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Chapter 6: Q6.2-36E (page 331)

Question: In parts (a)-(d), let U be the matrix formed by normalizing each column of matrix A in Exercise 35.

  1. Compute \({U^T}U\) and \(U{U^T}\). How do they differ?
  2. Generate a random vector y in\({\mathbb{R}^{\bf{8}}}\), and compute\({\bf{p}} = U{U^T}{\bf{y}}\)and\({\bf{z}} = {\bf{y}} - {\bf{p}}\). Explain why p is in col A. Verfiy that z is orthogonal to p.
  3. Verfiy that z is orthogonal to each column of U.
  4. Notice that \({\bf{y}} = {\bf{p}} + {\bf{z}}\), with p in Col A. Explain why z is in \({\left( {{\bf{Col}}A} \right)^ \bot }\). (The significance of this decomposition of y will be explained in the next section.)

Short Answer

Expert verified
  1. The matrices are \({U^T}U = \left( {\begin{array}{*{20}{c}}{1.0000}&0&0&0\\0&{1.0000}&0&0\\0&0&{1.0000}&0\\0&0&0&{1.0000}\end{array}} \right)\), and\(U{U^T} = \left( {\begin{array}{*{20}{c}}{0.8200}&0&{ - 0.2000}&{0.0800}\\0&{0.4200}&{0.2400}&0\\{ - 0.2000}&{0.2400}&{0.5800}&{0.2000}\\{0.0800}&0&{0.2000}&{0.8200}\\{0.0600}&{ - 0.2000}&0&{0.2400}\\{0.2000}&{0.0600}&{0.3200}&{ - 0.2000}\\{0.2400}&{0.2000}&0&{0.0600}\\0&{ - 0.3200}&{0.0600}&0\end{array}} \right)\). The matrix \({U^T}U\) is of order \(4 \times 4\) , and the matrix \(U{U^T}\) is of order \(8 \times 8\).
  1. It is verified that z and p are orthogonal to each other.
  2. It is verified that z is orthogonal to each column of U.
  3. z is in the \({\left( {{\rm{Col}}A} \right)^ \bot }\) because the product of vectors is zero.

Step by step solution

01

Find the answer for part (a)

Use the following MATLAB command to obtain the transpose of A.

\( > > A = \left( \begin{array}{l}\begin{array}{*{20}{c}}{ - 6}&{ - 3}&6&1\end{array};\,\begin{array}{*{20}{c}}{ - 1}&2&1&{ - 6}\end{array};\,\begin{array}{*{20}{c}}3&6&3&{ - 2;\,\begin{array}{*{20}{c}}6&{ - 3}&6&{ - 1}\end{array}}\end{array};\\\begin{array}{*{20}{c}}2&{ - 1}&2&{3;\,\begin{array}{*{20}{c}}{ - 3}&6&3&{2;\,\,\begin{array}{*{20}{c}}{ - 2}&{ - 1}&2&{ - 3;\,\,\begin{array}{*{20}{c}}1&2&1&6\end{array}}\end{array}}\end{array}}\end{array}\end{array} \right);\)

Find the matrix \(U\).

\( > > U = \frac{A}{{10}};\)

The matrix Uis:

\(U = \left( {\begin{array}{*{20}{c}}{ - 0.6000}&{ - 0.3000}&{0.6000}&{0.1000}\\{ - 0.1000}&{0.2000}&{0.1000}&{ - 0.6000}\\{0.3000}&{0.6000}&{0.3000}&{ - 0.2000}\\{0.6000}&{ - 0.3000}&{0.6000}&{ - 0.1000}\\{0.2000}&{ - 0.1000}&{0.2000}&{0.3000}\\{ - 0.3000}&{0.6000}&{0.3000}&{0.2000}\\{ - 0.2000}&{ - 0.1000}&{0.2000}&{ - 0.3000}\\{0.1000}&{0.2000}&{0.1000}&{0.6000}\end{array}} \right)\)

Compute the product \({U^T}U\) and \(U{U^T}\).

