Projection (linear algebra)

Projections on normed vector spaces[edit]

When the underlying vector space ${\displaystyle X}$ is a (not necessarily finite-dimensional) normed vector space, analytic questions, irrelevant in the finite-dimensional case, need to be considered. Assume now ${\displaystyle X}$ is a Banach space.


Many of the algebraic results discussed above survive the passage to this context. A given direct sum decomposition of ${\displaystyle X}$ into complementary subspaces still specifies a projection, and vice versa. If ${\displaystyle X}$ is the direct sum ${\displaystyle X=U\oplus V}$, then the operator defined by ${\displaystyle P(u+v)=u}$ is still a projection with range ${\displaystyle U}$ and kernel ${\displaystyle V}$. It is also clear that ${\displaystyle P^{2}=P}$. Conversely, if ${\displaystyle P}$ is projection on ${\displaystyle X}$, i.e. ${\displaystyle P^{2}=P}$, then it is easily verified that ${\displaystyle (1-P)^{2}=(1-P)}$. In other words, ${\displaystyle 1-P}$ is also a projection. The relation ${\displaystyle P^{2}=P}$ implies ${\displaystyle 1=P+(1-P)}$ and ${\displaystyle X}$ is the direct sum ${\displaystyle \operatorname {rg} (P)\oplus \operatorname {rg} (1-P)}$.


However, in contrast to the finite-dimensional case, projections need not be continuous in general. If a subspace ${\displaystyle U}$ of ${\displaystyle X}$ is not closed in the norm topology, then the projection onto ${\displaystyle U}$ is not continuous. In other words, the range of a continuous projection ${\displaystyle P}$ must be a closed subspace. Furthermore, the kernel of a continuous projection (in fact, a continuous linear operator in general) is closed. Thus a continuous projection ${\displaystyle P}$ gives a decomposition of ${\displaystyle X}$ into two complementary closed subspaces: ${\displaystyle X=\operatorname {rg} (P)\oplus \ker(P)=\ker(1-P)\oplus \ker(P)}$.


The converse holds also, with an additional assumption. Suppose ${\displaystyle U}$ is a closed subspace of ${\displaystyle X}$. If there exists a closed subspace ${\displaystyle V}$ such that X = U ⊕ V, then the projection ${\displaystyle P}$ with range ${\displaystyle U}$ and kernel ${\displaystyle V}$ is continuous. This follows from the closed graph theorem. Suppose x_n → x and Px_n → y. One needs to show that ${\displaystyle Px=y}$. Since ${\displaystyle U}$ is closed and {Px_n} ⊂ U, y lies in ${\displaystyle U}$, i.e. Py = y. Also, x_n − Px_n = (I − P)x_n → x − y. Because ${\displaystyle V}$ is closed and {(I − P)x_n} ⊂ V, we have ${\displaystyle x-y\in V}$, i.e. ${\displaystyle P(x-y)=Px-Py=Px-y=0}$, which proves the claim.


The above argument makes use of the assumption that both ${\displaystyle U}$ and ${\displaystyle V}$ are closed. In general, given a closed subspace ${\displaystyle U}$, there need not exist a complementary closed subspace ${\displaystyle V}$, although for Hilbert spaces this can always be done by taking the orthogonal complement. For Banach spaces, a one-dimensional subspace always has a closed complementary subspace. This is an immediate consequence of Hahn–Banach theorem. Let ${\displaystyle U}$ be the linear span of ${\displaystyle u}$. By Hahn–Banach, there exists a bounded linear functional ${\displaystyle \varphi }$ such that φ(u) = 1. The operator ${\displaystyle P(x)=\varphi (x)u}$ satisfies ${\displaystyle P^{2}=P}$, i.e. it is a projection. Boundedness of ${\displaystyle \varphi }$ implies continuity of ${\displaystyle P}$ and therefore ${\displaystyle \ker(P)=\operatorname {rg} (I-P)}$ is a closed complementary subspace of ${\displaystyle U}$.