It is well known that if $Q$ is a complex unitary matrix such that $I+Q$ is invertible (where $I$ is the identity matrix), that is, $-1$ is not an eigenvalue of $Q$, then $$ A:=(I-Q)(I+Q)^{-1} $$ is skew-Hermitian ($A=-A^*=-\bar{A}^T$).

It is easy to prove this by factoring out the inverse from $$ \begin{split} A+A^*&=(I-Q)(I+Q)^{-1}+(I+Q)^{-*}(I-Q)^*\\ &=(I+Q)^{-*}[\color{blue}{(I+Q)^*(I-Q)+(I-Q)^*(I+Q)}](I+Q)^{-1}. \end{split}\tag{1} $$ and showing that the blue term is equal to zero.

One can also use the spectral decomposition of $Q$. There is a unitary $U$ and a diagonal $D$ such that $Q=UDU^*$ with $D\bar{D}=I$. Then $$ A=U(I-D)(I+D)^{-1}U^*. $$ For each diagonal entry $\delta:=\alpha+i\beta$ of $D$, where $\alpha,\beta\in\mathbb{R}$ and $\delta\bar\delta=\alpha^2+\beta^2=1$, the corresponding entry of $(I-D)(I+D)^{-1}$ is imaginary $(1-\delta)(1+\delta)^{-1}=-2i\alpha\beta$ and we can use the fact that a matrix is skew-Hermitian if and only if it is unitarily similar to a diagonal matrix with imaginary diagonal entries.

However, the approach in (1) is more elementary (and, consequently, elegant) since it does not require to know anything about spectral decompositions of unitary and skew-Hermitian matrices.

Now the problem comes when one would like to drop the assumption that $I+Q$ is invertible while replacing the inverse with the Moore-Penrose pseudoinverse. Then $$\tag{2}A:=(I-Q)(I+Q)^{\dagger}$$ (where $\dagger$ denotes the pseudoinverse) is still skew-Hermitian as can be shown using the spectral decomposition. Indeed, $$ A=U(I-D)(I+D)^{\dagger}U^*, $$ and we can again show that the diagonal entries $\delta\neq -1$ of $D$ give imaginary entries while the diagonal entries $\delta=-1$ give $0$ in $(I-D)(I+D)^{\dagger}$.

However, to show this, it is not possible to use directly the same approach as in (1) since $(I+Q)^{\dagger}(I+Q)\neq I$ if $-1$ is the eigenvalue of $Q$. I was wondering whether there exists a more elementary way to show that (2) is skew-Hermitian for any unitary $Q$ (without spectral decompositions and, ideally, using only the four Penrose conditions of the pseudoinverse or some of their simple consequences).

EDIT: It might be maybe useful to consider the identities $X^\dagger=(X^*X)^\dagger X^*=X^*(XX^*)^\dagger$, which give $(I+Q)^\dagger=B(I+Q)^*=(I+Q)^*B$, where $B:=(2I+Q+Q^*)^\dagger$ is Hermitian (the operations $\dagger$ and $*$ "commute"). I don't know yet how to get the final result (e.g., the first identity gives that $A=-A^*$ iff $B=QBQ^*$).

  • $\begingroup$ Interesting...but why do you need to consider the degenerate case? Just curious. $\endgroup$
    – Troy Woo
    Jan 26, 2015 at 12:35
  • $\begingroup$ @TroyWoo Same as you. Just curious :-) $\endgroup$ Jan 26, 2015 at 12:38

1 Answer 1


A partial answer. I usually prove that $A=(I-Q)(I+Q)^{-1}$ is skew-Hermitian by showing that $A^\ast=-A$ rather than $A+A^\ast=0$. The merit of doing this is that we can factor out $Q^\ast$ and make the proof shorter: $$ A^\ast = \left[ (I-Q)(I+Q)^{-1} \right]^\ast \color{red}{=} (I+Q^\ast)^{-1}(I-Q^\ast) \color{blue}{=} \left[(Q+I)^{-1}Q\right]\left[Q^\ast(Q-I)\right] = -A.\tag{1} $$ Now, suppose the existence and uniqueness of Moore-Penrose pseudoinverse is also known. Then, by using the four defining properties of M-P pseudoinverse, it is straightforward to verify that $(X^+)^\ast\color{red}{=}(X^\ast)^+$ and $(UX)^+\color{blue}{=}X^+U^\ast$ for any square matrix $X$ and any unitary matrix $U$. In our case, we have $\left[(I+Q)^+\right]^\ast\color{red}{=}(I+Q^\ast)^+=[Q^\ast(Q+I)]^+\color{blue}{=}(Q+I)^+Q$. The equalities in $(1)$ are therefore almost preserved when the inverses is replaced by M-P pseudoinverses: $$ A^\ast = \left[ (I-Q)(I+Q)^+ \right]^\ast \color{red}{=} (I+Q^\ast)^+(I-Q^\ast) \color{blue}{=} \left[(Q+I)^+Q\right]\left[Q^\ast(Q-I)\right] = -(I+Q)^+(I-Q). $$ So, it remains to show that $(I+Q)^+$ commutes with $I-Q$. This, of course, is clear if one already knows that $XX^+=X^+X$ when $X$ is normal. Otherwise, some more work has to be done to complete the proof.

  • $\begingroup$ I see, thank you! The identity for normal matrices actually follows from the identities in my edit so that's quite easy to finish. Nice! $\endgroup$ Jan 26, 2015 at 14:47
  • $\begingroup$ @AlgebraicPavel But how to prove those identities in your edit? And to make use of them, apparently one also needs to prove that $HH^+=H^+H$ when $H=XX^\ast=X^\ast X$. It seems to me that a lot of work still has to be done if we assume only the four defining properties as well as the existence and uniqueness theorem. $\endgroup$
    – user1551
    Jan 26, 2015 at 15:06
  • $\begingroup$ Well, if $XX^*=X^*X$ then assuming the identities in the edit are true, $XX^+=XX^*(XX^*)^+=(XX^*)^+XX^*=(X^*X)^+X^*X=X^+X$. The first and last inequality follows from these identities, the "inner" equalities uses the normality and the symmetry $(YY^+)^*=YY^*$. $\endgroup$ Jan 26, 2015 at 15:15
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    $\begingroup$ The wording in my question is probably not very good. I meant "using the four identities and their consequences", which need not to be obtained trivially from the defining conditions. The problem was to avoid spectral decompositions (or SVD) at any cost :-) $\endgroup$ Jan 26, 2015 at 15:39

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