# The normal form of a skew symmetric matrix

In Werner Greub's book Linear Algebra, 4th ed. on p. 230, he gives this proof of the normal form for a skew transformation on a finite-dimensional real inner product space. (Note Greub's convention for the matrix of a transformation is the transpose of that normally used with left-hand notation.)

I believe this proof is incorrect because it is not true in general that the $$a_n$$ defined form an orthonormal basis of the space. For example in $$\mathbb{R}^4$$, if we define the transformation $$\psi$$ by $$e_1\mapsto e_2\qquad e_2\mapsto -e_1\qquad e_3\mapsto e_4\qquad e_4\mapsto -e_3$$ where $$e_i$$ is the $$i$$-th standard basis vector, then $$\psi$$ is skew and $$\varphi=\psi^2=-\iota$$ is diagonalized by the standard basis. If we follow the proof for this example, we get $$a_1=e_1$$, $$a_2=\psi e_1=e_2$$, $$a_3=e_2$$, and $$a_4=\psi e_2=-e_1$$, so the $$a_n$$ do not form a basis of $$\mathbb{R}^4$$.

Does anyone see a way to salvage this proof while still retaining its spirit (in particular, avoiding use of complex numbers)?

• It is perhaps worth noting that the proof presented does work in the case where each eigenvalue of $\psi^2$ has the minimal multiplicity, namely $2$. Oct 21 '19 at 3:40
• Actually, even this is incorrect: Greub seems to be making the tacit assumption that consecutive $e_j$ are taken from the same eigenspace when possible. Oct 21 '19 at 3:46
• Having tried googling a quick answer, I came across a growing list of errata on github that is either yours or a different blargoner's. Assuming that's you, best of luck! I would recommend that you send an email to the publisher asking whether such a list of errata already exists, if you have not done so already. Oct 21 '19 at 4:39
• @Omnomnomnom Yep, those are mine. Thanks for your help with this proof. Oct 21 '19 at 14:13

One fix is to be a bit more explicit with how we deal with each non-zero eigenspace in the following way.

Suppose that $$\lambda_1,\dots,\lambda_d$$ are the (distinct) negative eigenvalues of $$\varphi = \psi^2$$. Then by "the result of section 8.7" (presumably the spectral theorem for symmetric matrices), we can select eigenvectors $$e_{j,k}$$ such that $$\varphi \,e_{j,k} = \lambda_j \,e_{j,k}\quad k = 1,\dots,m_j$$ That is: $$m_j$$ is the multiplicity of $$\lambda_j$$, and $$e_{j,1},\dots,e_{j,m_{j}}$$ is a basis of the eigenspace.

For each $$\lambda_j$$, we produce a new basis $$\mathcal B_j$$ for the eigenspace via the following recursive process. Initially, we take $$S = \operatorname{span}\{e_{j,1},\dots,e_{j,m_j}\}$$. We then do the following to $$S$$:

• Select an arbitrary unit vector $$a_1 \in S$$ and define $$a_2 = \frac 1{\kappa_j}\psi a_1$$.
• Add $$a_1,a_2$$ to $$\mathcal B_j$$.
• Let $$S'$$ denote the orthogonal complement of $$\operatorname{span}\{a_1,a_2\}$$ relative to $$S$$. If $$S' = \{0\}$$, then we are done. Otherwise, $$S'$$ is a smaller eigenspace associated with $$\lambda_j$$; in this case we apply this process to $$S'$$.

In a proper writeup of the proof, we should prove that $$a_2 = \frac 1{\kappa_j}\psi a_1$$ (where $$\kappa_j = \sqrt{|\lambda_j|}$$) will necessarily be a unit vector from the same eigenspace, and that $$a_2$$ is orthogonal to $$a_1$$ (which Greub's text does not seem to mention); I will leave that to you.

• Incidentally, we can write the non-zero part of the normal form as $$\pmatrix{0 & \kappa_1 \\-\kappa_1 & 0 \\ && \ddots \\ &&& 0 & \kappa_p\\ &&&-\kappa_p & 0} = \pmatrix{\kappa_1 \\ & \ddots \\ && \kappa_p} \otimes \pmatrix{0 & 1\\-1 & 0}$$ where $\otimes$ denotes a Kronecker product. Oct 21 '19 at 4:21