Lecture 4: Vector Spaces
- \(u + v = v + u\) for all \(u,v \in V\).
- \((u + v) + w = u + (v + w)\) for all \(u, v, w \in V\).
- There exists an element \(\bar{0} \in V\) such that \(v + \bar{0} = v\) for all \(v \in V\).
- For all \(v \in V\), there exists \(w \in V\) such that \(v + w = \bar{0}\).
- \(1v = v\) for all \(v \in V\)
- \(a(bv) = a(bv)\) for all \(v \in V\) and for all \(a, b \in \mathbf{F}\)
- \(a(u + v) = au + av\) for all \(u, v \in V\) and for all \(a \in \mathbf{F}\)
- \((a + b)v = av + bv\) for all \(u, v \in V\) and for all \(a, b \in \mathbf{F}\)
For property (4), we don’t call it \(-v\) yet because we didn’t prove yet if it’s unique.
Example 1: \(\mathbf{R}\)
\(\mathbf{R}\) is a vector space equipped with the usual addition and scalar multiplication. The number 0 is the zero vector. We can additionally verify that all the 8 properties are true.
Example 2: A Set of Matrices
The set of \(m\) by \(n\) matrices (\(M_{m \times n}\)) equipped with component wise addition such that
and component wise scalar multiplication such that
is a vector space.
Example 3: Sets of Functions
let \(S\) be a nonempty set. For example \(S = \mathbf{R}\), or \(S = \{\pi, \pi^2\}\), \(S = \{\)atoms in the universe\(\}\). Basically any non-empty set.
Now consider \(F(S) = \{f: S \rightarrow \mathbf{R}\}\), the set of all functions or mappings from \(S\) to \(\mathbf{R}\). One way to think of this is the all the ways we can label the elements in the set \(S\) with real numbers.
Define addition as \((f+g)(s) = f(s) + g(s)\) for all \(s \in S\). So addition of functions works as addition of their values and produces a real number which is what we want. Define scalar multiplication as \(cf(s) =c(f(s))\) for all \(s \in S\).
\(F(S)\) is a vector space. It satisfies all 8 conditions. For example. The zero vector in this space is \(\bar{0}(s) = 0\) for all \(s \in S\). Note also that \(C^1(\mathbf{R})\) (the functions where with continues derivatives) is a subset of \(C^0(\mathbf{R})\) (the set of continuous functions) which is a subset of \(F(S)\).
Example 4: The Set of all Sequences
Consider the set of all natural numbers \(\mathbf{N}\) and the set of functions \(F(\mathbf{N}) = \{\sigma: \mathbf{N} \rightarrow \mathbf{R}\). \(\sigma\) is a function that takes a natural number and assigns it a real number. But
is a sequence. So we’re giving the set of sequences the structure of a vector space. Let \(\sigma(1) =a_1, \sigma(2) = a_2, ...\) and so on. So now we can write the sequence as
Let \(V = \{\) sequences \(\{a_n\}\) is a vector space. Define the addition of two sequences as \(\{a_n\} + \{b_n\} = \{a_n + b_n\}\). Adding the terms one by one. Define scalar multiplication as \(c\{a_n\} = \{ca_n\}\).
Example 5: The Set of Polynomials
Let \(P_n = \{\) polynomials \(f(x)\) of degree at most \(n\}\).
Define the addition operation as follows,
and define scalar multiplication as
\(P_n\) is a vector space. The zero vector is the function \(f\bar{0} = 0 = 0x^n + .... + 0\).
Question: why did we define the polynomials to have at most \(n\) and not just \(n\)? because take \((X^5 + 1)\) and \((-x^5 + 9)\). The addition of these two will generate a 0 and so we have to say at most.
Additional Vector Space Results
Proof:
Let \(V\) be a vector space and \(u, v, w\) be elements in \(V\). By property (4) there is a \(z \in V\) such that
We also know that by property (3) that
But \(\bar{0} = w+z\) and so
Therefore \(u = v\) as we wanted to show. \(\blacksquare\)
Proof:
Suppose \(V\) is a vector space. Now suppose for the sake of contradiction that \(\bar{0}\) and \(\bar{0}'\) are additive inverses of \(V\) where \(\bar{0} \neq \bar{0}'\). This means that
Therefore, \(\bar{0} = \bar{0}'\) which is a contradiction and the zero vector must be unique. \(\blacksquare\)
Proof:
Suppose \(V\) is a vector space. Let \(v \in V\). Suppose for the sake of contradiction that \(w\) is not unique and there exists two additive inverses \(w\) and \(w'\) such that \(w \neq w'\). Then
Since \(w = w'\), we can conclude that \(w\) is a unique additive inverse. \(\blacksquare\)
Two additional implications mentioned in the class is that \(w = (-1)v\) and \(0v = \bar{0}\).
Example 6: A Non Example
Consider the set \(\mathbf{R}^2\) equipped with a different set of operations. Let’s define addition as
and scalar multiplication.
Is this a vector space? No. why?
What is the zero vector is this space?
because for every vector \(v\), we want \(v + \bar{0} = v\). \((0, 1)\) works here because
The claim is that property 4 can’t be true. Let \(v = (0,0)\). There is no \((a_1, a_2) \in \mathbf{R}^2\) such that
\(v + (a_1, a_2) = \bar{0}\). To see this,
so it can never be equal to (0, 1).
References:
- Math416 by Ely Kerman
- Linear Algebra Done Right for the last two proofs