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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 \({{\rm{x}}^T}A{\rm{x}} = 3\).

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Most popular questions from this chapter

In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).

4. \(Q\left( x \right) = 3x_1^2 + 3x_2^2 + 5x_3^2 + 6x_1^{}x_2^{} + 2x_1^{}x_3^{} + 2x_2^{}x_3^{}\).

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix \(P\) and a diagonal matrix \(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17) \( - {\bf{4}}\), 4, 7; (18) \( - {\bf{3}}\), \( - {\bf{6}}\), 9; (19) \( - {\bf{2}}\), 7; (20) \( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

15. \(\left( {\begin{aligned}{{}}{\,3}&4\\4&9\end{aligned}} \right)\)

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

19. \(\left( {\begin{aligned}{{}}3&{ - 2}&4\\{ - 2}&6&2\\4&2&3\end{aligned}} \right)\)

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 \({P^T}SP\).

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.)

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