Let F be a subfield of the complex numbers. In F³ the vectors

$$\begin{aligned} \alpha_{1} & =(1,0,2) \\ \alpha_{2} & =(1,1,-1) \\ \alpha_{3} & =(2,-2,0) \\ \alpha_{4} & =(1,3,3) \end{aligned}$$

are linearly dependent, since

2α₁ + α₂ − α₃ − α₄ = 0.

Think back to arrows in three dimension, if we choose a coordinate system. We will need three coordinates to specify the direction of the arrow. This nicely corresponds to three 3-tuples in F³. Now when we try to add another 3-tuple, it will have to fall into the span of the three 3-tuples we already have. It is therefore linearly dependent with the original three 3-tuples. To make it linearly independent, we will have to make all the n-tuples 4-tuples, and the added 4-tuple needs to have a component perpendicular to the three dimensional space we have. We now have an intuition from geometry that if we have more element in the basis then the dimension of space, we necessarily have a linearly dependent basis.

Now obviously: $$\begin{aligned} & \epsilon_{1}=(1,0,0) \\ & \epsilon_{2}=(0,1,0) \\ & \epsilon_{3}=(0,0,1) \end{aligned}$$

are linearly independent: to linearly combine them into 0, they each have to be scaled by 0. In fact, {ϵ₁, ϵ₂} is linearly independent by consequence [consequence:subset_of_linearly_independent_set] above.

Let P be an invertible n × n matrix with entries in the field F. Let P₁, …, P_n be the columns of P, $$\begin{bmatrix} \begin{bmatrix}\\\\P_1\\\\\\ \end{bmatrix} \begin{bmatrix}\\\\P_2\\\\\\ \end{bmatrix} \cdots \begin{bmatrix}\\\\P_n\\\\\\ \end{bmatrix} \end{bmatrix}$$ They form a basis for the space of column matrices, F^n × 1. $$\begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \\ \end{bmatrix}$$To see that they indeed form a basis, we need them to be linearly independent and span the space F^n × 1. We use the properties of an inverse matrix here, first we write linear combination of column vectors P_k as the matrix P multiply by a column matrix X is a column matrix,

x₁P₁ + ⋯ + x_nP_n = PX.

But P is an invertible matrix, so PX = 0 has only the trivial solution X = 0, and {P₁,…,P_n} is a linearly independent set. To see that it spans F_n × 1, if let Y be any column matrix, we will want to write it as a linear combination of the set {P₁,…,P_n}. But P has a inverse, so we can always get another column matrix X such that X = P⁻¹Y. Then Y = PX, that is,

Y = x₁P₁ + ⋯ + x_nP_n.

So {P₁,…,P_n} is a basis for F^n × 1. This gives us the [cols_rows_independent] and [cols_rows_span] of our equivalence theorem.

Basis for solution space of a system of linear equations Back to solving m linear equations with n unknowns, we hope to find a basis for the solution space so we can express the solution as a linear combination of the bases elements. Let A be an m × n matrix and let S be the solution space for the homogeneous system AX = 0 (Example 7). Let R be a rowreduced echelon matrix which is row-equivalent to A. Then S is also the solution space for the system RX = 0. If R has r non-zero rows, then the system of equations RX = 0 simply expresses r of the unknowns x₁, …, x_n in terms of the remaining (n−r) unknowns x_j. Suppose that the leading non-zero entries of the non-zero rows occur in columns k₁, …, k_r. Let J be the set consisting of the n − r indices different from k₁, …, k_r :

J = {1, …, n} − {k₁,…,k_r}.

The system RX = 0 has the form

$$\begin{aligned} & x_{k 1}+\sum_{J} c_{1 j} x_{j}=0 \\ & \vdots \begin{array}{cc}\vdots & \vdots \\x_{k_{r}}+\sum_{J} c_{r j} x_{j}=0\end{array} \end{aligned}$$

where the c_ij are certain scalars. All solutions are obtained by assigning (arbitrary) values to those x_j ’s with j in J and computing the corresponding values of x_k1, …, x_kr. For each j in J, let E_j be the solution obtained by setting x_j = 1 and x_i = 0 for all other i in J. We assert that the (n−r) vectors E_j, j in J, form a basis for the solution space.

Since the column matrix E_j has a 1 in row j and zeros in the rows indexed by other elements of J, the reasoning of Example 13 shows us that the set of these vectors is linearly independent. That set spans the solution space, for this reason. If the column matrix T, with entries t₁, …, t_n, is in the solution space, the matrix

N = ∑_Jt_jE_j

is also in the solution space and is a solution such that x_j = t_j for each j in J. The solution with that property is unique; hence, N = T and T is in the span of the vectors E_j.