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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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