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Greetings to all !

Let me first confess that this question was mentionned to me by Bernard Dacorogna, who doesn't sail on MO.

Let $A\in M_{2n}(k)$ be an alternate matrix. Say that $A$ is non-singular. It is well-known that there exists an $M\in GL_{2n}(k)$ such that $A=M^TJM$, where $$J=\begin{pmatrix} 0_n & I_n \\\\ - I_n & 0_n \end{pmatrix}.$$ Of course, $M$ is not unique. Every product $M'=QM$ with $Q\in Sp_n(k)$ ($Q$ symplectic) works as well.

Is it always possible to choose $M$ symmetric ? In this case, we have $A=MJM$, but an identity $A=RJR$ does not imply that $R$ be symmetric.


Let $M\in GL_{2n}(k)$ be given. Is it true that $Sp_n(k)\cdot M$ meets $Sym_{2n}(k)$ non-trivially?

Notice that we must have $\det M=Pf(A)$. Therefore, if $k=\mathbb R$, the symmetric $M$ that we are looking for cannot always be positive definite.

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By "alternate" you mean skew-symmetric/anti-symmetric? –  Federico Poloni Jan 2 '12 at 11:21
@Federico. Alternate is a bit more precise than skew-symmetric. The latter means that $A^T+A=0$ ; if the field $k$ has characteristic $2$, this says nothing about the diagonal entries. For an alternate matrix, we always have $a_{jj}=0$ by definition. –  Denis Serre Jan 2 '12 at 11:25
So we have an alternate reason why $M$ cannot always be positive-definite: the field might have characteristic $2$. –  Tom Goodwillie Jan 2 '12 at 22:40
It is always possible to find a symmetric non-singular matrix $S\in\mathrm{Sym}_{2n}(k)$ such that $SAS=MJM$, i.e., $SAS$ can be expressed in the form $MJM$ with $M$ symmetric and non-singular. If this interests you, and is not already known to you, I can explain. –  Mahdi Majidi-Zolbanin Jan 3 '12 at 17:20
For an arbitrary field, my statement follows from an old theorem of Taussky and Zassenhaus, that *every matrix $M$ with entries in a field can be transformed into its transpose by a symmetric non-singular matrix $S$*, i.e., there exists $S$ symmetric and non-singular such that $MS=SM^T$ (see Theorem 1 in On the similarity transformation between a matrix and its transpose, Pacific J. Math, (1959)). Now, we can write $A=M^TJM$ and choose $S$ such that $MS=SM^T$. Then we get my statement if we multiply by $S$ on both sides of $A=M^TJM$. Note $S$ and $MS$ are symmetric. –  Mahdi Majidi-Zolbanin Jan 4 '12 at 13:39

3 Answers 3

up vote 24 down vote accepted

I feel that framing this question in terms of matrices rather than bilinear forms on a vector space obscures what is actually going on and makes it harder to understand what needs to be proved. Here is how I would describe the problem and the partial answer that results from this description:

Let $V$ be a finite dimensional vector space over a field $k$, and let $\mathsf{B}(V)$ denote the vector space over $k$ consisting of bilinear forms on $V$, i.e., an element $\beta\in \mathsf{B}(V)$ is a bilinear mapping $\beta: V\times V\to k$. An element $\alpha\in \mathsf{B}(V)$ is said to be alternating if $\alpha(x,x) = 0$ for all $x\in V$, and an element $\sigma\in \mathsf{B}(V)$ is said to be symmetric if $\sigma(x,y)=\sigma(y,x)$ for all $x,y\in V$. The subset $\mathsf{A}(V)\subset\mathsf{B}(V)$ of alternating forms is a subspace, as is the subset $\mathsf{S}(V)\subset\mathsf{B}(V)$ of symmetric forms. When the characteristic of $k$ is not $2$, there is a $GL(V)$-invariant direct sum decomposition $\mathsf{B}(V) = \mathsf{A}(V)\oplus\mathsf{S}(V)$. When the characteristic of $k$ is $2$, one has, instead, $\mathsf{A}(V)\subset \mathsf{S}(V)\subset\mathsf{B}(V)$, and, apparently, these inclusions have no $GL(V)$-invariant splittings.

