Direct sums of subspaces

Jul 19

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.