Let \(\mathbb{F}\) be a field and \(n\) a natural number. Then \(\mathbb{F}^n\) forms a vector space under tuple addition and scalar multplication where scalars are elements of \(\mathbb{F}\).

\(\mathbb{F}^n\) is probably the most common vector space studied, especially when \(\mathbb{F} = \mathbb{R}\) and \(n \leq 3\). For example, \(\mathbb{R}^2\) is often depicted by a 2-dimensional plane and \(\mathbb{R}^3\) by a 3-dimensional space.

The zero vector in \(\mathbb{F}^n\) is given by the \(n\)-tuple of all 0's.

More generally, \(\mathbb{F}^{m \times n}\) forms a vector space under matrix addition and scalar multiplication. The zero vector in \(\mathbb{F}^{m \times n}\) is given by the \(m\times n\) matrix of all 0's.

Another common vector space is given by the set of polynomials in \(x\) with coefficients from some field \(\mathbb{F}\) with polynomial addition as vector addition and multiplying a polynomial by a scalar as scalar multiplication. The zero vector is given by the zero polynomial.

The degree of the polynomials could be restricted or unrestricted. For example, one could consider the vector space of polynomials in \(x\) with degree at most \(2\) over the real numbers, which will be denoted by \(P_2\) from now on. In such a vector space, all vectors can be written in the form \(ax^2 + bx + c\) where \(a,b,c\in \mathbb{R}\).

If \(V,W\) are vector spaces such that

the set of vectors in \(W\) is a subset of the set of vectors in \(V\),

\(V\) and \(W\) have the same vector addition and scalar multiplication,

Given a matrix \(A\in \mathbb{F}^{m \times n}\) where \(\mathbb{F}\) is a field, the nullspace of \(A\) is a subspace of \(\mathbb{F}^n\). (Recall that the nullspace of \(A\) is given by \(\{x \in \mathbb{F}^n : Ax = 0\}.\))

Let \(V\) be a vector space. Let \(W\) be a set of vectors from \(V\). How do we know if \(W\) is a subspace of \(V\)?

First of all, that \(W\) is a subset of \(V\) does not automatically make it a subspace of \(V\). For instance, if \(W\) does not contain the zero vector, then it is not a vector space.

Of course, one can check if \(W\) is a vector space by checking the properties of a vector space one by one. But in this case, it is actually sufficient to check that \(W\) is closed under vector addition and scalar multiplication as they are defined for \(V\).

Let \(W =\left\{\begin{bmatrix} a\\b\end{bmatrix} : a, b \in \mathbb{Z}\right\}\). Clearly, \(W \subseteq \mathbb{R}^2\). However, \(W\) is not a subspace of \(\mathbb{R}^2\) because it is not closed under scalar multiplication. To see this, let \(\alpha\) be a real number that is not an integer. Then \(\begin{bmatrix} 1\\0 \end{bmatrix} \in W\) but \(\alpha \begin{bmatrix} 1\\0\end{bmatrix} = \begin{bmatrix} \alpha\\0\end{bmatrix}\notin W\) since the first entry is \(\alpha\) and it is not an integer.

Is \(U = \left\{ \begin{bmatrix} x\\y \end{bmatrix} \in \mathbb{R}^2 : \begin{array}{rl} x + y = 0\\ x-y = 1\end{array}\right\}\) a subspace of \(\mathbb{R}^2\)?

Note that in order for \(U\) to be a subspace of \(\mathbb{R}^2\), it must contain the zero vector, which in this case is \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\). However, \(\begin{bmatrix} x\\y\end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}\) is not a solution to the system \(\begin{array}{rl} x + y = 0\\ x-y = 1\end{array}\). Hence, \(\begin{bmatrix} 0 \\ 0 \end{bmatrix} \notin U\), implying that \(U\) is not a subspace.

Give an example of a proper subspace of the vector space of polynomials in \(x\) with real coefficients of degree at most \(2\).

Let \(W = \left \{ \begin{bmatrix} a & b \\ b & 0 \end{bmatrix} : a,b \in \mathbb{R}\right\}\). Show that \(W\) is a subspace of \(\mathbb{R}^{2\times 2}\).