Vector spaces
Notes
Note
Definition of vector space
A vector space $V$ is a collection of objects, called vectors along with two operations:
- Addition of vectors
- Multiplication by a number (scalar)
The vector space $V$ satisfies the following 8 properties (called axioms of a vector space):
- Commutativity: $\mathbf{v} + \mathbf{w} = \mathbf{w} + \mathbf{v}$ for all $\mathbf{v}, \mathbf{w} \in V$.
- Associativity: $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$ for all $\mathbf{u} + \mathbf{v} + \mathbf{w} \in V$.
- Zero vector: There exists a special vector, denoted by $\mathbf{0}$ such that $\mathbf{v} + \mathbf{0} = \mathbf{v}$ for all $\mathbf{v} \in V$.
- Additive inverse: For every $\mathbf{v} \in V$ there exists a vector $\mathbf{w} \in V$ such that $\mathbf{v} + \mathbf{w} = \mathbf{0}$. Such additive inverse is usually denoted as $-\mathbf{v}$.
- Multiplicative identity: $1\mathbf{v} = \mathbf{v}$ for all $\mathbf{v} \in V$.
- Multiplicative associativity: $(\alpha\beta)\mathbf{v} = \alpha(\beta\mathbf{v})$ for all $\mathbf{v} \in V$ and all scalars $\alpha, \beta$.
- $\alpha(\mathbf{u} + \mathbf{v}) =\alpha\mathbf{u} + \alpha\mathbf{v}$ for all $\mathbf{u}, \mathbf{v} \in V$ and all scalars $\alpha$.
- $(\alpha + \beta)\mathbf{v} = \alpha\mathbf{v} + \beta\mathbf{v}$ for all $\mathbf{v} \in V$ and all scalars $\alpha, \beta$.
If the scalars are the real numbers we call the space $V$ a real vector space. If the scalars are the complex numbers we call the space $V$ a complex vector space.
An $m \times n$ matrix is a rectangular array with $m$ rows and $n$ columns. Elements of the array are called entries of the matrix.
The entry in row $j$ and column $k$ is denoted by $A_{j,k}$.
The transpose of a matrix (or transposed matrix) $A^T$ is defined by transforming the rows of A into the columns.
$$ \begin{pmatrix} 1 & 2 & 3 \\\ 4 & 5 & 6 \end{pmatrix}^T = \begin{pmatrix} 1 & 4 \\\ 2 & 5 \\\ 3 & 6 \end{pmatrix} $$Note
Definition of the matrix transpose
The formal definition is $(A^T)_{j,k} = A_{k,j}$.
Exercises
1.1
Let $\mathbf{x} = (1, 2, 3)^T$, $\mathbf{y} = (y_1, y_2, y_3)^T$, $\mathbf{z} = (4, 2, 1)^T$. Compute $2\mathbf{x}$, $3\mathbf{y}$, $\mathbf{x} + 2\mathbf{y} − 3\mathbf{z}$.
$$ \begin{aligned} 2\mathbf{x} &= 2\begin{pmatrix} 1 \\\ 2 \\\ 3 \end{pmatrix} \\\ &= \begin{pmatrix} 2 \\\ 4 \\\ 6 \end{pmatrix} \end{aligned} $$$$ \begin{aligned} 3\mathbf{y} &= 3\begin{pmatrix} y_1 \\\ y_2 \\\ y_3 \end{pmatrix} \\\ &= \begin{pmatrix} 3y_1 \\\ 3y_2 \\\ 3y_3 \end{pmatrix} \end{aligned} $$$$ \begin{aligned} \mathbf{x} + 2\mathbf{y} − 3\mathbf{z} &= \begin{pmatrix} 1 \\\ 2 \\\ 3 \end{pmatrix} + 2\begin{pmatrix} y_1 \\\ y_2 \\\ y_3 \end{pmatrix} - 3\begin{pmatrix} 4 \\\ 2 \\\ 1 \end{pmatrix} \\\ &= \begin{pmatrix} 1 \\\ 2 \\\ 3 \end{pmatrix} + \begin{pmatrix} 2y_1 \\\ 2y_2 \\\ 2y_3 \end{pmatrix} - \begin{pmatrix} 12 \\\ 6 \\\ 3 \end{pmatrix} \\\ &= \begin{pmatrix} -11 + 2y_1 \\\ -4 + 2y_2 \\\ 2y_3 \end{pmatrix} \end{aligned} $$
1.2
Which of the following sets (with natural addition and multiplication by a scalar) are vector spaces? Justify your answer.
a) The set of all continuous functions on the interval $[0,1]$;
b) The set of all non-negative functions on the interval $[0,1]$;
c) The set of all polynomials of degree exactly n;
d) The set of all symmetric $n \times n$ matrices, i.e., the set of matrices $A = \{ a_{j,k} \}^n_{j,k=1}$ such that $A^T = A$.
a) This set is a vector space. The sum of all continuous functions is continuous and it can be easily demonstrated that they satisfy the 8 axioms of a vector space.
b) This set is not a vector space. The inverse of $\mathbf{v}$ must be a negative function and because of this axiom 4 is not fulfilled.
c) This set is not a vector space because there is not a zero vector. The zero vector would need to have all the coefficients equal to zero, but then the degree would not be n.
d) This set is a vector space because it satisfies the 8 axioms of a vector space.