An element $\beta\in \mathsf{B}(V)$ is nondegenerate if, for each $x\not=0$ in $V$, there exists a $y\in V$ such that $\beta(x,y)\not=0$. If $\alpha\in \mathsf{A}(V)$ is nondegenerate, then the dimension of $V$ over $k$ must be even. Conversely, if the dimension of $V$ over $k$ is even, then there exists a nondegenerate $\alpha\in\mathsf{A}(V)$, and, moreover, any other nondegenerate $\overline\alpha\in\mathsf{A}(V)$ is of the form $\overline\alpha = m^\ast(\alpha)$ for some $m\in GL(V)$, where, by definition, $$ \bigl(m^\ast(\alpha)\bigr)(x,y) := \alpha(mx,my) $$ for any $m:V\to V$. When $\dim_k(V) = 2n$, let $\mathsf{K}(V) = \Lambda^{2n}(V^\ast)$ denote the $1$-dimensional vector space consisting of $2n$-multilinear alternating functions on $V$. There exists a canonical polynomial mapping $\mathrm{Pf}:\mathsf{A}(V)\to \mathsf{K}(V)$ of degree $n$ that satisfies $\alpha^n = n!\ \mathrm{Pf}(\alpha)$ (a property that defines $\mathrm{Pf}$ when $n$ is less than the characteristic of $k$). This Pfaffian vanishes if and only if $\alpha$ is degenerate, and it satisfies $\mathrm{Pf}\bigl(m^\ast(\alpha)\bigr) = \det(m)\ \mathrm{Pf}(\alpha)$.

When $\alpha\in\mathsf{A}(V)$ is nondegenerate, let $Sp(\alpha)\subset GL(V)$ denote the subgroup consisting of those $m\in GL(V)$ such that $\alpha(mx,my)=\alpha(x,y)$ for all $x,y\in V$. Define two subspaces ${\frak{s}}(\alpha)\subset \mathrm{End}(V) \simeq V\otimes V^\ast$ and ${\frak{a}}(\alpha)\subset \mathrm{End}(V)$, by saying that $s\in{\frak{s}}(\alpha)$ if $s^\flat(x,y) := \alpha(x,sy)$ is symmetric, while $a\in{\frak{a}}(\alpha)$ if $a^\flat(x,y) := \alpha(x,ay)$ is alternating. Note that ${\frak{s}}(\alpha)$ is a subalgebra of $V\otimes V^*$ under the commutator bracket; in fact, it is the Lie algebra of $Sp(\alpha)$. The subspaces ${\frak{s}}(\alpha)$ and ${\frak{a}}(\alpha)$ are invariant under conjugation by elements of $Sp(\alpha)$.

Now, there is a natural map $S:{\frak{s}}(\alpha)\to {\frak{a}}(\alpha)$, given by $S(s) = s^2$. In other words, if $\alpha(x,sy)$ is symmetric, then $\alpha(x,s^2x) = \alpha(sx,sx) = 0$, so $\alpha(x,s^2y)$ is alternating.

Here, then, is the question: What is the image of $S$? (The OP is actually asking whether the image of $S$ contains the invertible elements of ${\frak{a}}(\alpha)$.)

Note that the dimension of ${\frak{s}}(\alpha)$ is $2n^2{+}n$, while the dimension of ${\frak{a}}(\alpha)$ is $2n^2{-}n$, so it's conceivable that $S$ is actually surjective. Also, the map $S$ is $Sp(\alpha)$-equivariant, so it's a question that can be studied by looking at the orbits of this group acting on ${\frak{a}}(\alpha)$.

