Linear Independence and Dependence Examples 5

# Linear Independence and Dependence Examples 5

Recall from the Linear Independence and Dependence page that a set of vectors $\{ v_1, v_2, ..., v_n \}$ is said to be Linearly Independent in $V$ if the vector equation $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$ implies that $a_1 = a_2 = ... = a_n = 0$, that is, the zero vector is uniquely expressed as a linear combination of the vectors in $\{ v_1, v_2, ..., v_n \}$ with the coefficients all being zero.

If a set of vectors $\{ v_1, v_2, ..., v_n \}$ is not linearly independent then we say the set if Linearly Dependent and that there exists scalars $a_1, a_2, ..., a_n \in \mathbb{F}$, not all zero, such that $a_1v_1 + a_2v_2 + ... + a_nv_n = 0$.

We will now look at some more examples to regarding the linear independence / dependence of a set of vectors.

## Example 1

Prove or disprove: If $\{ v_1, v_2, ..., v_n \}$ and $\{ w_1, w_2, ..., w_n \}$ are both linearly independent in $V$ then $\{ v_1 + w_1, v_2 + w_2, ..., v_n + w_n \}$ is linearly independent in $V$.

Consider the vector space $\mathbb{R}$. The set $\{ 1 \}$ is linearly independent in $\mathbb{R}$ because for $a_1 \in \mathbb{F}$, $a_1 * 1 = 0$ implies that $a_1 = 0$. Furthermore, the set $\{ -1 \}$ is also linearly independent in $\mathbb{R}$ because for $b_1 \in \mathbb{F}$, $b_1 * -1 = 0$ implies that $b_1 = 0$.

Note though that $\{ 1 + (-1) \} = \{ 0 \}$ is not linearly independent in $V$ though.

Thus the statement above is false in general.

## Example 2

Let $\{ p_0(x), p_1(x), ..., p_n(x) \}$ be a set of $n + 1$ polynomials in $\wp (\mathbb{R})$ where $p_j(1) = 0$ for each $j = 1, 2, ..., n$. Show that this set of polynomials is linearly dependent in $\wp (\mathbb{R})$.

Let $\{ p_0(x), p_1(x), ..., p_n(x) \}$ be a set of $n + 1$ polynomials in $\wp_n (\mathbb{R})$ such that $p_j(1) = 0$ for each $j = 1, 2, ..., n$, and for $a_0, a_1, a_2, ..., a_n \in \mathbb{F}$, consider the vector equation:

(1)
\begin{align} \quad a_0p_0(x) + a_1p_1(x) + ... + a_np_n(x) = 0 \\ \end{align}

For $x = 1$ we have that:

(2)
\begin{align} \quad a_0p_0(1) + a_1p_1(1) + ... + a_np_n(1) = 0 \\ \quad a_0(0) + a_1(0) + ... + a_n(0) = 0 \end{align}

Note that the above equation can be satisfied by ANY set of scalars $a_0, a_1, ..., a_n \in \mathbb{F}$, and so $\{ p_0(x), p_1(x), ..., p_n(x) \}$ is linearly dependent in $\wp (\mathbb{R})$.