Let the four vectors #k_1,k_2,k_3# and #k_4# form a basis of the vector space #K#. Since #K# is a subspace of #V#, these four vectors form a linearly independent set in #V#. Since #L# is a subspace of #V# different from #K#, there must be at least one element, say #l_1# in #L#, which is not in #K#, i.e, which is not a linear combination of #k_1,k_2,k_3# and #k_4#.

So, the set #{k_1,k_2,k_3,k_4,l_1}# is a linear independent set of vectors in #V#. Thus the dimensionality of #V# is at least 5!

In fact, it is possible for the span of #{k_1,k_2,k_3,k_4,l_1}# to be the entire vector space #V# - so that the minimum number of basis vectors must be 5.

Just as an example, let #V# be #RR^5# and let #K# and #V# consists of vectors of the forms

#((alpha),(beta),(gamma),(delta),(0)) # and #((mu),(nu),(lambda),(0),(phi)) #

It is easy to see that the vectors

#((1),(0),(0),(0),(0))#,#((0),(1),(0),(0),(0))#,#((0),(0),(1),(0),(0))#and #((0),(0),(0),(0),(0))#

form a basis of #K#. Append the vector #((0),(0),(0),(0),(0) )#, and you will get a basis for the entire vector space,