Remark: It took me a while to recognize that this is what is going on because the question, as asked, sneaks in an extraneous quadratic form that breaks the symplectic symmetry. A 'reference' alternating form on $k^{2n}$ has been specified by the formula $\alpha_0(x,y) = x^TJy$ for $x,y\in k^{2n}$. Note that the matrix $J$ satisfies $J^2 = -I$, an identity that has no meaning for an alternating form. The only way one can interpret an alternating form as a linear transformation (so that squaring makes sense) is to have some other way of identifying $V$ with $V^\ast$. Of course, this is supplied by the linear map $x\mapsto x^T$ in the formula. In other words, a (symmetric) bilinear form $\beta(x,y) = x^Ty$ has been introduced into the picture, and it breaks the symplectic symmetry. Anyway, writing $\alpha(x,y) = x^TAy$ and asking whether one can write $A = MJM$ for $M$ symmetric can be re-interpreted as follows: Note that $M=Js$ where $s\in{\frak{s}}(\alpha_0)$ and that $\alpha(x,y) = x^TAy = x^TJJ^{-1}Ay = \alpha_0(x,J^{-1}Ay) = \alpha_0(x,ay)$ where $a = J^{-1}A$ lies in ${\frak{a}}(\alpha_0)$. Putting this together says that $$ Ja = A = MJM = (Js)J(Js) = -Js^2, $$ so showing that the equation $A = MJM$ can be solved is equivalent to showing that the equation $a = -s^2$ for a given $a\in{\frak{a}}(\alpha_0)$ can be solved for some $s\in{\frak{s}}(\alpha_0)$. (It's off by a minus sign, but that's OK because the goal is to characterize the image of $S$, so characterizing its negative is just as good.)

Anyway, back to the question: One approach is to look at the orbits of $Sp(\alpha)$ acting on ${\frak{s}}(\alpha)$ and see what their squares look like. This may be easier because the adjoint orbits of $Sp(\alpha)$ on its Lie algebra have been much studied.

As an example, suppose that $k$ is algebraically closed and (for my comfort) that it has characteristic zero. Say that a pair of nondegenerate alternating $2$-forms $(\alpha_0,\alpha)$ on $k^{2n}$ is generic if the $n$ roots of the equation $\textrm{Pf}(\alpha - \lambda\ \alpha_0) = 0$ are all distinct. Then one can prove (see below) that a basis of $1$-forms on $k^{2n}$ exists so that $$ \alpha_0 =\theta^1\wedge\theta^2+\theta^3\wedge\theta^4+\cdots +\theta^{2n-1}\wedge\theta^{2n} $$ while $$ \alpha =\lambda_1\ \theta^1\wedge\theta^2+\lambda_2\ \theta^3\wedge\theta^4+\cdots +\lambda_n\ \theta^{2n-1}\wedge\theta^{2n}. $$ Thus, the problem uncouples into $n$ separate problems that are each trivially solvable, so, the problem is solvable for the generic pair in this case.

As another example, in the case $n=2$, for an arbitrary field (even one of characteristic $2$), one can, by hand, classify the pairs $(\alpha_0,\alpha)$ with $\alpha_0$ nondegenerate and show that $S:{\frak{s}}(\alpha_0)\to{\frak{a}}(\alpha_0)$ is surjective. (I'll put in the details if someone asks. Note, by the way, that the claimed counterexample when $n=2$ in the 'answer' below does not actually work.)

To prove surjectivity for all $n$, one may need to understand the orbits of $Sp(\alpha)$ acting on ${\frak{a}}(\alpha)$. I think that this is a classical problem (I'm not an algebraist, so I'm not completely sure), so maybe it's time to look at the literature. The classification of the possible $Sp(\alpha)$-orbit types in ${\frak{a}}(\alpha)$ gets more complicated as $n$ increases, so maybe some other approach needs to be tried. One would expect the orbits of $Sp(\alpha)$ in ${\frak{a}}(\alpha)$ to be somewhat simpler than the orbits of $Sp(\alpha)$ in ${\frak{s}}(\alpha)$, just because the dimension is lower. However, I note that the rings of $Sp(\alpha)$-invariant polynomials on each of the vector spaces ${\frak{s}}(\alpha)$ and ${\frak{a}}(\alpha)$ are each free polynomial rings on $n$ generators, so it may be that the complexity of the orbit structures are (at least roughly) comparable in the two cases.

