Rank of $I_n + M$ where M is skew-symmetric? Apparently the answer is $n$. I know that the rank of $M$ is always even, but how does this help?
Many thanks.
 A: Hint: If $\lambda$ is an eigenvalue and $v$ corresponding eigen-vector, so that $Mv=\lambda v$, what is the nature of $\lambda$? what are the eigenvalues of $I_n+M$? what do you know about the relation between eigenvalues and determinant of a matrix? what is the relation between rank and the determinant?
A: Eigenvalues of any skew-Hermitian matrix (or a real skew-symmetric matrix) are pure imaginary. Eigenvalues of $I+M$ are $1+\lambda$, where $\lambda$ is an eigenvalue of $M$. Since $\lambda$ is pure imaginary, it can never be zero. Since $I+M$ does not have a zero eigenvalue, it is non-singular and has hence rank $n$.

Why $M=-M^*$ implies that $M$ has pure imaginary eigenvalues? Let $Mx=\lambda x$ for a nonzero $x$. Then $\lambda x^*x=x^*Mx=-x^*M^*x=-\overline{x^*Mx}$. Since $x^*x$ is real positive, this implies that $\lambda\in\mathbb{C}$ satisfies $\lambda=-\overline{\lambda}$. This is possible only if $\mathfrak{Re}(\lambda)=0$.
A: That the rank of $M$ is even doesn't help, because there are plenty of even-ranked matrices whose sums with $I$ are singular. What is crucial here is the fact that $M$ is skew-symmetric.
Hint. Use the skew-symmetry of $M$ to show that for any real vector $x$, the scalar $x^TMx$ is always zero. Hence argue that $x^T(I+M)x$ and in turn $(I+M)x$ are always nonzero whenever $x\ne0$.
