, u 2 =, u 3 = Row reducing these equations leads to the system


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1 Solutions HW 5.. Consider the vectors u =, u =, u = in R 4. Can you find a vector u 4 such that u, u, u, u 4 are orthonormal? If so, how many such vectors are there? Note that u, u, u are already orthonormal, so we just need to find u 4, which must satisfy u u 4 = u u 4 = u u 4 = and u 4 u 4 =. If u 4 = (a, b, c, d) T, then the first three conditions give the linear equations a + b + c + d = a + b c d = a b + c d = Row reducing these equations leads to the system a + b + c + d = b + d = c + d = which has general solution u 4 = (a, b, c, d) T = (t, t, t, t) T. Any such vector has dot product zero with u, u, u. For an orthonormal basis, we want u 4 u 4 =, which gives t = ±, so there are choices for u 4, namely or 5.. (See the book for the full statement of the problem, which involves finding a least squares estimate for a line y = mx). Let x be the vector of x coordinates and y be the vector of y coordinates. Geometrically, {mx)} is a dimension vector space. We which to minimize the distance from y into this space; this is accomplished when mx is the projection x of y into this space. Note that x is a unit vector in this space, so the corresponding projection should be (y x x ) x x = y x x x. In particular, we should take m = y x x = =.....
2 r = m x y from our formula. (Note that y / x is the expected slope if all of the points did lie on a line, so r can be thought of as a ratio of two slopes.) If you draw the relevant line in the picture, most points will be near the line (even if none are on the line). 5.. Consider a basis v, v,..., v m of a subspace V of R n. Show that a vector x in R n is orthogonal to v if and only if it is orthogonal to all the vectors v,..., v n. If x is orthogonal to all the vectors v,..., v n, then for any v V, we can write v = c v +... c m v m. Then x v = x (c v +... c m v m ) = c (x v ) + +c m (x v m ) = (since each x v i = ). So x is orthogonal to every vector in V. If x v = for all v V, then this applies in particular to the vectors v,... v m V Complete the proof of Fact 5..4: Orthogonal Projection are linear transformations. We must show that proj V (x + y) = proj V (x) + proj V (y) and proj V (cx) = cproj V (x). From page 89, if u,... u m are an orthonormal basis for V, then proj V (x + y) = (u (x + y))u + + (u m (x + y))u m = ((u x) + (u y))u + + ((u m x)+(u m y))u m = [(u x)u + +(u m x)u m ]+[(u y)u + +(u m y)u m ] = proj V (x) + proj V (y). Similarly proj V (cx) = (u cx)u + + (u m cx)u m = c[(u x)u + + (u m x)u m ] = cproj V (x). 5.. Consider a subspace V or R n and a vector x in R n. Let y = proj V (x). What is the relation ship between y and y x? The two quantities are equal. One can see this by writing x = y + w; then y is the parallel component to V and w is orthogonal to V, and in particular y w =. Then x y = (y + w) y = y y = y as desired. 5.. Perform GramSchmidt on the vectors We have
3 u = v = u = / / / / / / / / = / 5..4 Perform GramSchmidt on the vectors We have 8 u = v = u = v = u = / / / / / / 8 / / / / = / / / / / 5.. Find the QR decomposition of the matrix / / =
4 From the work in problem, we can take Q to be / / / / / / R is calculated by computing various dot products of the column vectors of Q with the column vectors of the original matrix, so we get R to be ( ) 5..8 Find the QR decomposition of the matrix 8 From the work in problem, we can take Q to be / / / / / / / / R is calculated by computing various dot products of the column vectors of Q with the column vectors of the original matrix, so we get R to be 5.. Find an orthonormal basis of the plane x + x + x =. First we find a basis for the plane by backsolving the equation. For example, one such basis is v = v = Next we apply GramSchmidt to this basis to make it orthonormal.
5 u = v = u = / / / / / / / / = / / u, u form an orthonormal basis for the plane x + x + x =. 5.. Is orthogonal? (.8 )...8 Yes. Either one can note that the columns are orthogonal vectors, or one can compute A T A and see that you get the identity matrix. 5.. If A and B are orthogonal matrices, is B AB orthogonal also? Yes. By fact 5..4 a, B is also orthogonal, and then applying Fact 5..4 b several times shows that the product B AB of three orthogonal matrices is an orthogonal matrix. (Alternatively, one can show (B AB) T = (B AB) directly.) 5.. If A and B are symmetric, is AB A symmetric? Yes. (AB A) T = A T (B ) T A T = AB A as required. 5.. If B is an n n matrix, is BB T symmetric? Yes. (BB T ) T = (B T ) T B T = BB T as desired Consider the subspace W of R 4 spanned by the vectors v = v = 9 5 Find the matrix of orthogonal projection onto W.
6 We start by finding an orthonormal basis of W using GramSchmidt. We get u = v = u = / / / / 9 5 / / / / 4 / / / = / So setting Q = We have that the matrix of orthogonal projection onto W is QQ T or Find the dimension of the space of all skew symmetric matrices. The dimension is n n, since such a matrix must have all s on the diagonal, and then is determined by the values in the strictly upper triangular part of the matrix.
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