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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

Question: [M] The covariance matrix below was obtained from a Landsat image of the Columbia River in Washington, using data from three spectral bands. Let \({x_1},{x_2},{x_3}\) denote the spectral components of each pixel in the image. Find a new variable of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2} + {c_3}{x_3}\) that has maximum possible variance, subject to the constraint that \(c_1^2 + c_2^2 + c_3^2 = 1\). What percentage of the total variance in the data is explained by \({y_1}\)?

\[S = \left[ {\begin{array}{*{20}{c}}{29.64}&{18.38}&{5.00}\\{18.38}&{20.82}&{14.06}\\{5.00}&{14.06}&{29.21}\end{array}} \right]\]

Question: 13. Exercises 12–14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Suppose the equation\(A{\rm{x}} = {\rm{b}}\)is consistent, and let\({{\rm{x}}^ + } = {A^ + }{\rm{b}}\). By Exercise 23 in Section 6.3, there is exactly one vector\({\rm{p}}\)in Row\(A\)such that\(A{\rm{p}} = {\rm{b}}\). The following steps prove that\({{\rm{x}}^ + } = {\rm{p}}\)and\({{\rm{x}}^ + }\)is the minimum length solution of\(A{\rm{x}} = {\rm{b}}\).

  1. Show that \({{\rm{x}}^ + }\) is in Row \(A\). (Hint: Write \({\rm{b}}\) as \(A{\rm{x}}\) for some \({\rm{x}}\), and use Exercise 12.)
  2. Show that\({{\rm{x}}^ + }\)is a solution of\(A{\rm{x}} = {\rm{b}}\).
  3. Show that if \({\rm{u}}\) is any solution of \(A{\rm{x}} = {\rm{b}}\), then \(\left\| {{{\rm{x}}^ + }} \right\| \le \left\| {\rm{u}} \right\|\), with equality only if \({\rm{u}} = {{\rm{x}}^ + }\).

Find the matrix of the quadratic form. Assume x is in \({\mathbb{R}^2}\) .

a. \(3x_1^2 - 4{x_1}{x_2} + 5x_2^2\) b. \(3x_1^2 + 2{x_1}{x_2}\)

Suppose A is invertible and orthogonally diagonalizable. Explain why \({A^{ - {\bf{1}}}}\) is also orthogonally diagonalizable.

(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.

26. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{ - {\bf{18}}}&{{\bf{13}}}&{ - {\bf{4}}}&{\bf{4}}\\{\bf{2}}&{{\bf{19}}}&{ - {\bf{4}}}&{{\bf{12}}}\\{ - {\bf{14}}}&{{\bf{11}}}&{ - {\bf{12}}}&{\bf{8}}\\{ - {\bf{2}}}&{{\bf{21}}}&{\bf{4}}&{\bf{8}}\end{array}} \right)\)
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