ladya Posted December 5, 2007 Posted December 5, 2007 Can anyone help me? Let U be a subspace of R3 that coincides with the plane throuogh the origin that is perpendicular to the vector n=(1,1,1) 1. Find an orthonormal basis for U 2. Find the matrix with respect to the canonical basis on R3 of the orthogonal projection (P a linear map from R3 to R3) onto U such that range (P) = U Any help would be greatly appreciated! I dont exactly understand what to do. Thanks!!!
CPL.Luke Posted December 5, 2007 Posted December 5, 2007 think about what your subspace is in this case you have a plane such that the vectors of your subpace have to be orthogonal to the original vector (dot product=0) so you subspace is defined by the eqution c1(1)+c2(1)+c3(1)=0 as you can see c1+c2=-c3, so your talking about vectors where the third component is a linear combination of the first two (of course you could easily say that c1, or c2 were linear combinations of the other two as well, but the main point is that every vector in your subspace is defined by two parameters) now in order to right an orthonormal basis for this you have to find a way of writing the subspace in terms of two vectors which are orthonormal, there are two ways of doing this, one you could chose two arbitrary vecotrs (as long as one wasn't a constant multiple of the other) and use the gram-shmidt orthonormalization procedure, or you could just find two vectors which are naturally perpendicular to each other and normalize them. personally I like (1,1,2) and (-1,1,0) (for this particular instance) which are orthogonal and in the subspace. now you just need to normalize them and you have an orthonormal basis. to find a matrix that maps the vectors into the space just find the orthogonal projections of each basis vector in R^3 into you subspace, these three vectors will form the columns of your matrix.
ladya Posted December 5, 2007 Author Posted December 5, 2007 to find a matrix that maps the vectors into the space just find the orthogonal projections of each basis vector in R^3 into you subspace, these three vectors will form the columns of your matrix. Thank you!! That was a really nice explanation, one more question: About the matrix, which three vectors in R3 are you refering to? Must I map the two vectors that make up my orthonormal basis through the canonical basis? That only leaves me with two columns in my basis. Thank you! That only leaves me with two columns in my basis. Thank you! *** Matrix! not basis... thanks!****
Country Boy Posted December 9, 2007 Posted December 9, 2007 Thank you!! That was a really nice explanation, one more question: About the matrix, which three vectors in R3 are you refering to? Must I map the two vectors that make up my orthonormal basis through the canonical basis? That only leaves me with two columns in my basis. Thank you! *** Matrix! not basis... thanks!**** The standard way of writing a linear transformation from space U to space V, with U having basis {u1, u2, ..., um} and V having basis {v1, v2, ..., vn} is to apply the linear transformation to each of u1, u2, ..., um in order. Write the resulting vector in terms of v1, v2, ..., mn. The coefficients in that linear combination are the columns of the matrix. Here, U= V= R3 and the "canonical basis" is <1, 0, 0>, <0, 1, 0>, <0, 0, 1>. Project <1, 0, 0> into this plane, and write that as <a, b, c> . Then "a b c" will be the first column. Do the same with <0, 1, 0> and <0, 0, 1> to get the second and third columns. Those, the projections of <1, 0, 0>, <0, 1, 0> and <0, 0, 1> in the subspace perpendicular to <1, 1, 1>, are the "3 vectors" Cpl. Luke is talking about. Notice that a projection operator is clearly not invertible (it is not one-to-one) so your matrix will not be invertible and may well have two or three identical columns.
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