Let's say we have a set of linear vectors \(V\) in a 3-dimensional space, where \(V = \{v_1, v_2, v_3, v_4\}\). The vectors are defined as follows: \(v_1 = (1, 0, 0)\), \(v_2 = (0, 1, 0)\), \(v_3 = (0, 0, 1)\) and \(v_4 = (1, 1, 1)\). What is the cardinality of the base of vector space that these vectors span?

Step 1 ：First, we need to find out if the vectors \(v_1, v_2, v_3, v_4\) are linearly independent. This can be done by setting up the following equation: \(a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 = 0\), where \(a_1, a_2, a_3, a_4\) are scalars.

Step 2 ：Substituting the vectors into the equation gives us: \(a_1(1, 0, 0) + a_2(0, 1, 0) + a_3(0, 0, 1) + a_4(1, 1, 1) = (0, 0, 0)\). This simplifies to: \((a_1 + a_4, a_2 + a_4, a_3 + a_4) = (0, 0, 0)\).

Step 3 ：Setting each component of the vector equal to 0 gives us the following system of equations: \(a_1 + a_4 = 0\), \(a_2 + a_4 = 0\), \(a_3 + a_4 = 0\). Solving this system, we find that \(a_4 = 0\), \(a_1 = 0\), \(a_2 = 0\), and \(a_3 = 0\).

Step 4 ：Since the only solution to this system of equations is the trivial solution where all \(a_i\) are 0, we can conclude that the vectors \(v_1, v_2, v_3, v_4\) are linearly independent.

Step 5 ：The base of a vector space is a set of linearly independent vectors that span the space. Therefore, the cardinality of the base of the vector space that these vectors span is 4.