Linear Independence and Dependence Examples 1

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

Consider the set of vectors $\left \{ \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} \right \}$ from the vector space $M_{22}$ of $2 \times 2$ matrices. Determine if this set is linearly independent or linearly dependent.

We first consider the following vector equation:

(1)

\begin{align} a_1 \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} + a_2 \begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix} + a_3 \begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix} = 0 \\ \begin{bmatrix} a_1 & 0\\ 0 & a_1 \end{bmatrix} + \begin{bmatrix} 0 & -a_2\\ 0 & 0 \end{bmatrix} + \begin{bmatrix} a_3 & 0\\ 0 & 0 \end{bmatrix} = 0 \\ \begin{bmatrix} a_1 + a_3 & -a_2\\ 0 & a_1 \end{bmatrix} = 0 \end{align}

This equation equals zero if and only if $a_1 + a_3 = 0$, $-a_2 = 0$, and $a_1 = 0$, and we deduce that therefore the only set of scalars for which the above vector equation is true is $a_1 = a_2 = a_3 = 0$. We can verify this by showing the following matrix representing this system is invertible, $A = \begin{bmatrix}1 & 0 & 1\\ 0 & -1 & 0\\ 1 & 0 & 0 \end{bmatrix}$ is invertible by cofactor expansion along the third row to get that $\mathrm{det} A = 1 \cdot \begin{vmatrix}0 & 1\\ -1 & 0 \end{vmatrix} = 1$, and so the homogenous system $\begin{bmatrix} 1 & 0 & 1\\ 0 & -1 & 0\\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} a_1\\ a_2\\ a_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$ has only the trivial solution $a_1 = a_2 = a_3 = 0$.

Example 2

Consider the set of vectors $\{ 1 + x - x^2, 2 + x^2, x^3 - 2x^4 \}$ from the vector space $\wp_{4} (\mathbb{F})$ of polynomials of degree $4$ or less. Determine if this set is linearly independent or linearly dependent.

Let's first consider the following vector equation:

(2)

\begin{align} a_1(1 + x - x^2) + a_2(2 + x^2) + a_3(x^3 - 2x^4) = 0 \\ \quad a_1 + a_1x - a_1x^2 + a_2 \cdot 2 + a_2x^2 + a_3x^3 - a_3 \cdot 2x^4 = 0 \\ \quad (a_1 + 2a_2)1 + (a_1)x + (-a_1 + a_2)x^2 + (a_3)x^3 + (-2a_3)x^4 = 0 \end{align}

This equation is zero if and only if:

(3)

\begin{align} a_1 + 2a_2 = 0 \\ a_1 = 0 \\ -a_1 + a_2 = 0 \\ a_3 = 0 \\ a_4 = 0 \end{align}

And the only solution to this system is $a_1 = a_2 = a_3 = a_4 = 0$, and so this set of vectors is linearly independent.

Example 3

Consider the set of vectors $\{ (1, 2, 0, 0), (0, 4, 4, 0), (2, 0, 3, 0) \}$ from the vector space $\mathbb{R}^4$ of 4-component standard vectors. Determine if this set is linearly independent or linearly dependent.

Let's first look at the following vector equation:

(4)

\begin{align} \quad a_1(1, 2, 0, 0) + a_2(0, 4, 4, 0) + a_3(2, 0, 3, 0) = 0 \\ \quad (a_1, 2a_1, 0, 0) + (0, 4a_2, 4a_2, 0) + (2a_3, 0, 3a_3, 0) = 0 \\ \quad (a_1 + 2a_3, 2a_1 + 4a_2, 4a_2 + 3a_3, 0) = (0, 0, 0, 0) \end{align}

We must check to see if the following homogenous system has more than just the trivial solution.

(5)

\begin{align} a_1 + 2a_3 = 0 \\ 2a_1 + 4a_2 = 0 \\ 4a_2 + 3a_3 = 0 \end{align}

We will do this, once again, by determining if the following matrix representing this system is invertible $A = \begin{bmatrix}1 & 0 & 2\\ 2 & 4 & 0\\ 0 & 4 & 3 \end{bmatrix}$. Using cofactor expansion along row 1 we have that $\det A = 1 \cdot \begin{vmatrix}4 & 0\\ 4 & 3\ \end{vmatrix} + 2 \cdot \begin{vmatrix}2 & 4\\ 0 & 4\ \end{vmatrix} = 12 + 16 = 28$, and so $A$ is invertible while implies that the system $\begin{bmatrix}1 & 0 & 2\\ 2 & 4 & 0\\ 0 & 4 & 3 \end{bmatrix} \begin{bmatrix}a_1\\ a_2\\ a_3\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ has only the trivial solution $a_1 = a_2 = a_3 = 0$.