In $\mathbb{R}^n$, the standard basis vectors $e_1 = (1, 0, \ldots, 0)$, $e_2 = (0, 1, \ldots, 0)$, $\ldots$, $e_n = (0, 0, \ldots, 1)$ are linearly independent.

If $a_1 e_1 + \cdots + a_n e_n = 0$, then $(a_1, a_2, \ldots, a_n) = (0, 0, \ldots, 0)$.

$\{(1, 2), (3, 4)\}$ is linearly independent. Suppose $a(1, 2) + b(3, 4) = (0, 0)$:

$a + 3b = 0, \quad 2a + 4b = 0.$

From the first equation, $a = -3b$. Substituting: $-6b + 4b = -2b = 0$, so $b = 0$ and $a = 0$.

$\{(1, 2, 3), (4, 5, 6), (5, 7, 9)\}$ is linearly dependent because

$(1, 2, 3) + (4, 5, 6) = (5, 7, 9).$

Equivalently, $1 \cdot (1,2,3) + 1 \cdot (4,5,6) + (-1) \cdot (5,7,9) = (0,0,0)$, a nontrivial relation.

$\{1, x, x^2, \ldots, x^n\}$ is linearly independent in $P(F)$. If $a_0 + a_1 x + \cdots + a_n x^n = 0$ as a polynomial function, then all coefficients must be zero (when $F$ is infinite, or by comparing coefficients formally).

$\{\sin x, \cos x\}$ is linearly independent in $C(\mathbb{R})$. If $a\sin x + b\cos x = 0$ for all $x$:

• Setting $x = 0$: $b = 0$.
• Setting $x = \pi/2$: $a = 0$.

$\{e^x, e^{2x}\}$ is linearly independent in $C^\infty(\mathbb{R})$. If $ae^x + be^{2x} = 0$ for all $x$, then dividing by $e^x$ gives $a + be^x = 0$ for all $x$, which forces $b = 0$ (let $x \to \infty$) and then $a = 0$.

More generally, $\{e^{\lambda_1 x}, e^{\lambda_2 x}, \ldots, e^{\lambda_n x}\}$ is linearly independent whenever $\lambda_1, \ldots, \lambda_n$ are distinct.

In $M_{2 \times 2}(\mathbb{R})$, the matrices

$A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad A_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad A_3 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

are linearly independent: $aA_1 + bA_2 + cA_3 = \begin{pmatrix} a & b \\ b & c \end{pmatrix} = 0$ implies $a = b = c = 0$.

$\{1, |x|\}$ is linearly independent in $F(\mathbb{R}, \mathbb{R})$. If $a \cdot 1 + b|x| = 0$ for all $x$:

• $x = 0$: $a = 0$.
• $x = 1$: $b = 0$.

If $0 \in \{v_1, \ldots, v_n\}$, the set is linearly dependent. Say $v_1 = 0$. Then $1 \cdot v_1 + 0 \cdot v_2 + \cdots + 0 \cdot v_n = 0$ is a nontrivial relation.

If $\{v_1, \ldots, v_n\}$ is linearly dependent and $w \in V$, then $\{v_1, \ldots, v_n, w\}$ is also linearly dependent: the existing nontrivial relation extends with coefficient $0$ for $w$.

If $\{v_1, \ldots, v_n\}$ is linearly independent, then every nonempty subset is linearly independent. (Any nontrivial relation in the subset would extend to a nontrivial relation in the full set.)

$\{1, i\}$ is linearly independent in $\mathbb{C}$ viewed as a vector space over $\mathbb{R}$, but linearly dependent in $\mathbb{C}$ over $\mathbb{C}$ (since $i = i \cdot 1$). The field matters.

For $n$ times differentiable functions $f_1, \ldots, f_n$, the Wronskian is

$W(f_1, \ldots, f_n)(x) = \det \begin{pmatrix} f_1(x) & \cdots & f_n(x) \\ f_1'(x) & \cdots & f_n'(x) \\ \vdots & \ddots & \vdots \\ f_1^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{pmatrix}.$

If $W \neq 0$ at some point, then $f_1, \ldots, f_n$ are linearly independent.

For example, $W(\sin x, \cos x) = \sin x \cdot (-\sin x) - \cos x \cdot \cos x = -1 \neq 0$.