## Ask Singapore Homework?

Upload a photo of a Singapore homework and someone will email you the solution for free.

### See 1 Answer

*done*{{ upvoteCount }} Upvotes

*clear*{{ downvoteCount * -1 }} Downvotes

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:
11 months ago