\(\begin{array}{c} > > P\_1 = U'*U\\ > > P\_2 = U*U'\end{array}\)

The product \({U^T}U\) and \(U{U^T}\) is shown below:

\({U^T}U = \left( {\begin{array}{*{20}{c}}{1.0000}&0&0&0\\0&{1.0000}&0&0\\0&0&{1.0000}&0\\0&0&0&{1.0000}\end{array}} \right)\)

And,

\(U{U^T} = \left( {\begin{array}{*{20}{c}}{0.8200}&0&{ - 0.2000}&{0.0800}\\0&{0.4200}&{0.2400}&0\\{ - 0.2000}&{0.2400}&{0.5800}&{0.2000}\\{0.0800}&0&{0.2000}&{0.8200}\\{0.0600}&{ - 0.2000}&0&{0.2400}\\{0.2000}&{0.0600}&{0.3200}&{ - 0.2000}\\{0.2400}&{0.2000}&0&{0.0600}\\0&{ - 0.3200}&{0.0600}&0\end{array}} \right)\)

The matrix \({U^T}U\) is of order \(4 \times 4\) , and the matrix \(U{U^T}\) is of order \(8 \times 8\).

02

Find an answer for part (b)

Consider the following column vector \(y = {\left( {\begin{array}{*{20}{c}}1&1&1&1&1&1&1&1\end{array}} \right)^T}\).

Use the following command in the MATLAB to obtain the random vector:

\(\begin{array}{l} > > {\bf{y}} = {\rm{ones}}\left( {1,8} \right)\\ > > {\rm{p}} = {\rm{U}}*{\rm{U}}*{\rm{y}}\\ > > {\rm{z}} = {\rm{y}} - {\rm{p}}\end{array}\)

So, the vector z is shown below:

\(z = \left( {\begin{array}{*{20}{c}}{ - 0.2000}\\{0.6000}\\{ - 0.2000}\\{ - 0.2000}\\{0.6000}\\{ - 0.2000}\\{0.6000}\\{0.6000}\end{array}} \right)\)

As \({\bf{p}} = U{U^T}{\bf{y}}\), therefore, pis the column space of U.

Find the product of z and p.

\( > > {\rm{z}}*{\bf{p}}\)

The product \({\rm{z}}*{\rm{p}}\) gives the result as zero. Therefore z and p are orthogonal to each other.

03

Find an answer for part (c)

Multiply each column of U with zby using the following MATLAB commands:

\(\begin{array}{l} > > {\rm{z}}*{\rm{U}}\left( {;1} \right)\\ > > {\rm{z}}*{\rm{U}}\left( {;2} \right)\\ > > {\rm{z}}*{\rm{U}}\left( {;3} \right)\\ > > {\rm{z}}*{\rm{U}}\left( {;4} \right)\end{array}\)

The result of all the products is zero. Therefore z is orthogonal to each column of U.

04

Find an answer for part (d)

As it is proved in step 3, that zis orthogonal to each column of A. Therefore z must be orthogonal to every vector in the column space of A.

Thus, z is in the column space of A.

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Most popular questions from this chapter

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)

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17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram–Schmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Suppose radioactive substance A and B have decay constants of \(.02\) and \(.07\), respectively. If a mixture of these two substances at a time \(t = 0\) contains \({M_A}\) grams of \(A\) and \({M_B}\) grams of \(B\), then a model for the total amount of mixture present at time \(t\) is

\(y = {M_A}{e^{ - .02t}} + {M_B}{e^{ - .07t}}\) (6)

Suppose the initial amounts \({M_A}\) and are unknown, but a scientist is able to measure the total amounts present at several times and records the following points \(\left( {{t_i},{y_i}} \right):\left( {10,21.34} \right),\left( {11,20.68} \right),\left( {12,20.05} \right),\left( {14,18.87} \right)\) and \(\left( {15,18.30} \right)\).

a.Describe a linear model that can be used to estimate \({M_A}\) and \({M_B}\).

b. Find the least-squares curved based on (6).

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

a. If \(W = {\rm{span}}\left\{ {{x_1},{x_2},{x_3}} \right\}\) with \({x_1},{x_2},{x_3}\) linearly independent,

and if \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is an orthogonal set in \(W\) , then \(\left\{ {{v_1},{v_2},{v_3}} \right\}\) is a basis for \(W\) .

b. If \(x\) is not in a subspace \(W\) , then \(x - {\rm{pro}}{{\rm{j}}_W}x\) is not zero.

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independent columns), the columns of \(Q\) form an

orthonormal basis for the column space of \(A\).
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