Basis of a Vector Space Examples 3

# Basis of a Vector Space Examples 3

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

We will now look at some more problems regarding bases of vector spaces.

## Example 1

Suppose that $\{ v_1, v_2, v_3, v_4 \}$ is a basis of the vector space $V$. Determine whether or not the set of vectors $\{ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 + v_1 \}$ is a basis of $V$ or not.

Let $a_1, a_2, a_3, a_4 \in \mathbb{F}$ and consider the following vector equation:

(1)
\begin{align} \quad a_1(v_1 + v_2) + a_2(v_2 + v_3) + a_3(v_3 + v_4) + a_4(v_4 + v_1) = 0 \\ \quad (a_4 + a_1)v_1 + (a_1 + a_2)v_2 + (a_2 + a_3)v_3 + (a_3 + a_4)v_4 = 0 \end{align}

Since $\{ v_1, v_2, v_3, v_4 \}$ is a basis of $V$, then $\{ v_1, v_2, v_3, v_4 \}$ is linearly independent in $V$ which implies that:

(2)
\begin{align} \quad a_4 + a_1 = 0 \\ \quad a_1 + a_2 = 0 \\ \quad a_2 + a_3 = 0 \\ \quad a_3 + a_4 = 0 \end{align}

Let $a_1 = 1$. Then $a_2 = -1$, and $a_3 = 1$ and $a_4 = -1$ and so:

(3)
\begin{align} \quad 1(v_1 + v_2) -1(v_2 + v_3) + 1(v_3 + v_4) - 1(v_4 + v_1) \\ \quad = v_1 + v_2 - v_2 - v_3 + v_3 + v_4 - v_4 - v_1 \\ \quad = 0 \end{align}

Therefore $\{ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 + v_1 \}$ is not a linearly independent set and so $\{ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 + v_1 \}$ is not a basis of $V$.

## Example 2

Suppose that $\{ v_1, v_2, v_3, v_4 \}$ is a basis of the vector space $V$. Determine whether or not the set of vectors $\{ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \}$ is a basis of $V$ or not.

Let $a_1, a_2, a_3, a_4 \in \mathbb{F}$ and consider the following equation:

(4)
\begin{align} \quad a_1(v_1 + v_2) + a_2(v_2 + v_3) + a_3(v_3 + v_4) + a_4v_4 = 0 \\ \quad a_1v_1 + (a_1 + a_2)v_2 + (a_2 + a_3)v_3 + (a_3 + a_4)v_4 = 0 \end{align}

Since $\{ v_1, v_2, v_3, v_4 \}$ is a basis of $V$ we have that $\{ v_1, v_2, v_3, v_4 \}$ is linearly independent in $V$ and so this implies that:

(5)
\begin{align} \quad a_1 = 0 \\ \quad a_1 + a+2 = 0 \\ \quad a_2 + a_3 = 0 \\ \quad a_3 + a_4 = 0 \end{align}

Using forward substitution we have that $a_1 = a_2 = a_3 = a_4 = 0$, so $\{ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \}$ is a linearly independent set of vectors in $V$. We now only need to show that this set of vectors spans $V$. This is easy to do. Since $\{ v_1, v_2, v_3, v_4 \}$ is a basis of $V$ then for scalars $b_1, b_2, b_3, b_4 \in \mathbb{F}$ we have that:

(6)
\begin{align} \quad v = b_1v_1 + b_2v_2 + b_3v_3 + b_4v_4 \end{align}

Let $a_1 = b_1$, $a_1 + a_2 = b_2$, $a_2 + a_3 = b_3$ and $a_3 + a_4 = b_4$.

Then we have that:

(7)
\begin{align} \quad a_1 = b_1 \\ \quad a_2 = b_2 - b_1 \\ \quad a_3 = b_3 - b_2 + b_1 \\ \quad a_4 = b_4 - b_3 + b_2 - b_1 \end{align}

So indeed $\{ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \}$ spans $V$ and so $\{ v_1 + v_2, v_2 + v_3, v_3 + v_4, v_4 \}$ is a basis of $V$ as well.