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Let \(A\) and \(B\)be symmetric \(n \times n\) matrices whose eigenvalues are all positive. Show that the eigenvalues of\(A + B\) are all positive.

Short Answer

Expert verified

It is proved that the eigenvalues of \(A + B\) are all positive.

Step by step solution

01

Symmetric Matrices and Quadratic Forms

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

\(\begin{aligned}{}{x^T}Ax = {y^T}Dy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as }}x = Py} \right\}\\{\rm{and}}\\\left\| x \right\| = \left\| {Py} \right\| = \left\| y \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\forall y \in \mathbb{R}} \right\}\end{aligned}\)

02

Show that the eigenvalues of \(A + B\) are all positive

As per the question, we have:

The positive \(n \times n\)matrices\(A{\rm{ and }}B\)have all positive eigenvalues. So, their quadratic forms \({x^T}Ax{\rm{ and }}{x^T}Bx\) will also be positive.

So, we have:

\(\begin{aligned}{}{x^T}Ax > 0{\rm{ and }}{x^T}Bx > 0\\{x^T}Ax + {x^T}Bx > 0\\{x^T}\left( {A + B} \right)x > 0\end{aligned}\)

Here, we see a matrix \(A + B\)whose quadratic form is positive.

Hence proved, the eigenvalues of \(A + B\) are all positive.

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Most popular questions from this chapter

Question: Find the principal components of the data for Exercise 2.

Question: 13. The sample covariance matrix is a generalization of a formula for the variance of a sample of \(N\) scalar measurements, say \({t_1},................,{t_N}\). If \(m\) is the average of \({t_1},................,{t_N}\), then the sample variance is given by

\(\frac{1}{{N - 1}}\sum\limits_{k = 1}^n {{{\left( {{t_k} - m} \right)}^2}} \)

Show how the sample covariance matrix, \(S\), defined prior to Example 3, may be written in a form similar to (1). (Hint: Use partitioned matrix multiplication to write \(S\) as \(\frac{1}{{N - 1}}\) times the sum of \(N\) matrices of size \(p \times p\). For \(1 \le k \le N\), write \({X_k} - M\) in place of \({\hat X_k}\).)

Compute the quadratic form \({{\bf{x}}^T}A{\bf{x}}\), when \(A = \left( {\begin{aligned}{{}}5&{\frac{1}{3}}\\{\frac{1}{3}}&1\end{aligned}} \right)\) and

a. \({\bf{x}} = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

b. \({\bf{x}} = \left( {\begin{aligned}{{}}6\\1\end{aligned}} \right)\)

c. \({\bf{x}} = \left( {\begin{aligned}{{}}1\\3\end{aligned}} \right)\)

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,4}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,4}&{ - 1}\\{ - 1}&{ - 1}&{\,\,\,4}\end{aligned}} \right)\), and\({\rm{v}} = \left( {\begin{aligned}{{}}1\\1\\1\end{aligned}} \right)\). Verify that 5 is an eigenvalue of \(A\) and \({\rm{v}}\)is an eigenvector. Then orthogonally diagonalize \(A\).

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?

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