The uncoupling step: In the general case, for $a\in{\frak{a}}(\alpha_0)$, one has $\textrm{Pf}(a^\flat - \lambda\ \alpha_0) = p_a(\lambda)\ \textrm{Pf}(\alpha_0)$, where $\det(a - \lambda I) = p_a(\lambda)^2$. Letting $f_1(\lambda),\ldots,f_k(\lambda)$ denote the distinct irreducible factors of $p_a(\lambda)$, one has $$ p_a(\lambda) = f_1(\lambda)^{d_1}\cdots f_k(\lambda)^{d_k}. $$ There is a direct sum decomposition $V = V_1\oplus\cdots\oplus V_k$ into the corresponding generalized eigenspaces of $a$, and one sees without difficulty (using the identity $\alpha_0(ax,y) = \alpha_0(x,ay)$) that $\alpha_0(V_i,V_j) = 0$ for $i\not=j$. Moreover, any solution $s\in{\frak{s}}(\alpha_0)$ of $a = \pm s^2$ must commute with $a$ and therefore preserve its generalized eigenspaces. Thus, generalizing the 'uncoupled' situation given in the first example, one sees that the problem reduces to the case in which $p_a(\lambda)$ is a power of a single irreducible polynomial.

Unfortunately, it turns out that, even in this uncoupled case, the minimal polynomial of $a$ can fail to be irreducible (i.e., $a$ need not be semi-simple), and I do not know a simple way to handle all of these cases. When $n=2$, this can be handled 'by hand', but even for $n=3$, it seems to be a little tricky (although, I think that I have correctly handled the cases there and shown surjectivity in that case as well).

The uncoupled semi-simple case: Here is how one can complete the proof of solvability in the uncoupled, semi-simple case, i.e., when the minimal polynomial of $a$ is irreducible. Let $p(\lambda)=0$ be the (irreducible) minimal polynomial of $a$, say of degree $m$. We can assume that $p(0)\not=0$, since, otherwise, $a=0$, and there is nothing to prove (i.e., one can just take $s=0$, and the problem is solved). Let $K\subset End(V)$ denote the field generated by $a$, so that $[K:k]=m$.

Now, for any nonzero $x\in V$, one has $\alpha_0(K{\cdot}x, K{\cdot}x) = 0$, since $\alpha_0(a^ix,a^jx)=0$ for all $i$ and $j$. If one chooses $x,y\in V$ such that $\alpha_0(x,y)\not=0$, then it is easy to see (using $p(0)\not=0$) that $\alpha_0$ is nondegenerate on the $a$-invariant subspace $W = K{\cdot}x\oplus K{\cdot}y$. One then can write $V = W\oplus W^\perp$ (where the $\perp$ is taken with respect to $\alpha_0$) and see that the problem uncouples into separate problems on $W$ and $W^\perp$. By induction, it then suffices to consider the case $W = V$ (and, hence, $n=m$).

In this case, the general element of $V = W = K{\cdot}x\oplus K{\cdot}y$ can be written uniquely in the form $z = f_1(a)x + f_2(a)y$ where $f_1$ and $f_2$ are polynomials (with coefficients in $k$) of degree at most $m{-}1$. Let $q(a)$ and $r(a)$ be any polynomials in $a$ (i.e., elements of $K$) and define a map $s:V\to V$ by $$ s\bigl(f_1(a)x + f_2(a)y\bigr) = f_1(a)q(a)y + f_2(a)r(a)x. $$ One checks that $s\in{\frak{s}}(\alpha_0)$ and notes that, by construction, one has the identity $s^2 = q(a)r(a)$. By setting $q(a) = a$ and $r(a) = \pm1$, one sees that one may arrange $s^2 = \pm a$. This $s$ solves the problem.

