Lately I encountered a situation when particular property of a real square matrix depended on the parity of matrix dimension, namely:

  • for even dimension from equality of adjugates of invertible matrices we can infer about equality of matrices, for odd dimension there is no equivalence ( i.e. matrix function $B=\text{adj}(A)$ is bijective in the set of non-singular matrices only for even dimensions)

The other well known situation when the parity of matrix dimension also matters is the situation for skew-symmetric matrices:

  • only even-dimensional skew-symmetric matrices can have full rank.

What are the other situations when some property of a real square matrix is alternating due to the changing dimension parity?

Maybe somewhere is a list of such properties?
If there is no general list single examples will be also good as the answer ... at least one additional example...

I'm adding the next property

  • only for even dimensions the equation $A^2=kI$ where $k <0$ has a solution in real numbers (more accurately: infinite number of solutions) - lack of solution for odd $n$ follows immediately from $\det(A^2) \geq 0 $ and $det(kI)=k^n$.
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    $\begingroup$ I don't know if this is really what you're looking for, but the groups of matrices $$\textrm{SO}_n(\mathbb{R}) = \{ x \in \textrm{GL}_n(\mathbb{R}) : xx^t = I \}$$ are quite different for $n$ even or odd. $\endgroup$ – D_S Jun 10 '17 at 13:34
  • $\begingroup$ @D_S Yes, if they are quite different that is what I'm looking for, But what do you regarding as 'quite different' ? There is a difference between $n=2$ and $n=3$, but what is the general difference between odd and even $n$ ...? $\endgroup$ – Widawensen Jun 10 '17 at 13:46
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    $\begingroup$ Their root systems are in different classes. The "four classical semisimple groups" are $$\textrm{SL}_n, \textrm{Sp}_{2n}, \textrm{SO}_{2n}, \textrm{SO}_{2n+1}$$ so the odd and even orthogonal groups can be considered as different from each other as they are from $\textrm{SL}_n$. I can post a more detailed answer if you want. $\endgroup$ – D_S Jun 10 '17 at 13:58
  • $\begingroup$ @D_S Very interested in.. all details are welcome.. $\endgroup$ – Widawensen Jun 10 '17 at 14:02
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    $\begingroup$ "only even-dimensional skew-symmetric matrices can have full rank" - this of course extends to all matrices with eigenvalues appearing in $\pm$ pairs, e.g. skew-centrosymmetric matrices. $\endgroup$ – ekkilop Jul 15 '17 at 15:29

Let $K$ be an involutory matrix, i.e. $K^2 = I$, and let the matrix $A$ be a skew-centrosymmetric $K$-matrix, i.e. $A$ anti-commutes with $K$: $$ AK = -KA. $$ If $(\lambda,v)$ is an eigenpair of $A$ s.t. $$ A v = \lambda v, $$ then $$ \lambda K v = K A v = K A K K v = - A K v, $$ i.e. $(-\lambda, Kv)$ is also an eigenpair of A. Thus, the eigenvalues of $A$ come in $\pm$ pairs, so if the dimension of $A$ is odd, then $A$ must have a zero eigenvalue. Hence, only even-dimensional skew-centrosymmetric $K$-matrices can have full rank.


The same argument may be applied in the case where $AK = -KA^\top$ by noting that $A$ and $A^\top$ have the same eigenvalues. This includes skew-symmetric ($K=I$) and skew-persymmetric ($K=J$, where $J$ is the exchange matrix) matrices as special cases.

