If $\det P=-1$ and $P$ is an orthogonal matrix. Show that $P+I_n$ is singular matrix.

Please help it with only matrix algebra.

  • 3
    $\begingroup$ I think you may need to stipulate that $\text{size}(P)$ is odd. $\endgroup$ Jan 2, 2018 at 18:47

2 Answers 2


Using the fact that $P^TP=I$ and $\det(P)=-1$, we obtain $$ \det(I+P)=\det(P^TP+P)=\det(P)\det(P^T+I)=-\det((I+P)^T)=-\det(I+P)$$ which implies that $\det(I+P)=0$ and hence that $I+P$ is singular.

  • $\begingroup$ Why is det ((I+P)^T)=-det (I+P) $\endgroup$ Jan 3, 2018 at 4:11
  • $\begingroup$ @SMukherjee Because the determinant of a matrix is equal to the determinant of its transpose. It is just a property of determinants. $\endgroup$
    – Geoffrey
    Jan 3, 2018 at 6:54
  • $\begingroup$ @SMukherjee It's $-\det((I+P)^T) = -\det(I+P)$. You missed the first minus sign. $\endgroup$
    – Arthur
    Jan 3, 2018 at 7:31

Let $\lambda_1,\dots,\lambda_n$ the eigenvalues. Because $P$ is an orthogonal matrix, then $\lambda_j=\pm 1$ for all real eigenvalues. Note that if $\lambda\in\mathbb{C}$ is a complex eigenvalue, then $\overline{\lambda}$ too and $\lambda\overline{\lambda}=\mid \lambda\mid^2=1$

Also, we have $\prod\limits_{j=1}^n\lambda_j=\det P =-1$ and exist $j_0$ such that $\lambda_{j_0}=-1$.

Now, if $J$ is then Jordan form of $P$, then $\det(P+I_n)=\det(J+I_n)=0$ because $J$ has a $-1$ in the diagonal and is inf triang.

  • 1
    $\begingroup$ $P$ orthogonal $\not \Longrightarrow \lambda_j = \pm 1$. See my comment below. $\endgroup$ Jan 2, 2018 at 18:41
  • $\begingroup$ @RobertLewis yes, i fixed it $\endgroup$ Jan 2, 2018 at 18:43

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