Vector decomposition involves breaking down a vector into two or more vectors whose sum equals the original vector. When talking about the **direct sum** of subspaces as in \( R^3 = U \oplus W \), every vector in the space can be uniquely written as a sum of vectors from each subspace.
This can be a very powerful concept in linear algebra, enabling easier calculations and deeper insights.In this exercise:
- Any vector \( v = (x, y, z) \) in \( R^3 \) can be decomposed uniquely into vectors \( u \) and \( w \), such that:
- Vector \( u = (a, a, a) \) belongs to subspace \( U \).
- Vector \( w = (0, b, c) \) belongs to subspace \( W \).
One main aspect of any decomposition is uniqueness, ensuring that no other combinations of values for \( a \), \( b \), and \( c \) will satisfy the conditions for another set \( v = u + w \). Our calculations reveal how specific operations within these rules create that unique decomposition for any vector in the specified space.