What are vector spaces and subspaces?

Vector spaces

A set of vectors $V$ is defined as a vector space if, and only if, the vectors in the set$V$ follow the 10 axioms defined for a vector space.

Axioms

Let $V$ be an arbitrary set of vectors defined under addition and scalar multiplication. These two operations can be performed on the set $V$. Also, let's consider that the vectors $\vec{u}, \vec{v}, and \, \vec{w}$ exist in the set $V$. The set can be defined as a vector space if it follows all the 10 axioms mentioned below:

1. If the vectors $\vec{u}$ and $\vec{v}$ belong to the set $V$, then $\vec{u} + \vec{v}$ should also be in the set $V$

2. $\vec{u} + \vec{v}$ = $\vec{v} + \vec{u}$

3. $\vec{u} + (\vec{v} + \vec{w}) = (\vec{u} + \vec{v}) + \vec{w}$

4. $\vec{0} + \vec{u} = \vec{u} + \vec{0} = \vec{u}$ for all vectors $\vec{u}$ in the set $V$

5. For each $\vec{u}$, there must exist a negative of $\vec{u}$ such that $\vec{u} + (-\vec{u}) = (-\vec{u}) + \vec{u} = \vec{0}$

6. If $\vec{u}$ exists in $V$, then $k\vec{u}$ should also exist in $V$, where $k$ is any scalar

7. $k(\vec{u} + \vec{v}) = k\vec{u} + k\vec{v}$

8. $(k+m) \vec{u} = k\vec{u} + m\vec{u}$

9. $k(m\vec{u}) = (km)\vec{u}$

10. $1\vec{u} = \vec{u}$

Note: If the set $V$ violates any of the axioms, it is not a vector space.

Example 1

Let's consider $V$ a set of $2 \times 2$ matrices with real entries. How do we prove that $V$ is a vector space?

Solution

The set $V$ is tested for all the 10 axioms and proved as a vector space. Some of the proofs are explained below:

Axiom 1

If $\vec{u}$ and $\vec{v}$ belong to $V$, $\vec{u} + \vec{v}$ also belongs to the set$V$.

Axiom 2

This axiom holds true as shown below:

Axiom 4

A zero vector in the set $V$ is defined as follows:

Using the definition of Axiom 4, the following holds:

Axiom 5

A negative of a vector $\vec{u}$ in the set $V$ is defined as follows:

According to the definition of Axiom 6, the following holds:

Axiom 6

Scalar multiplication holds for the set $V$ as follows:

Axiom 10

For the set $V$, Axiom 10 holds true as follows:

Note: Since all of the axioms provide solutions that exist in the set $V$ defined for all $2 \times 2$ matrices, we can define the set $V$ as a vector space.

Example 2

Let $V = R^2$ with the following addition and multiplication operations:

How can we state whether $V$ is a vector space or not?

Solution

We can easily prove that the set $V$ follows all the first 9 axioms. Let's see if we can prove Axiom 10:

Axiom 10

$V$ violates the following axiom as follows:

Since the last axiom is violated, we cannot define $V$ as a vector space.

Subspaces

A subset $W$ of a vector space $V$ is defined as a subspace if it is closed under addition and multiplication.

To test for a subspace, we consider $W$ a nonempty set containing vectors from the vector space $V$. $W$ is a subspace of $V$ if it fulfills the three conditions:

1. The set $W$ contains a zero vector.

2. If $\vec{u}$ and $\vec{v}$ are defined $W$, then $\vec{u} + \vec{v}$ must also be in $W$.

3. If $k$ is any scalar and $\vec{u}$ is contained in $W$, then $k\vec{u}$ must also be in $W$.

Example 1

Given that $M_{nn}$ is a vector space, let's mention one subset for $M_{nn}$ that is a subspace and another subset of $M_{nn}$ that is not a subspace.

Solution

The addition between two symmetric matrices provides a symmetric matrix in return. Scalar multiplication with a symmetric matrix also returns a symmetric matrix. Zero $n \times n$ matrix is also a symmetric matrix. Hence, a symmetric $n \times n$ matrix is a subspace of $M_{nn}$.

Note: Upper triangularThe matrix contains zero elements below the diagonal., lower triangularThe matrix contains zero elements above the diagonal., and diagonalThe matrix has nonzero elements on the diagonal only. matrices are also subspaces of $M_{nn}$.

Invertible matrices are a subset of $M_{nn}$. However, they don't form a subspace because the subset of invertible matrices does not contain a zero matrix.

Applications

The concept of vector spaces is fundamental in linear algebra because, together with the concept of matrices, it allows the manipulation of the system of linear equations.

Vector spaces generalize vectors and enable the modeling of physical quantities that have a direction and magnitude attached to them, such as forces.