30
Jul 19

Direct sums of subspaces

category: Matrix algebra, MT2175 Further linear algebra No Comments

Direct sums of subspaces

The definition of an orthogonal sum L=L_{1}\oplus L_{2} requires two things: 1) every element l\in L can be decomposed as l=l_{1}+l_{2} with l_{i}\in L_{i}, i=1,2, and 2) every element of L_{1} is orthogonal to every element of L_{2}. Orthogonality of L_{1} to L_{2} implies L_{1}\cap L_{2}=\{0\} which, in turn, guarantees uniqueness of the representation l=l_{1}+l_{2}. If we drop the orthogonality requirement but retain 1) and L_{1}\cap L_{2}=\{0\}, we get the definition of a direct sum.

Definition. Let L_{1},L_{2} be two subspaces such that L_{1}\cap L_{2}=\{0\}. The set L=\{l_{1}+l_{2}: l_{i}\in L_{i}, i=1,2\} is called a direct sum of L_{1},L_{2} and denoted L=L_{1}\dotplus L_{2}.  The condition L_{1}\cap L_{2}=\{0\} provides uniqueness of the representation l=l_{1}+l_{2}.

Exercise 1. Let L=L_{1}\dotplus L_{2}. If l\in L is decomposed as l=l_{1}+l_{2} with l_{i}\in L_{i}, i=1,2, define P_{1}l=l_{1}. Then P_{1} is linear and satisfies P_{1}^{2}=P_{1}, \text{Img}(P_{1})=L_{1}, N(P_{1})=L_{2}.

Under conditions of Exercise 1, P_{1} is an oblique projector of L onto L_{1} parallel to L_{2}.

Exercise 2. Prove that dimension additivity extends to direct sums: if L=L_{1}\dotplus L_{2}, then \dim L=\dim L_{1}+\dim L_{2}.

Exercise 3. Let L_{1},L_{2} be two subspaces of C^{n} and suppose n=\dim L_{1}+\dim L_{2}. Then to have C^{n}=L_{1}+L_{2} it is sufficient to check that L_{1}\cap L_{2}=\{0\}, in which case L=L_{1}\dotplus L_{2}

Proof. Denote L=L_{1}\dotplus L_{2}. By Exercise 2, \dim L=\dim L_{1}+\dim L_{2}=n. If C^{n}\setminus L is not empty, then we can complete a basis in L with a nonzero vector from C^{n}\setminus L to see that \dim C^{n}\geq n+1 which is impossible.

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