Let
Show that .
Essentially, this problem is asking us to show the distributive law by definition. This is usually a theorem in the textbook, so it is good enough to understand the proof first and then prove it again by yourself.
Fortunately, there are some special cases that is not too difficult.
The first special case is when is linearly dependent. In this case, we have , which easily make the equality hold.
The other special case is when both and are multiple of a vector . We may assume and . By definition, , , and , where
Thus, the equality holds again.
It turns out these two special cases with some extra steps are enough for us to prove the whole statement.
First, when is linearly dependent, , so obvsiouly .
Suppose is linearly independent. Then we may extend it into a basis of [1]. Thus, every vector in can be written as a linear combination of [2]. We may assume
for some real numbers and . Consequently, we have
Applying appropriate row operations to , , and , we have
and
By writing
we have
This completes the proof.
This note can be found at Course website > Learning resources.
Every independent set can be extended to a basis such that . Review Hefferon Two.III for the definition and properties of a basis, in particular, Corollary 2.12. ↩︎
If is a basis of the vector space , then every vector in can be uniquely written as a linear combination of . Review Hefferon Two.III for the definition and properties of a basis, in particular, Theorem 1.12. ↩︎