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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

## (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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