# If $S$ is real skew-symmetric matrix, then prove that $\det(I+S) \ge 1+\det(S)$,The equality holds if and only if $n\le 2$ or $n\ge3$ and $S=O$

If $$S$$ is real skew-symmetric matrix, then prove that $$\det(I+S) \ge 1+\det(S)$$,The equality holds if and only if $$n\le 2$$ or $$n\ge3$$ and $$S=O$$

Actually I only have trouble in proving that if and only if all the eigenvalues of $$S$$ is $$0$$ , then $$S=O$$.

• Actually, your last claim is not true, see here. May 19, 2021 at 16:56
• I can prove till that if $n\ge 3$, then all the eigenvalues of $S$ is 0(if and only if), i have no idea what to do next. May 19, 2021 at 17:05
• @Dietrich Burde, i saw that "Specifically, every $2n\times 2n$ real skew-symmetric matrix can be written in the form $Q \Lambda Q^T$ where $Q$ is orthogonal" from wikipedia. May 20, 2021 at 8:05

I hope this helps :

Since skew-symmetric is normal, i.e commute with adjoint, you know that by complexification of the vector space, we have canonical form for those operator. For example for $$f$$ such that $$f = -f^{*}$$ (as for skew-symmetric) $$\exists P^{-1}= P^t$$ such that $$A = P^{-1}S P = \begin{pmatrix}\textbf{0} & 0 \\ 0 & \Box\end{pmatrix}$$

Where the first $$0$$ are $$k$$ zero relative to the eigenvectors of $$0$$ eigenvalue of $$f$$, and the other part are $$j$$ (indeces are not very important here) block $$\Box_\mu = \begin{pmatrix}0 & \mu \\ -\mu & 0 \end{pmatrix}$$ relative to the purely imaginary eigenvalues of $$f$$ (Here $$\mu \in \mathbb{R}$$ and this is very important).

That being sad, since $$\det(P)^2 = 1$$ (being $$P$$ orthogonal) playing with Binet :

$$\text{det}(Id+S) = 1\cdot \text{det}(Id+S) = \text{det}(P)^2 \text{det}(Id+S) = \text{det}(P^{-1})\text{det}(Id+S)\text{det}(P) = \text{det}(P^{-1}(Id+S)P) = det(P^{-1}Id P + P^{-1}SP) = \text{det}(Id+ A)$$

It should clear that $$I_d+A = \begin{pmatrix}Id_k & 0 \\ 0 & \Box\end{pmatrix}$$ where the blocks now turned into $$B_\mu = \begin{pmatrix}1 & \mu \\ -\mu &1 \end{pmatrix}$$

And since the matrix is block diagonal we have that $$\text{det}(Id+A) = 1^k \prod\limits_{i=1}^j B_{\mu_i}$$.

Note that having $$\text{det}( B_{\mu_i}) = 1 + \mu_i^2$$, we end up with $$\text{det}(I_d+A) = \prod\limits_{i=1}^j (1+\mu_i^2)$$.

It should be clear now that

$$\prod\limits_{i=1}^j (1+\mu_i^2) \geq 1+ \prod\limits_{i=1}^j \mu_i^2 = 1 + \prod\limits_{i=1}^j \text{det}(\Box_{\mu_i}) = 1 + 1^k\prod\limits_{i=1}^j \text{det}(\Box_{\mu_i}) = 1 + \text{det}(P^{-1}SP) = 1 + \text{det}(P)^2\text{det}(S) = 1+\text{det}(S)$$

Edit : I think if this is right you can easily find the special cases for $$n$$ by yourself.

$$\textbf{Addendum :}$$ Rewatching this "proof" it seems to me that the equality holds if and only if holds $$\prod\limits_{i=1}^j (1+\mu_i^2) = 1+ \prod\limits_{i=1}^j \mu_i^2$$. If you are familiar with the first product, you should know or you could prove that $$\prod\limits_{i=1}^j (1+\mu_i^2) = \sum\limits_{J \subset[j]}\prod\limits_{i \in J} \mu_i^2$$. From this equality you see that holds if and only if $$\mu_i^2 = 0$$ for every $$i$$ since we're dealing with real numbers, i.e $$\mu_i = 0$$ for all $$i$$. But if we remember who $$\mu_i$$ was, we get $$S = 0$$ since $$0$$ is the only matrix similar to the $$0$$ matrix.

• Thanks a lot for your help. Actually your last formulas are enough to cover the proof and I've already solved that.The problem is, if the equality holds and $n\ge 2$, then all the $\mu_i=0$,which is all I have gone so far. Could you help me with the last step that $S=O$? Your assistance will be highly appreciated. May 20, 2021 at 10:08
• @RossRen I'm sorry, could you rephrase the question ? I don't understand what you are trying to prove May 20, 2021 at 10:10
• OK. Using $AX=\lambda X$ and your last formulas, I proved the main proposition.I only have trouble in the special case when the equality is achieved. I mean, how to prove $S=O$ when I only knew all the eigenvalues are zeros. May 20, 2021 at 10:16
• @RossRen Look at the Addendum and tell me what you think May 20, 2021 at 10:37
• I've got your points.Thanks a lot. And sorry again for my ambitious problem. I only knew that using congruent transformation, we can get $diag${$S,...,S,0,...,0$} where $S$ denotes\begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} , can you only use this to solve th e problem so that I could finish the proof before learning about orthogonal matrices. May 20, 2021 at 10:54

Actually I only have trouble in proving that if and only if all the eigenvalues of 𝑆 is 0 , then 𝑆=𝑂.

since OP is stuck on this point and does not want to use orthogonality, here's a different finish.

It is immediate that $$S=\mathbf 0\implies$$ all eigenvalues of $$S$$ are 0.

We need to prove the other direction, i.e. having all eigenvalues (considered in $$\mathbb C$$) of real skew symmetric $$S$$ be 0 $$\implies S=\mathbf 0$$

$$\text{trace}\big(S^2\big) = \sum_{k=1}^n \lambda_k^2 = \sum_{k=1}^n 0^2=0$$
but $$S^2=-S^TS\implies 0= \text{trace}\big(S^TS\big)=\Big\Vert S\Big \Vert_F^2 \implies S=\mathbf 0$$

• Thanks a lot. Your answer completely solves my problem. May 21, 2021 at 1:21