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Consider a symmetric 3x3matrix Awith eigenvalues 1,2and 3how many different orthogonal matricessare there such thatS-1ASis diagonal?

Short Answer

Expert verified

Thus, the different orthogonal matrices are 48

Step by step solution

01

The Orthogonal Matrix

An orthogonal matrix, also known as an orthonormal matrix, is a real square matrix whose columns and rows are orthonormal vectors in linear algebra.

Where QTis the transpose of Q and l is the identity matrix is one approach to describe this.

02

Determine the orthogonal Matrices

Since A has three distinct eigenvalues, then we have possible distinct orthonormal eigenvectors corresponding to these eigenvalues:

±v⇶Ä1,±v⇶Ä2,and±v⇶Ä3

Now, when obtaining the orthogonal matrix S, the first column would have 6 choices, moving to the second column we will have 4 choices (since we can't have a corresponding eigenvector to the same eigenvalue used in the first column), and finally moving to the third column we will have 2 choices left.

Therefore 6x4x2=48

The different orthogonal matrices are 48.

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Most popular questions from this chapter

Consider a singular value decomposition A=UΣVTof an n×mmatrix Awith rankA=m. Let u→1,…,u→nbe the columns of U. Without using the results of Chapter 5 , computeA(ATA)-1ATu→i. Explain your result in terms of Theorem 5.4.7.

A Cholesky factorization of a symmetric matrix A is a factorization of the formA=LLTwhere L is lower triangular with positive diagonal entries.

Show that for a symmetricn×nmatrix A, the following are equivalent:

(i) A is positive definite.

(ii) All principal submatricesrole="math" localid="1659673584599" A(m)of A are positive definite. See

Theorem 8.2.5.

(iii)det(Am)>0form=1,....,n

(iv) A has a Cholesky factorization A=LLT

Hint: Show that (i) implies (ii), (ii) implies (iii), (iii) implies (iv), and (iv) implies (i). The hardest step is the implication from (iii) to (iv): Arguing by induction on n, you may assume that A(n-1)) has a Cholesky factorization A(n-1)=BBT. Now show that there exist a vector x→inRn-1and a scalar t such that

A=[An-1v→v→Tk]=[B0x→T1][BTx→0t]

Explain why the scalar t is positive. Therefore, we have the Cholesky factorization

A=[B0x→Tt][BTx→0t]

This reasoning also shows that the Cholesky factorization of A is unique. Alternatively, you can use the LDLT factorization of A to show that (iii) implies (iv).See Exercise 5.3.63.

To show that (i) implies (ii), consider a nonzero vector, and define

role="math" localid="1659674275565" y→=[x→0M0]

In Rn(fill in n − m zeros). Then

role="math" localid="1659674437541" x→A(m)x→=y→TAy→>0

Show that an SVDA=UΣVT

can be written asA=σ1u→1v→1T+…+σru→rv→rT.

Consider an invertible n × n matrix A. What is the relationship between the matrix R in the QR factorization of A and the matrix L in the Cholesky factorization ofATA?

Determine the definiteness of the quadratic forms in Exercises 4 through 7.

q(x1,x2)=x12+4x1x2+x22

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