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For action of the symplectic group on its Lie algebra see mathoverflow.net/questions/31391/… and links there. –  Vít Tuček Jan 8 '12 at 1:18
@r0b0t: Unfortunately, this reference is not really all that useful. What is written there in the answer is fairly obvious (and can actually be deduced from what I have written above with very little trouble), and it only applies to the finite fields $\mathbb{F}_p$, not the general field of characteristic $p$ (or characteristic $0$). Moreover, what we really need to know are the orbits of $Sp(\alpha)$ on ${\frak{a}}(\alpha)$, especially the orbits of nondegenerate elements, not the orbits of $Sp(\alpha)$ on the Lie algebra, ${\frak{s}}(\alpha)$, if we are to solve the problem in general. –  Robert Bryant Jan 8 '12 at 2:34
@Robert: I think you meant to say that $a(\alpha)$ is a Lie algebra, not $s(\alpha)$. Fantastic answer, by the way. –  Spiro Karigiannis Jan 8 '12 at 12:56
@Spiro: Thanks for the compliment. It turned out to be more a interesting problem than I thought at first, once I figured out what was going on, and I now think that it should be possible to completely answer the question, even a generalization of the question. Actually, I do mean ${\frak{s}}(\alpha)$ is the Lie algebra of $Sp(\alpha)$, and not ${\frak{a}}(\alpha)$. It's counterintuitive for us geometers because, for a nondegenerate symmetric bilinear form $g$, the Lie algebra of the group $SO(g)$ is ${\frak{a}}(g)$, not of ${\frak{s}}(g)$ (once the appropriate definitions are made). –  Robert Bryant Jan 8 '12 at 21:10
Maybe I got the actions wrong, but I think that $\mathfrak{a}(\alpha)$ is a reducible representation of $\mathrm{Sp}(\alpha)$ as $\langle \alpha^\sharp \rangle$ is an invariant subspace. –  Vít Tuček Jan 9 '12 at 9:29

I claim that if $k$ is algebraically closed and $\alpha$ is nondegenerated then the natural map $S: \mathfrak{s}(\alpha) \to \mathfrak{a}(\alpha)$, given by $S(s) = s^2$ is surjective. As Robert Bryant showed above it then suffices to consider the case when $p_a(\lambda) = f(\lambda)^d$ where $f \in k[\lambda]$ is irreducible. So $f(\lambda) = \lambda - t$, $t \in k$. Now I am going to use the Jordan normal form of $a$. Let $v_1,\cdots,v_r,v_{r+1},\cdots$ be the cyclic generators of $V$, where the first $r$ are those of maximal order i.e. $(a - t)^{d-1} v_j \neq 0 $, $j = 1,\cdots,r$ and $(a - t)^{d-1} v_j = 0$ if $j > r$. Observe that:

i) there is $j_0 \in \{ 2,\cdots,r \}$ such that $$ \alpha((a - t)^{d-1} v_1, v_{j_0}) \neq 0 \, $$ otherwise $\alpha$ must be degenerated i.e. $\alpha((a - t)^{d-1} v_1, \cdot) = 0$.

ii) $\alpha$ is nondegenerated on the $a$-invariant subspace $W = K \cdot v_1 \oplus K\cdot v_{j_0}$ where $K = k[a] \subset End(V)$. Indeed, by computing the matrix $\mathrm{M}$ of $\alpha$ w.r.t. the base $$ ((a-t)^{d-1}v_1, (a-t)^{d-2}v_1\cdots,v_1,(a-t)^{d-1}v_{j_0},(a-t)^{d-2}v_{j_0},\cdots,v_{j_0}) $$ one gets that $\mathrm{det}(\mathrm{M}) = \pm \alpha((a - t)^{d-1} v_1, v_{j_0})^{2d} \neq 0$.

So $V = W \oplus W^{\perp}$ (where $\perp$ is taken w.r.t. $\alpha$) and by induction it is enough to consider $V = W = K \cdot v_1 \oplus K\cdot v_{j_0} $. In this case, the same idea as in the uncoupled semi-simple case works since the minimal polynomial of $v_1,v_{j_0}$ are the same i.e. $(\lambda - t)^d$. Namely, any element $z \in K \cdot v_1 \oplus K\cdot v_{j_0}$ can be written uniquely in the form $z = f_1(a) v_1 + f_2(a)v_{j_0}$ where $f_1,f_2$ are polynomials of degree at most $d-1$. Define the map $s : V \to V$ by $$ s(f_1(a) v_1 + f_2(a)v_{j_0}) = f_1(a)q(a)v_{j_0} + f_2(a)r(a)v_1 \, . $$ One checks that $s \in \mathfrak{s}(\alpha)$ and notes that, by construction, one has the identity $s^2=q(a)r(a)$. By setting $q(a)=a$ and $r(a)=\pm 1$ one may arrange that $s^2 = \pm a$ solving the problem. Keeping in mind Robert Bryant's answer this shows that the OP problem has a positive answer when $k = \mathbb{C}$.

