L(X,Y)$
Theorem 11.3.3. If $X$ and $Y$ are finite-dimensional vector spaces, then $L(X,Y)$ is also finite-dimensional.