Bases and Dimension Review

# Bases and Dimension Review

We will now review some of the recent content regarding bases and the dimension of a finite-dimensional vector space.

- Recall from the Basis of a Vector Space that if $V$ is a finite-dimensional vector space and the set of vectors $\{ v_1, v_2, ..., v_n \}$ is both linearly independent and spans $V$ then $\{ v_1, v_2, ..., v_n \}$ is said to be a
**Basis**of $V$.

- For example, consider the vector space $\mathbb{F}^n$. The
**Standard Basis**for $\mathbb{F}^n$ is the set of vectors $\{ e_1, e_2, ..., e_n \}$ where $e_j$ is the $n$-component vector whose components are all zero except the $j^{\mathrm{th}}$ component is $1$. For example, the standard basis for $\mathbb{F}^3$ is $\{ (1, 0, 0), (0, 1, 0), (0, 0, 1) \}$.

- We saw that a basis for the vector space $M_{22}$ of all $2 \times 2$ matrices is $\left \{ \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} \right \}$.

- We also saw that $\{ 1, i \}$ and $\{ 1 + i, 1 - i \}$ are both bases for the vector space $\mathbb{C}$ of complex numbers.

- On the Theorems Regarding a Basis of a Vector Space we saw that a set of vectors $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$ if and only if each vector $v \in V$ can be written
*uniquely*as a linear combination for $a_1, a_2, ..., a_n \in \mathbb{F}$ of the vectors in $\{ v_1, v_2, ..., v_n \}$:

\begin{align} \quad v = a_1v_1 + a_2v_2 + ... + a_nv_n \end{align}

- We then noted that every spanning set of vectors can be reduced to a basis of $V$. This can be achieved systematically. Take a spanning set of vectors $\{ v_1, v_2, ..., v_n \}$. If the first vector is zero, then throw it out. If the first vector is nonzero, then keep it. If the $j^{\mathrm{th}}$ vector is in the span of the previous vectors, then throw it out, and if not, keep it and move on. After the $n^{\mathrm{th}}$ step we will have produced a set of linearly independent vectors that still spans the whole space.

- We also noted that every linearly independent set of vectors from a finite-dimensional vector space $V$ can be extended to a basis of $V$. Once again, this can be achieved systematically. Take a linearly independent set of vectors $\{ v_1, v_2, ..., v_n \}$ and some largest set of vectors that spans $V$, say $\{ w_1, w_2, ..., w_m \}$. If $w_1$ is a linear combination of the vectors in $\{ v_1, v_2, ..., v_n \}$ then throw it out, and if not, add it in. If the $j^{\mathrm{th}}$ vector, $w_j$ is in the span of $\{ v_1, v_2, ..., v_n \}$ and the kept $w$ vectors, then throw it out, and if not, add it in. After the $n^{\mathrm{th}}$ step we will have produced a spanning set of vectors that contains all of the linearly independent vectors from the linearly independent set $\{ v_1, v_2, ..., v_n \}$.

- We also noted that every finite-dimensional vector space must then have a basis.

- On the Dimension of a Vector Space page we saw that if $V$ is a finite-dimensional vector space, then the length of any basis of $V$ is the same, and we defined the
**Dimension**of $V$ to be this length.

- The dimension of $\mathbb{F}^n$ is $n$ since $\{ (1, 0, 0, ..., 0), (0, 1, 0, ..., 0), ..., (0, 0, 0, ..., 1) \}$ is a basis of $n$ vectors for $\mathbb{F}^n$. The dimension of the set of all polynomials of degree $n$ or less, $\wp (\mathbb{R})_{n}$, is $n + 1$ since $\{ 1, x, x^2, ..., x^n \}$ is a basis of this vector space.

- We noted that the underlying field for the vector space $V$ is important in determining its dimension. For the vector space $\mathbb{C}$ of complex numbers whose scalar field is $\mathbb{R}$ we have that $\mathrm{dim}_{\mathbb{R}} (\mathbb{C}) = 2$ and $\{ 1, i \}$ is a basis of $\mathbb{C}$. If instead we consider the vector space $\mathbb{C}$ of complex numbers whose scalar field is $\mathbb{C}$, then we have that $\mathrm{dim}_{\mathbb{C}} (\mathbb{C}) = 1$.