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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 $x = x', y = y'$.

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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