STAT714 LINEAR STATISTICAL MODELS Fall Session 2009
Handout 2: More on Projection matrices
Result M.15. If Mis a perpendicular projection matrix onto C(X), then C(M) = C(X).
Proof. We need to show that C(M)โ C(X) and C(X)โ C(M). Suppose that vโ C(M).
Then, v=Mb, for some b. Now, write b=b1+b2, where b1โ C(X) and b2โฅC(X).
Thus, v=Mb =M(b1+b2) = Mb1+Mb2=b1โ C(X). Thus, C(M)โ C(X). Now
suppose that vโ C(X). Since Mis a perpendicular projection matrix onto C(X), we know
that v=Mv =M(v1+v2), where v1โ C(X) and v2โฅC(X). But, M(v1+v2) = Mv1,
showing that vโ C(M). Thus, C(X)โ C(M) and the result follows. ๎
Result M.16. Perpendicular projection matrices are unique.
Proof. Suppose that M1and M2are both perpendicular projection matrices onto any
arbitrary subspace S โ Rn. Let vโ Rnand write v=v1+v2, where v1โ S and v2โฅS .
Since vis arbitrary and M1v=v1=M2v, we have M1=M2.๎
Result M.17. If Mis the perpendicular projection matrix onto C(X), then IโMis the
perpendicular projection matrix onto N(XT).
Sketch of Proof.IโMis symmetric and idempotent so that IโMis the perpendicular
projection matrix onto C(IโM). Show that C(IโM) = N(XT) and use Result M.16. ๎