Math 307: Problems for section 3.2 November 9, 2010 1.Show that any2×2orthogonal matrix is either a rotation matrix or a reflection matrix. To obtain all possible 2×2 orthogonal matrices, we need to find all possible orthonormal bases to use as their columns. The first vector in an orthonormal basis is a vector of length one that can be written bracketleftbiggcos(θ) sin(θ) bracketrightbigg for someθ∈[0,2π). The next vector is one of the two vectors of unit length orthogonal to this one, eitherbracketleftbigg−sin(θ)cos(θ)bracketrightbiggorbracketleftbiggsin(θ) −cos(θ) bracketrightbigg . This implies that every orthogonal matrix is either of the formbracketleftbiggcos(θ)−sin(θ)sin(θ)cos(θ)bracketrightbigg(a rotation matrix) orbracketleftbiggcos(θ)sin(θ) sin(θ)−cos(θ) bracketrightbigg (a relection matrix). 2.LetQ=bracketleftBiggq1 |q2| · · · |qkbracketrightBigg where q1,q2, . . .qk ∈Rnform an orthonormal set.(That is, they satisfybardblqibardbl= 1fori= 1, . . . , kand qi·qj = 0ifinegationslash =j, but there might not be enough vectors to form a basis, i.e., possiblyk < n). Show thatQQT is the projection onto the subspace spanned by q1,q2, . . .qk. What isQTQ? Show that the projection p of a vector v onto the subspace spanned by q1,q2, . . .qkcan be written p=∑ki=1qiqT iv=∑ k i=1 (qi·v)qi . QTQ=Ik (thek×kidentity matrix).ThusQQTis the same asQ(QTQ)-1QT.But this is the formula for the projection onto the range ofQ, that is, onto the subspace spanned byq1,q2, . . .qk . The formulap=QQTvis equivalent top=∑ki=1qiqTi v=∑k i=1(qi·v)qi .
