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Chapter 7: Q7.4-15E (page 395)

Question: Suppose the factorization below is an SVD of a matrix A, with the entries in U and Vrounded to two decimal places.

\(A = \left( {\begin{array}{*{20}{c}}{.{\bf{40}}}&{ - .{\bf{78}}}&{.{\bf{47}}}\\{.{\bf{37}}}&{ - .{\bf{33}}}&{ - .{\bf{87}}}\\{ - .{\bf{84}}}&{ - .{\bf{52}}}&{ - .{\bf{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\bf{7}}{\bf{.10}}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{{\bf{3}}{\bf{.10}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}{.{\bf{30}}}&{ - .{\bf{51}}}&{ - .{\bf{81}}}\\{.{\bf{76}}}&{.{\bf{64}}}&{ - .{\bf{12}}}\\{.{\bf{58}}}&{ - .{\bf{58}}}&{.{\bf{58}}}\end{array}} \right)\)

a. What is the rank of A?

b. Use this decomposition of A, with no calculations, to write a basis for ColA and a basis for NulA. (Hint: First write the columns of V.)

Short Answer

Expert verified

a. The rank of matrix A is 2.

b. The basis of ColA is \(\left\{ {\left( {\begin{array}{*{20}{c}}{.40}\\{.37}\\{ - .84}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - .78}\\{ - .33}\\{ - .52}\end{array}} \right)} \right\}\) and the basis of Null space is \(\left( {\begin{array}{*{20}{c}}{.58}\\{ - .58}\\{.58}\end{array}} \right)\).

Step by step solution

01

Compare the given SVD to find the matrices

(a). On comparing the given SVD with the equation \(A = U\Sigma {V^T}\).

\(U = \left( {\begin{array}{*{20}{c}}{.40}&{ - .78}&{.47}\\{.37}&{ - .33}&{ - .87}\\{ - .84}&{ - .52}&{ - .16}\end{array}} \right)\), \(\Sigma = \left( {\begin{array}{*{20}{c}}{7.10}&0&0\\0&{3.10}&0\\0&0&0\end{array}} \right)\) and \(V = \left( {\begin{array}{*{20}{c}}{.30}&{.76}&{.58}\\{ - .51}&{.64}&{ - .58}\\{ - .81}&{ - .58}&{.58}\end{array}} \right)\)

As the matrix \(\Sigma \) has two nonzero singular values, therefore the rank of the matrix is 2.

02

Find the basis for NulA

(b). The orthogonal basis of column space of A is:

\(\left( {{{\bf{u}}_1},{{\bf{u}}_2}} \right) = \left\{ {\left( {\begin{array}{*{20}{c}}{.40}\\{.37}\\{ - .84}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - .78}\\{ - .33}\\{ - .52}\end{array}} \right)} \right\}\)

The basis for Nullspace is:

\(\begin{array}{c}V = \left( {{{\bf{v}}_3}} \right)\\ = \left( {\begin{array}{*{20}{c}}{.58}\\{ - .58}\\{.58}\end{array}} \right)\end{array}\)

Thus, the basis of \({\rm{Col}}A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}{.40}\\{.37}\\{ - .84}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - .78}\\{ - .33}\\{ - .52}\end{array}} \right)} \right\}\) and the basis of Null space is \(\left( {\begin{array}{*{20}{c}}{.58}\\{ - .58}\\{.58}\end{array}} \right)\).

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Most popular questions from this chapter

Construct a spectral decomposition of A from Example 2.

(M) Compute an SVD of each matrix in Exercises 26 and 27. Report the final matrix entries accurate to two decimal places. Use the method of Examples 3 and 4.

27. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{6}}&{ - {\bf{8}}}&{ - {\bf{4}}}&{\bf{5}}&{ - {\bf{4}}}\\{\bf{2}}&{\bf{7}}&{ - {\bf{5}}}&{ - {\bf{6}}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{1}}}&{ - {\bf{8}}}&{\bf{2}}&{\bf{2}}\\{ - {\bf{1}}}&{ - {\bf{2}}}&{\bf{4}}&{\bf{4}}&{ - {\bf{8}}}\end{array}} \right)\)

In Exercises 17鈥�24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) ,

\(\)

where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) 鈥渄iagonal鈥� matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

18. Suppose \(A\) is square and invertible. Find a singular value decomposition of \({A^{ - 1}}\)

Question 7: Prove that an \(n \times n\) A is positive definite if and only if A admits a Cholesky factorization, namely, \(A = {R^T}R\) for some invertible upper triangular matrix R whose diagonal entries are all positive. (Hint; Use a QR factorization and Exercise 26 in Section 7.2.)

Construct a spectral decomposition of A from Example 3.
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