# Let A be skew-symmetric, and denote its singular values by $\sigma_1\geq \sigma_2\geq \dots \sigma_n\geq0$.

Let A be skew-symmetric, and denote its singular values by $\sigma_1\geq \sigma_2\geq \dots \sigma_n\geq0$. Show that

a) If n is even, then $\sigma_{2k}=\sigma_{2k-1}\geq 0, k= 1,2,\dots n/2.$ If n is odd, then the same relationship holds up to $k=(n-1)/2$ and also $\sigma_n=0$.

b) The eigenvalues $\lambda_j=(-1)^ji\sigma_j$, $j=1,2,\dots,n$.

I know that skew symmetric means $-A=A^T$ and I know that the eigenvalues of a skew-symmetric matrix are either purely imaginary or zero. I am not able to get this though and I have been trying all week...

Hint: The singular values of $$A$$ are the (positive) square roots of the eigenvalues of the matrix $$A^TA = -A^2$$. So, if $$\lambda$$ is an eigenvalue of $$A$$, then $$\sqrt{-\lambda^2} = |\lambda|$$ is a singular value.
Because $$A$$ is real, its complex (i.e. non-real) eigenvalues must come in complex-conjugate pairs.
• Yes, that's true, but it does not mean that they don't come in conjugate pairs. For every non-zero eigenvalue $ai$, $-ai$ must also be an eigenvalue. Oct 5, 2016 at 20:20
• Isn't $\sqrt{-\lambda^2} = i|\lambda| \not= |\lambda|$? I feel that I am missing something obvious. Jan 26, 2019 at 4:15
• Oh OK, if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$, since (given the corresponding eigenvector of $A$) $A^2 u = A(Au) = A (\lambda u) = \lambda Au = \lambda^2 u$. Jan 28, 2019 at 16:53
• @hasManyStupidQuestions Right. Regarding your first comment, note that $\lambda$ is actually an imaginary number in this context Jan 28, 2019 at 17:50