  • $\begingroup$ Very interesting property, thank you.. what about skew-persymmetric?, I have checked by example calculations that probably the case may be similar ( if zeros are on the antidiagonal) - the eigenvalues also come in $\pm $ pairs.. $\endgroup$ – Widawensen Jul 17 '17 at 11:18
  • $\begingroup$ Since the eigenvalues of $A$ and $A^\top$ are the same, we can use an identical argument as above to obtain the same result for any matrix satisfying $AK = -KA^\top$. Then, skew-persymmetry corresponds to $K = J$, where $J$ is the exchange matrix, doesn't it? $\endgroup$ – ekkilop Jul 17 '17 at 12:44
  • $\begingroup$ yes, you are right. If you would like you may add your comment to your answer, it would make it richer.. unfortunately as I've already upvoted your answer I can't do this second time as it deserves.. $\endgroup$ – Widawensen Jul 17 '17 at 12:51
  • $\begingroup$ Thank you for your kind words. I have updated the answer as suggested. $\endgroup$ – ekkilop Jul 17 '17 at 12:56

From another thread:

If $p$ is a polynomial without real roots, and $A$ is a real matrix with odd dimension, then $p(A) = 0 $ never holds.

This follows mainly from the fact that odd degree real polynomials always have a root.

  • $\begingroup$ Thank You Raul for the answer, If it would happen that you know also other properties which depend on parity of matrix dimension, don't hesitate - add them here.. $\endgroup$ – Widawensen Jun 21 '17 at 10:02

The response here can be regarded as a follow-up to ekkilop’s fine answer. It can be viewed as a generalization of that answer, and remains pertinent to the question being asked.

Let the term $m$-involution mean any complex square matrix $K$ for which $K^m=I$, where $m$ is an integer greater than $1.$ The usual involutory matrix case corresponds to the situation where $m=2.$ One question we may ask is, for a given square matrix $A$, what can we say about the set of $m$-involutory matrices that anti-commute with it? Let’s denote this set as $\tilde S_m\left( A \right)$.

Given $A \in \mathbb{C}^{n \times n},$ it is possible to establish the following result:

  1. Let $n=1$. If $A$ is the number zero, then $\tilde S_m\left( A \right)$ contains the $m$-th roots of unity. Otherwise, $\tilde S_m\left( A \right)$ is empty.
  2. Suppose $n>1$ and let $m>1$ be an odd integer. If $A$ is the zero matrix, then $\tilde S_m\left( A \right)$ is non-denumerable. Otherwise, $\tilde S_m\left( A \right)$ is empty.
  3. Suppose $n>1$ and let $m>1$ be an even integer. If the non-zero eigenvalues of $A$ come in pairs of opposite sign where the corresponding pairs have Jordan blocks of equal size, then the cardinality of the set $\tilde S_m\left( A \right)$ is non-denumerable. Otherwise $\tilde S_m\left( A \right)$ is empty.

Characterizing the sets to be empty or non-denumerable for $n>1$ may seem to be a rather weak statement, but note that this result contrasts, for example, with the set of $m$-involutions that commute with a fixed diagonalizable matrix $A\in \mathbb{C}^{n \times n}$ with distinct eigenvalues in which the case the cardinality of the set is $m^n.$

Note that the first and third cases together imply ekkilop’s observation. For the third case with $m=2$, if $n$ is odd then if $A$ is non-singular the Jordan blocks corresponding to nonzero eigenvalues of $A$ cannot all be paired up as positive - negative pairs of equal size. In other words, if $n$ is odd and $A$ is non-singular, then $\tilde S_2\left( A \right)$ must be empty: i.e., there are no involutory matrices $K$ that can anti-commute with $A$. Therefore, if $n$ is odd and $A$ anti-commutes with some involutory matrix $K$, then $A$ must be singular.

The theorem quoted above is trivial to prove for the first two cases. For the second case, we can simply observe that for an $m$-involutory matrix $K$ that anti-commutes with $A,$ we have that $A = K^mA = (-1)^mAK^m = (-1)^mA$, so if $m$ is odd it then follows that $\tilde S_m\left( A \right)$ is empty except when $A$ is the zero matrix.

For the third case, the argument is somewhat more involved. For that demonstration, the interested reader can find a proof in the paper "Some properties of commuting and anti-commuting m-involutions. Perhaps more interesting is that this paper's Lemma 5.1, which is the main observation used in the proof, can be used to construct $m$-involutory matrices $K$ that anti-commute with a given matrix $A$, provided such $K$ exist.


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