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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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