* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Answers to Even-Numbered Homework Problems, Section 6.2 20
System of linear equations wikipedia , lookup
Rotation matrix wikipedia , lookup
Symmetric cone wikipedia , lookup
Determinant wikipedia , lookup
Matrix (mathematics) wikipedia , lookup
Non-negative matrix factorization wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Gaussian elimination wikipedia , lookup
Eigenvalues and eigenvectors wikipedia , lookup
Vector space wikipedia , lookup
Euclidean vector wikipedia , lookup
Jordan normal form wikipedia , lookup
Perron–Frobenius theorem wikipedia , lookup
Matrix multiplication wikipedia , lookup
Principal component analysis wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Cayley–Hamilton theorem wikipedia , lookup
Singular-value decomposition wikipedia , lookup
Covariance and contravariance of vectors wikipedia , lookup
Matrix calculus wikipedia , lookup
Answers to Even-Numbered Homework Problems, Section 6.2 20. Let u= − 23 1 3 2 3 and v = 1 3 2 3 0 . Since u · v = 0, {u, v} is an orthogonal set. However, kuk2 = u · u = 1 and kvk2 = v · v = 5 6= 1, so {u, v} is not an orthonormal set. The vector v can be normalized, with 9 ṽ = 3 1 v= √ v= kvk 5 √1 5 √2 5 0 , so that {u, ṽ} is an orthonormal set. (Note that u is already a unit vector.) 26. A set of n nonzero orthogonal vectors must be linearly independent by Theorem 4, so if such a sets spans W , it is a basis for W . Since W is therefore an n-dimensional subspace of Rn , it must be equal to Rn itself. 28. If U is an n × n orthogonal matrix, then In = U U −1 = U U T . Since U is the transpose of U T , that is, since (U T )T = U , Theorem 6 applied to U T says that U T has orthogonal columns. In particular, the columns of U T are linearly independent and hence form a basis for Rn , by the Invertible Matrix Theorem. That is, the rows of U form an orthonormal basis for Rn .