3
$\begingroup$

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.

Question:
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...

Edit
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 $det(kI)=k^n$.
$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ "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
1
$\begingroup$

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.

Edit

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.