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# Tracing the Trace | ISI MStat 2016 PSB Problem 3 This ISI MStat 2016 problem is an application of the ideas of tracing the trace and Eigen values of a matrix and using a cute sum of squares identity.

## Problem- Tracing the Trace

Suppose A is an $n × n$ real symmetric matrix such that
$Tr(A^2) = T r(A) = n$. Show that all the eigenvalues of A are equal to 1.

This problem is from ISI MStat 2016 PSB ( Problem #3)

### Prerequisites

• Trace of a Matrix
• Eigen values of $A^n$ w.r.t to the eigen values of $A$.
• Sum of Squares $\geq 0$.

## Solution

$Av = {\lambda}v \Rightarrow A^nv = {\lambda}^nv$.

Since, A is a real symmetric matrix, then all the eigen values of the matrix A are real say {${\lambda}_1, {\lambda}_2, ..., {\lambda}_n$}.

$Tr(A^2) = \sum_{i=1}^{n} {{\lambda}_i}^2 = Tr(A) = \sum_{i=1}^{n} {{\lambda}_i} = n$

$\Rightarrow n\sum_{i=1}^{n} {{\lambda}_i}^2 = (\sum_{i=1}^{n} {{\lambda}_i})^2$

$\Rightarrow (n-1)\sum_{i=1}^{n} {{\lambda}_i}^2 = \sum_{i, j = 1, i \neq j }^{n} 2{\lambda}_i{\lambda}_j$

$\Rightarrow \sum_{i=1}^{n} ({{\lambda}_i - {\lambda}_j })^2 = 0$

$\Rightarrow {\lambda}_i = {\lambda}_j = \lambda \forall i \neq j$

$\Rightarrow Tr(A) = n\lambda = n \Rightarrow \lambda = 1$.

This ISI MStat 2016 problem is an application of the ideas of tracing the trace and Eigen values of a matrix and using a cute sum of squares identity.

## Problem- Tracing the Trace

Suppose A is an $n × n$ real symmetric matrix such that
$Tr(A^2) = T r(A) = n$. Show that all the eigenvalues of A are equal to 1.

This problem is from ISI MStat 2016 PSB ( Problem #3)

### Prerequisites

• Trace of a Matrix
• Eigen values of $A^n$ w.r.t to the eigen values of $A$.
• Sum of Squares $\geq 0$.

## Solution

$Av = {\lambda}v \Rightarrow A^nv = {\lambda}^nv$.

Since, A is a real symmetric matrix, then all the eigen values of the matrix A are real say {${\lambda}_1, {\lambda}_2, ..., {\lambda}_n$}.

$Tr(A^2) = \sum_{i=1}^{n} {{\lambda}_i}^2 = Tr(A) = \sum_{i=1}^{n} {{\lambda}_i} = n$

$\Rightarrow n\sum_{i=1}^{n} {{\lambda}_i}^2 = (\sum_{i=1}^{n} {{\lambda}_i})^2$

$\Rightarrow (n-1)\sum_{i=1}^{n} {{\lambda}_i}^2 = \sum_{i, j = 1, i \neq j }^{n} 2{\lambda}_i{\lambda}_j$

$\Rightarrow \sum_{i=1}^{n} ({{\lambda}_i - {\lambda}_j })^2 = 0$

$\Rightarrow {\lambda}_i = {\lambda}_j = \lambda \forall i \neq j$

$\Rightarrow Tr(A) = n\lambda = n \Rightarrow \lambda = 1$.

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