Linear Independence and Spanning Sets Review

# Linear Independence and Spanning Sets Review

We will now review some of the recent content regarding linear independent and spanning sets.

- Recall from the Linear Combinations of a Set of Vectors page that if $\{ v_1, v_2, ..., v_n \}$ is a set of vectors, then for scalars $a_1, a_2, ..., a_n \in \mathbb{F}$, a
**Linear Combination**of the vectors in $\{ v_1, v_2, ..., v_n \}$ is of the form:

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

- For example, if we have the set of vectors $\{ (1, 2, 3), (4, 5, 6) \}$ from $\mathbb{R}^3$ then $(6, 9, 12)$ is a linear combination of these vectors since:

\begin{equation} (6, 9, 12) = 2(1, 2, 3) + (4, 5, 6) = (2, 4, 6) + (4, 5, 6) = (6, 9, 12) \end{equation}

- From the Span of a Set of Vectors we defined the
**Span**of the set of vectors $\{ v_1, v_2, ..., v_n \}$ to be the set of all linear combinations of the vectors in $\{ v_1, v_2, ..., v_n \}$, that is:

\begin{align} \quad \mathrm{span} (v_1, v_2, ..., v_n) = \{ a_1v_1 + a_2v_2 + ... + a_nv_n : a_1, a_2, ..., a_n \in \mathbb{F} \} \end{align}

- By convention we say that $\mathrm{span} (\{ 0 \}) = \emptyset$.

- To show that a vector $v$ is in the span of a set of vectors $\{ v_1, v_2, ..., v_n \}$, we must find a set of scalars $a_1, a_2, ..., a_n \in \mathbb{F}$ such that:

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

- If no such set of scalars exist, then we say that $v \not \in \mathrm{span} (v_1, v_2, ..., v_n)$.

- On the Spanning Set of a Vector Space we said that if $V$ is a vector space and if a set of vectors $\{ v_1, v_2, ..., v_n\}$ is such that $V = \mathrm{span} (v_1, v_2, ..., v_n)$ then $\{ v_1, v_2, ..., v_n \}$ is a
**Spanning Set of $V$**.

- In $\mathbb{R}^3$, the set of vectors $\{ (1, 0, 0), (0, 1, 0), (0, 0, 1) \}$ spans $\mathbb{R}^3$ since for any vector $(a, b, c) \in \mathbb{R}^3$ where $a, b, c \in \mathbb{R}$ we have that:

\begin{align} \quad (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) \end{align}

- Furthermore, the set of vectors $\{ (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 2, 3) \}$ also spans $\mathbb{R}^3$ since once again, for any vector $(a, b, c) \in \mathbb{R}^3$ where $a, b, c \in \mathbb{R}$ we have that:

\begin{align} \quad (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) + 0(1, 2, 3) \end{align}

- On the Finite and Infinite-Dimensional Vector Spaces page we said that if a vector space $V$ can be spanned by a finite set of vectors $\{ v_1, v_2, ..., v_n \}$ then $V$ is said to be
**Finite-Dimensional**, and we said that a vector space that is not finite-dimensional is**Infinite-Dimensional**.

- The vector spaces $\mathbb{F}^n$ and $\wp (\mathbb{R})$ are both finite-dimensional vector spaces, while the vector space $\mathbb{F}^{\infty}$ of infinite sequences whose terms are in $\mathbb{F}$ is infinite-dimensional.

- On the Linear Independence and Dependence we said that a set of vectors $\{ v_1, v_2, ..., v_n \}$ were
**Linearly Independent**in $V$ if the vector $0$ can be written*uniquely*as a linear combination of the vectors in $\{ v_1, v_2, ..., v_n \}$. This is equivalent to $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ implying that $a_1 = a_2 = ... = a_n = 0$ (the "obvious" way to write $0$ as a linear combination of these vectors), as the existence of any other different set of scalars for which this equation is true implies that $\{ v_1, v_2, ..., v_n \}$ is**Linearly Dependent**in $V$.

- From the Theorems Regarding Linear Independence and Dependence we saw a lot of important results regarding linear independence/dependence.

- We first saw that if the zero vector was contained in a set of vectors, then that set of vectors cannot be linearly independent in $V$. To show this, suppose that $\{ v_1, v_2, ..., v_{n-1}, 0 \}$ is a set of vectors in $V$. Then $a_1v_1 + a_2v_2 + ... + a_{n-1}v_{n-1} + a_n 0 = 0$ can be satisfied by $a_1 = a_2 = ... = a_{n-1} = 0$ and for $a_n \in \mathbb{F}$ which gives us infinitely many linear combinations to represent the zero vector. Therefore, $\{ v_1, v_2, ..., v_{n-1}, 0 \}$ is a linearly dependent set of vectors.

- We also saw that a set of vectors that contains one nonzero vector is automatically linearly independent.

- Furthermore, if we have a set of two nonzero vectors $\{ v_1, v_2 \}$ then this set of vectors is linearly independent if and only if $v_1$ is not a scalar multiple of $v_2$.

- We then extended this idea to a set of $n$ vectors. We saw that $\{ v_1, v_2, ..., v_n \}$ is a linearly independent set of vectors in $V$ if and only if for each vector $v_i$, $i = 1, 2, ..., n$ we have that $v_i$ is not a linear combination of the remaining vectors $\{ v_1, v_2, ..., v_{i-1}, v_{i+1}, ..., v_n \}$.

- We then looked at the extremely important Linear Dependence Lemma which says that if $\{ v_1, v_2, ..., v_m \}$ is a linearly dependent set of vectors then there exists an index $j \in \{ 2, 3, ..., m \}$ such that $v_j \in \mathrm{span} (v_1, v_2, ..., v_{j-1})$ and such that $\mathrm{span} (v_1, v_2, ..., v_{j-1}, v_{j+1}, ..., v_m) = \mathrm{span} (v_1, v_2, ..., v_m)$. In showing this, we let $j$ be the largest index such that $a_j \neq 0$, and then we had that:

\begin{align} \quad v_j = -\frac{a_1}{a_j}v_1 -\frac{a_2}{a_j}v_2 - ... - \frac{a_{j-1}}{a_j}v_{j-1} \end{align}

- We then showed that $\mathrm{span} (v_1, v_2, ..., v_{j-1}, v_{j+1}, ..., v_m) = \mathrm{span} (v_1, v_2, ..., v_m)$ by noting that for any vector $v \in V$ we have that:

\begin{align} \quad v = a_1v_1 + a_2v_2 + ... + a_jv_j + ... + a_nv_n \\ \quad v = a_1v_1 + a_2v_2 + ... + a_j \left [ -\frac{a_1}{a_j}v_1 -\frac{a_2}{a_j}v_2 - ... - \frac{a_{j-1}}{a_j}v_{j-1} \right ] + ... + a_nv_n \end{align}

- From the Finite Dimensional Linearly Independent Set of Vectors Theorem we saw that every linearly independent set of vectors is smaller or equal to every spanning set of vectors for a vector space $V$.

- We also saw from the Infinite Finite Dimensional Vector Space Comparison Theorem that if $V$ is a finite-dimensional vector space then any subspace $U$ of $V$ will also be finite-dimensional, and if $U$ is an infinite-dimensional subspace of $V$ then $V$ must also be infinite-dimensional.