# Show that this matrix is singular [closed]

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

• I think you may need to stipulate that $\text{size}(P)$ is odd. – Robert Lewis Jan 2 '18 at 18:47

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.

• $P$ orthogonal $\not \Longrightarrow \lambda_j = \pm 1$. See my comment below. – Robert Lewis Jan 2 '18 at 18:41
• @RobertLewis yes, i fixed it – Martín Vacas Vignolo Jan 2 '18 at 18:43

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.

• Why is det ((I+P)^T)=-det (I+P) – S Mukherjee Jan 3 '18 at 4:11
• @SMukherjee Because the determinant of a matrix is equal to the determinant of its transpose. It is just a property of determinants. – Geoffrey Jan 3 '18 at 6:54
• @SMukherjee It's $-\det((I+P)^T) = -\det(I+P)$. You missed the first minus sign. – Arthur Jan 3 '18 at 7:31