Log In Start studying!

Select your language

Suggested languages for you:
Deutsch (DE)
Deutsch (UK)
Deutsch (US)

Americas

Europe

Answers without the blur. Sign up and see all textbooks for free! → Illustration

Q7.5-12E

Expert-verified
Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later
Illustration

Short Answer

Question: Let \({\bf{X}}\) denote a vector that varies over the columns of a \(p \times N\) matrix of observations, and let \(P\) be a \(p \times p\) orthogonal matrix. Show that the change of variable \({\bf{X}} = P{\bf{Y}}\) does not change the total variance of the data. (Hint: By Exercise 11, it suffices to show that \(tr\left( {{P^T}SP} \right) = tr\left( S \right)\). Use a property of the trace mentioned in Exercise 25 in Section 5.4.)

It is verified that the total variance would not change when variables change as \({\bf{X}} = P{\bf{Y}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Mean Deviation form and Covariance Matrix.

The Mean Deviation form of any \(p \times N\) is given by:

\(B = \left( {\begin{array}{*{20}{c}}{{{{\bf{\hat X}}}_1}}&{{{{\bf{\hat X}}}_2}}&{........}&{{{{\bf{\hat X}}}_N}}\end{array}} \right)\)

Whose \(p \times p\) covariance matrix is:

\(S = \frac{1}{{N - 1}}B{B^T}\)

Step 2: The Variance

From exercise 11, we have:

\({S_Y} = {P^T}SP\)

When variable changes as: \({\bf{X}} = P{\bf{Y}}\)

The traces of the covariance matrices \({S_Y}{\rm{ and }}S\) will be the same.

The total variance of the data is given by \({\bf{Y}}\) is \({\rm{tr}}\left( {{P^T}SP} \right)\).

For two similar matrices \(A,B\) are such that, \({\bf{B}} = P{\bf{A}}{P^{ - 1}}\) which implies \({\rm{tr}}\left( {\bf{B}} \right) = {\rm{tr}}\left( {P{\bf{A}}{P^{ - 1}}} \right)\).

In the obtained equation, if \(P\) is an orthogonal matrix, then \({P^T} = {P^{ - 1}}\).

Apply trace on both sides of \({P^T} = {P^{ - 1}}\) and simplify.

\(\begin{array}{c}{\rm{tr}}\left( {{P^T}SP} \right) = {\rm{tr}}\left( {{P^{ - 1}}SP} \right)\\ = {\rm{tr}}\left( {{P^{ - 1}}PS} \right)\\ = {\rm{tr}}\left( {\left( {{P^{ - 1}}P} \right)S} \right)\\ = {\rm{tr}}\left( {IS} \right)\\ = {\rm{tr}}\left( S \right)\end{array}\)

Thus, the total variance would not change when variables change as: \({\bf{X}} = P{\bf{Y}}\).

Hence, this is the required proof.

Most popular questions for Math Textbooks

Determine which of the matrices in Exercises 1–6 are symmetric.

\(\left( {\begin{array}{{}{}}1&2&2&1\\2&2&2&1\\2&2&1&2\end{array}} \right)\)

Question 11: Prove that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\), where P is a \(n \times n\) positive semidefinite matrix with the same rank as A and where Q is an \(n \times n\) orthogonal matrix. (Hint: Use a singular value decomposition, \(A = U\sum {V^T}\), and observe that \(A = \left( {U\sum {U^T}} \right)\left( {U{V^T}} \right)\).) This decomposition is used, for instance, in mechanical engineering to model the deformation of a material. The matrix P describe the stretching or compression of the material (in the directions of the eigenvectors of P), and Q describes the rotation of the material in space.

Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.

12. \(P = \left( {\begin{aligned}{{}}{.5}&{.5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{.5}\\{.5}&{ - .5}&{ - .5}&{.5}\\{.5}&{ - .5}&{.5}&{ - .5}\end{aligned}} \right)\)

Classify the quadratic forms in Exercises 9–18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

13. \({\bf{ - }}x_{\bf{1}}^{\bf{2}}{\bf{ - 6}}{x_{\bf{1}}}{x_{\bf{2}}} + {\bf{9}}x_{\bf{2}}^{\bf{2}}\)

Let B be an \(n \times n\) symmetric matrix such that \({B^{\bf{2}}} = B\). Any such matrix is called a projection matrix (or an orthogonal projection matrix.) Given any y in \({\mathbb{R}^n}\), let \({\bf{\hat y}} = B{\bf{y}}\)and\({\bf{z}} = {\bf{y}} - {\bf{\hat y}}\).

a) Show that z is orthogonal to \({\bf{\hat y}}\).

b) Let W be the column space of B. Show that y is the sum of a vector in W and a vector in \({W^ \bot }\). Why does this prove that By is the orthogonal projection of y onto the column space of B?
Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

Decision Maths

Read explanation

Calculus

Read explanation

Probability and Statistics

Read explanation

Statistics

Read explanation

Mechanics Maths

Read explanation

Geometry

Read explanation

Pure Maths

Read explanation

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.