Why is the orthogonal matrix of C divied by the square root of 2? In other quest
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Why is the orthogonal matrix of C divied by the square root of 2? In other questions it's divied by the square root of 5, why? Where does this number come from?
Here's the link: http://www.chegg.com/homework-help/Linear-Algebra-with-Applications-8th-edition-chapter-5.3-problem-6E-solution-9781449679545
Chapter 5.3, Problem 6E Show all steps ON K Chapter A Chapter E Chapter D.1 Chapter D.10 Chapter D.11 Chapter D.12 Chapter D.13 Chapter D.14 Chapter D.15 Chapter D.16 Chapter D.17 Chapter D.18 Chapter D.19 Chapter D.2 Chapter D.20 Chapter D.21 Chapter D.22 Chapter D.23 Chapter D.24 Chapter D.25 Chapter D.26 (a) Let us consider the following symmetric matrix A square matrix A is said to be orthogonally diagonalizable if it is a symmetric matrix If matrix A has two linearly independent eigenvectors, it is said that it is diagonalizable Therefore it is required to find the eigenvectors of matrix A Let us derive the characteristic polynomial of A We get 21110 1- 21- The characteristic polynomial of A is -14-(1-1)(1-1)-2x2 = -22+1-4 =2-22-3 Comment Step 2 of 24 A Now, we solve the characteristic polynomial of A 2-22-3-0 2-32+2-3-0 This implies,Explanation / Answer
A square matrix A or order n is said to be orthogonally diagonalizable if there exists an orthogonal matrix P such that D = PT A P is a diagonal matrix. Further, since PT = P-1 for orthogonal matrix P, hence we also have D = P-1 A P. The entries on the leading diagonal of D are eigenvalues of A, and the columns of P are corresponding eigenvectors. The only difference is the additional requirement that P be orthogonal, which is equivalent to the fact that those eigenvectors i.e. the columns of P form an orthonormal basis of Rn.
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