Raising the bar

Direct sums of subspaces

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

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

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

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

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

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

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

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