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I.S.I. and C.M.I. Entrance ISI M.Stat PSB Linear Algebra

The Unique Decomposition | ISI MStat 2015 PSB Problem 3

The solution plays with eigen values and vectors to solve this cute and easy problem in Linear Algebra from the ISI MStat 2015 problem 3.

The solution plays with Eigen values and vectors to solve this cute and easy problem in Linear Algebra from the ISI MStat 2015 problem 3.

Problem

Let \(A\) be a real valued and symmetric \(n \times n\) matrix with entries such that \(A \neq \pm I\) and \(A^{2}=I\).
(a) Prove that there exist non-zero column vectors \(v\) and \(w\) such that
\(A v=v\) and \(A w=-w\).
(b) Prove that every vector \(z\) has a unique decomposition \(z=x+y\)
where \(A x=x\) and \(A y=-y\).

This problem is from ISI MStat 2015 PSB ( Problem #3).

Prerequisites

  • Eigen values and Eigen vectors

Solution

(a)

Let’s say \(\lambda\) is an eigenvalue of \(A\). Let’s explore the possibilities of \(\lambda\).

\(Av= \lambda v \Rightarrow A^2v= {\lambda}^2 v \Rightarrow Iv= {\lambda}^2 v \Rightarrow v= {\lambda}^2 v \). Since, \( v\) is arbitrary, we get \({\lambda}^2 = 1 \Rightarrow \lambda = \pm 1\).

Since \(A\) is real symmetric, it has real eigenvalues, and the possibilities are 1 and -1. Since, \(A \neq \pm I\), there exists non-zero column vectors \(v\) and \(w\) such that \(A v=1.v\) and \(A w=-1.w\).

(b)

Suppose \(z\) has two decompositions \(z= x+y = x’+y’\) where \(A x=x\) and \(A y=-y\) and \(A x’=x’\) and \(A y’=-y’\).

Tberefore, \( A(x+y) = A(x’+y’) \Rightarrow Ax+Ay = Ax’+Ay’ \Rightarrow x – y = x’ – y’\).

But, we also have \( x+y = x’+y’\). Thus, by adding and subtracting, we get \(x = x’, y = y’ \).

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