Bases of vector spaces in n dimensions

We say that the vectors $\vec{v}_1$ and $\vec{v}_2$ are linearly dependent if there exists non-zero coefficients $\alpha_1, \alpha_2 \in \mathbb{C}$ such that:

$$\alpha_1 \vec{v}_1 + \alpha_2 \vec{v}_2 = \vec{0}.$$

If the only solutions to the above equation are $\alpha_1 = \alpha_2 = 0$, then we say the vectors are linearly independent.

Geometrically, it can be helpful to think of linearly independent vectors as pointing in different directions, whereas dependent vectors lie along the same line or plane, meaning one can be written in terms of the others.

For example, the vectors 

$$\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

are linearly independent since:

$$\alpha_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

is only satisfied if $\alpha_1 = \alpha_2 = 0$.

However, the vectors 

$$\vec{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} i \\ i \end{bmatrix}$$

are linearly dependent since:

$$\alpha_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + \alpha_2 \begin{bmatrix} i \\ i \end{bmatrix} = \begin{bmatrix} \alpha_1 + i\alpha_2 \\ \alpha_1 + i\alpha_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

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