A^2,is a orthogonal matrix,is A a orthogonal matrix? If this is not true ,in which case it is true?
M muzhiqingfeng New member Joined Dec 25, 2009 Messages 8 Dec 27, 2009 #1 A^2,is a orthogonal matrix,is A a orthogonal matrix? If this is not true ,in which case it is true?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Dec 28, 2009 #2 The product of orthogonal matrices is orthogonal They preserve orthogonality and thus form a group. So, if A is orthogonal, then A^2 is also orthogonal. Assume that ∣∣Ax∣∣=∣∣x∣∣\displaystyle ||Ax||=||x||∣∣Ax∣∣=∣∣x∣∣ for all x in R^n. So, we have Ax⋅Ay=14∣∣Ax+Ay∣∣2−14∣∣Ax−Ay∣∣2=14∣∣A(x+y)∣∣2−14A(x−y)∣∣2=14∣∣x+y∣∣2−14∣∣x−y∣∣2=x⋅y\displaystyle Ax\cdot Ay=\frac{1}{4}||Ax+Ay||^{2}-\frac{1}{4}||Ax-Ay||^{2}=\frac{1}{4}||A(x+y)||^{2}-\frac{1}{4}A(x-y)||^{2}=\frac{1}{4}||x+y||^{2}-\frac{1}{4}||x-y||^{2}=x\cdot yAx⋅Ay=41∣∣Ax+Ay∣∣2−41∣∣Ax−Ay∣∣2=41∣∣A(x+y)∣∣2−41A(x−y)∣∣2=41∣∣x+y∣∣2−41∣∣x−y∣∣2=x⋅y ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Let's say Rm→W\displaystyle R^{m}\rightarrow WRm→W be the orthogonal projection of Rm\displaystyle R^{m}Rm onto a subspace W. Let's say we have [P]=A(ATA)−1AT\displaystyle [P]=A(A^{T}A)^{-1}A^{T}[P]=A(ATA)−1AT where A is any matrix formed using a set of basis vectors for W as its column vectors. Therefore, [P]2=[A(ATA)−1AT][A(ATA)−1AT]\displaystyle [P]^{2}=[A(A^{T}A)^{-1}A^{T}][A(A^{T}A)^{-1}A^{T}][P]2=[A(ATA)−1AT][A(ATA)−1AT] =A[(ATA)−1(ATA)](ATA)−1AT\displaystyle =A[(A^{T}A)^{-1}(A^{T}A)](A^{T}A)^{-1}A^{T}=A[(ATA)−1(ATA)](ATA)−1AT =A(ATA)−1AT\displaystyle =A(A^{T}A)^{-1}A^{T}=A(ATA)−1AT =[P]\displaystyle =[P]=[P] Does that help?.
The product of orthogonal matrices is orthogonal They preserve orthogonality and thus form a group. So, if A is orthogonal, then A^2 is also orthogonal. Assume that ∣∣Ax∣∣=∣∣x∣∣\displaystyle ||Ax||=||x||∣∣Ax∣∣=∣∣x∣∣ for all x in R^n. So, we have Ax⋅Ay=14∣∣Ax+Ay∣∣2−14∣∣Ax−Ay∣∣2=14∣∣A(x+y)∣∣2−14A(x−y)∣∣2=14∣∣x+y∣∣2−14∣∣x−y∣∣2=x⋅y\displaystyle Ax\cdot Ay=\frac{1}{4}||Ax+Ay||^{2}-\frac{1}{4}||Ax-Ay||^{2}=\frac{1}{4}||A(x+y)||^{2}-\frac{1}{4}A(x-y)||^{2}=\frac{1}{4}||x+y||^{2}-\frac{1}{4}||x-y||^{2}=x\cdot yAx⋅Ay=41∣∣Ax+Ay∣∣2−41∣∣Ax−Ay∣∣2=41∣∣A(x+y)∣∣2−41A(x−y)∣∣2=41∣∣x+y∣∣2−41∣∣x−y∣∣2=x⋅y ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Let's say Rm→W\displaystyle R^{m}\rightarrow WRm→W be the orthogonal projection of Rm\displaystyle R^{m}Rm onto a subspace W. Let's say we have [P]=A(ATA)−1AT\displaystyle [P]=A(A^{T}A)^{-1}A^{T}[P]=A(ATA)−1AT where A is any matrix formed using a set of basis vectors for W as its column vectors. Therefore, [P]2=[A(ATA)−1AT][A(ATA)−1AT]\displaystyle [P]^{2}=[A(A^{T}A)^{-1}A^{T}][A(A^{T}A)^{-1}A^{T}][P]2=[A(ATA)−1AT][A(ATA)−1AT] =A[(ATA)−1(ATA)](ATA)−1AT\displaystyle =A[(A^{T}A)^{-1}(A^{T}A)](A^{T}A)^{-1}A^{T}=A[(ATA)−1(ATA)](ATA)−1AT =A(ATA)−1AT\displaystyle =A(A^{T}A)^{-1}A^{T}=A(ATA)−1AT =[P]\displaystyle =[P]=[P] Does that help?.