A modification of the above argument shows that $S$ is surjective for $k=\mathbb{R}$ and $\alpha$ nondegenerated. Indeed, at the light of the above it is enough to consider the case $p_a(\lambda) = (\lambda^2 + t \lambda + s)^d$ with $s,t \in \mathbb{R}, \, \, s - \frac{t^2}{4} > 0$. Set $\tilde{a} := \frac{a + \frac{t}{2}}{\sqrt{s - \frac{t^2}{4}}}$. Then $\tilde{a} \in \mathfrak{a}(\alpha)$ and $p_{\tilde{a}}(\lambda) = (\lambda^2 + 1)^d$. Now I use the generalized Jordan normal form, or primary rational canonical of $\tilde{a}$ (e.g. http://en.wikipedia.org/wiki/Frobenius_normal_form). Let $v_1,\cdots,v_r,v_{r+1},\cdots$ be the cyclic generators of $V$, where the first $r$ are those of maximal order i.e. $(\tilde{a}^2 + 1)^{d-1} v_j \neq 0 $, $j = 1,\cdots,r$ and $(\tilde{a}^2 + 1)^{d-1} v_j = 0$ if $j > r$.

I claim that there is $j_{0} \in \{2,\cdots,r\}$ such that the $2 \times 2$ determinant $$ \begin{vmatrix} \alpha(\tilde{a} (\tilde{a}^2 + 1)^{d-1} v_1 , \tilde{a} v_{j_0}) & \alpha(\tilde{a}(\tilde{a}^2 + 1)^{d-1} v_1 , v_{j_0} )\\ \alpha((\tilde{a}^2 + 1)^{d-1} v_1 , \tilde{a} v_{j_0}) & \alpha(\tilde{a}^2 + 1)^{d-1} v_1 , v_{j_0} ) \end{vmatrix} \neq 0 \, .$$ Indeed, the above determinant is equal to $$ -\left(\alpha(\tilde{a}^2 + 1)^{d-1} v_1 , v_{j_0} )\right)^2 - \left(\alpha(\tilde{a}(\tilde{a}^2 + 1)^{d-1} v_1 , v_{j_0} ) \right)^2 \, . $$ If for all $j \in {2,\cdots,r}$ the above determinants are zero then $$ \alpha(\tilde{a}^2 + 1)^{d-1} v_1 , v_{j} ) = \alpha((\tilde{a}^2 + 1)^{d-1} v_1 , \tilde{a}v_{j} ) = 0 $$ and $\alpha$ must be degenerated i.e. $\alpha((\tilde{a}^2 + 1)^{d-1} v_1, \cdot)=0$.

Thus there is $j_0$ such that the above determinant is not zero. Then it is not difficult to see that $\alpha$ is nondegenerated on the $\tilde{a}$-invariant subspace $W = K \cdot v_1 \oplus K\cdot v_{j_0}$. Then as in the uncoupled semi-simple case by setting $q(\lambda) = \lambda \sqrt{s - \frac{t^2}{4}} - \frac{t}{2} ,r(\lambda) = \pm 1$ ones gets $s \in \mathfrak{s}(\alpha)$ and $s^2 = q(\tilde{a})r(\tilde{a}) = \pm a$ solving the problem. Keeping in mind Robert Bryant's answer this shows that the OP problem has a positive answer when $k = \mathbb{R}$.

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If the field $k$ has characteristic 2 the following alternate matrix $A$ can not be written as $A = MJM$ with $M$ symmetric:
$$ A = \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right) $$

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Actually, in characteristic 2, the symmetric matrix $$ M = \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ \end{array} \right) $$ satisfies $A = MJM$. –  Robert Bryant Mar 31 at 14:03
I made a mistake. –  Holonomia May 18 at 12:31

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