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Since all the u vectors in S and v vectors in T must be linearly independent to one another for set R (which is the union of set S and set T) to be linearly independent, it means the span of S and T must not have any common elements and the intersection is the null set.
It also implies that if l+k>n where n is the dimension number, then there MUST be a vector in S that can be written as a linearly combination of the vectors in T and vice versa by the pigeonhole arguement.
It also implies that if l+k>n where n is the dimension number, then there MUST be a vector in S that can be written as a linearly combination of the vectors in T and vice versa by the pigeonhole arguement.
Date Posted:
1 year ago