1.3
True or false:
a) Every vector space contains a zero vector;
b) A vector space can have more than one zero vector;
c) An $m \times n$ matrix has $m$ rows and $n$ columns;
d) If $f$ and $g$ are polynomials of degree $n$, then $f + g$ is also a polynomial of degree $n$;
e) If $f$ and $g$ are polynomials of degree at most $n$, then $f + g$ is also a polynomial of degree at most $n$.
a) True, by axiom 3.
b) False, the zero vector is unique.
c) True, by the definition of a matrix.
d) False, because when adding polynomials, the coefficients of like terms are added. When the coefficient of $x^n$ in $f$ is $\alpha$ and the coefficient of $x^n$ in $g$ is $-\alpha$, the resultant polynomial is of degree less than $n$.
e) True, because when adding polynomials, the coefficients of like terms are added. The sum of two polynomials of degree at most $n$ will have degree at most $n$.
1.4
Prove that a zero vector $\mathbf{0}$ of a vector space $V$ is unique.
Let’s suppose that $\mathbf{0}$ is not unique. There is a vector $\mathbf{x}$ such that $\mathbf{v} + \mathbf{x} = \mathbf{v}$ for all $\mathbf{v} \in V$.
$$ \begin{aligned} \mathbf{v} + \mathbf{0} &= \mathbf{v} + \mathbf{x} \\\ \mathbf{v} + \mathbf{0} - \mathbf{v} &= \mathbf{v} + \mathbf{x} - \mathbf{v} \\\ \mathbf{0} + \mathbf{0} &= \mathbf{x} + \mathbf{0} \\\ \mathbf{0} &= \mathbf{x} \end{aligned} $$For $\mathbf{x}$ to have the property $\mathbf{v} + \mathbf{x} = \mathbf{v}$, the equation above shows that $\mathbf{0} = \mathbf{x}$ and therefore $\mathbf{0}$ is unique.
1.5
What matrix is the zero vector of the space $M_{2 \times 3}$?
$$ \mathbf{0} = \begin{pmatrix} 0 & 0 & 0 \\\ 0 & 0 & 0 \end{pmatrix} $$
1.6
Prove that the additive inverse, defined in Axiom 4 of a vector space, is unique.
If there is a vector $\mathbf{x}$ such that $\mathbf{v} + \mathbf{x} = \mathbf{0}$ other than $-\mathbf{v}$, then:
$$ \begin{aligned} \mathbf{v} + \mathbf{x} &= \mathbf{0} \\\ -\mathbf{v} + \mathbf{v} + \mathbf{x} &= -\mathbf{v} + \mathbf{0} \\\ \mathbf{0} + \mathbf{x} &= -\mathbf{v} + \mathbf{0} \\\ \mathbf{x} &= -\mathbf{v} \end{aligned} $$Given that $\mathbf{x} = -\mathbf{v}$, then $-\mathbf{v}$ is unique.
1.7
Prove that $0\mathbf{v} = \mathbf{0}$ for any vector $\mathbf{v} \in V$.
$$ \begin{aligned} 0\mathbf{v} &= (1-1)\mathbf{v} \\\ &= \mathbf{v} - \mathbf{v} \\\ &= \mathbf{0} \\\ \end{aligned} $$
1.8
Prove that for any vector $\mathbf{v}$ its additive inverse $-\mathbf{v}$ is given by $(-1)\mathbf{v}$.
By axiom 8, we can see that $\mathbf{v} + (-1)\mathbf{v} = (1 - 1)\mathbf{v}$.
Using this fact, we can follow the equalities:
$$ \begin{aligned} \mathbf{v} + (-1)\mathbf{v} &= (1 - 1)\mathbf{v} \\\ &= 0\mathbf{v} \\\ &= \mathbf{0} \\\ &= \mathbf{v} + (-\mathbf{v}) \end{aligned} $$By the last equality, and remembering that $-\mathbf{v}$ is unique, we can see that $-\mathbf{v} = (-1)\mathbf{v}$.