Suppose \({\rm{x}}\)is a unit eigenvector of a matrix \(A\) corresponding to an eigenvalue 3. What is the value of \({{\rm{x}}^T}A{\rm{x}}\)?

Short Answer

Expert verified

The required value is: \({{\rm{x}}^T}A{\rm{x}} = 3\).

Step by step solution

01

Symmetric Matrices and Quadratic Forms 

When any Symmetric Matrix\(A\)is diagonalized orthogonallyas \(PD{P^{ - 1}}\), we have:

\(\begin{array}{l}{{\rm{x}}^T}A{\rm{x}} = {{\rm{y}}^T}D{\rm{y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as x}} = Py} \right\}\\\\\left\| {\rm{x}} \right\| = \left\| {P{\rm{y}}} \right\| = \left\| {\rm{y}} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\forall {\rm{y}} \in \mathbb{R}} \right\}\end{array}\)

02

Find the maximum value

As per the question, we have:

The matrix\(A\)has eigenvalue 3 with eigenvector\({\rm{x}}\)which is a unit vector, then:

\(\begin{array}{c}A{\rm{x}} = 3{\rm{x}}\\{{\rm{x}}^T}A{\rm{x}} = {{\rm{x}}^T}3{\rm{x}}\\ = 3\left( {{{\rm{x}}^T}{\rm{x}}} \right)\\ = 3{\left\| {\rm{x}} \right\|^2}\\ = 3\end{array}\)

Hence, the required value is \({{\rm{x}}^T}A{\rm{x}} = 3\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: 11. Given multivariate data \({X_1},................,{X_N}\) (in \({\mathbb{R}^p}\)) in mean deviation form, let \(P\) be a \(p \times p\) matrix, and define \({Y_k} = {P^T}{X_k}{\rm{ for }}k = 1,......,N\).

  1. Show that \({Y_1},................,{Y_N}\) are in mean-deviation form. (Hint: Let \(w\) be the vector in \({\mathbb{R}^N}\) with a 1 in each entry. Then \(\left( {{X_1},................,{X_N}} \right)w = 0\) (the zero vector in \({\mathbb{R}^p}\)).)
  2. Show that if the covariance matrix of \({X_1},................,{X_N}\) is \(S\), then the covariance matrix of \({Y_1},................,{Y_N}\) is \({P^T}SP\).

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

a. \(5x_1^2 + 16{x_1}{x_2} - 5x_2^2\)

b. \(2{x_1}{x_2}\)

Question: In Exercises 1 and 2, convert the matrix of observations to mean deviation form, and construct the sample covariance matrix.

\(1.\,\,\left( {\begin{array}{*{20}{c}}{19}&{22}&6&3&2&{20}\\{12}&6&9&{15}&{13}&5\end{array}} \right)\)

(M) Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

38. \(\left( {\begin{aligned}{{}}{.{\bf{63}}}&{ - .{\bf{18}}}&{ - .{\bf{06}}}&{ - .{\bf{04}}}\\{ - .{\bf{18}}}&{.{\bf{84}}}&{ - .{\bf{04}}}&{.{\bf{12}}}\\{ - .{\bf{06}}}&{ - .{\bf{04}}}&{.{\bf{72}}}&{ - .{\bf{12}}}\\{ - .{\bf{04}}}&{.{\bf{12}}}&{ - .{\bf{12}}}&{.{\bf{66}}}\end{aligned}} \right)\)

Let u be a unit vector in \({\mathbb{R}^n}\), and let \(B = {\bf{u}}{{\bf{u}}^T}\).

  1. Given any x in \({\mathbb{R}^n}\), compute Bx and show that Bx is the orthogonal projection of x onto u, as described in Section 6.2.
  2. Show that B is a symmetric matrix and \({B^{\bf{2}}} = B\).
  3. Show that u is an eigenvector of B. What is the corresponding eigenvalue?
See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free