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The concept of linear independence is foundational in understanding vector spaces, such as \(R^2\). A set of vectors is considered to be linearly independent if no vector in the set can be written as a linear combination of the others. In practical terms, this means that in a linearly independent set, no vector is 'redundant'—each vector adds a unique direction to the space.

To find out if a set of vectors is linearly independent, we can utilize the method of calculating the determinant for a matrix comprised of those vectors. For instance, in our exercise, we looked at subsets of two vectors and computed their determinants to determine linear independence. A non-zero determinant indicates that the vectors are indeed linearly independent.

Understanding the determinant of a matrix is vital for various areas in linear algebra. Specifically, in the context of our exercise, it helps us identify linear independence among vectors. The determinant is a scalar value that can be computed from the elements of a square matrix and has important implications for the matrix properties.

To calculate the determinant of a 2x2 matrix, which is the case for our subsets in \(R^2\), we use the formula \(ad - bc\), where \(a\), \(b\), \(c\), and \(d\) are the elements of the matrix in positions \(1,1\), \(1,2\), \(2,1\), and \(2,2\) respectively. When the determinant is non-zero, as seen with the subsets \(\{(1,0),(0,1)\}\), \(\{(1,0),(1,1)\}\), and \(\{(0,1),(1,1)\}\), the vectors form a linearly independent set.

A spanning set of a vector space is a specific collection of vectors that provides coverage for the entire space. In other words, any vector in that space can be expressed as a linear combination of the vectors within the spanning set. For our exercise involving \(R^2\), a set of vectors must have at least two linearly independent vectors to span this two-dimensional space.

This implies that each point in \(R^2\) can be reached by scaling and combining these two vectors in various ways. The subsets we verified as linearly independent in the prior step are also capable of spanning \(R^2\), as any vector in this space can be represented using these subsets, thus meeting both criteria for a basis of the vector space.