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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.

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 \(x = x', y = y' \).

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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