# Real matrix without real eigenvalues commutes with some matrix of square $-I$.

Let $A\in M_n(\mathbb R)$ be such that, the minimal polynomial of $A$, has not any real root. Prove that there exist some $B\in M_n(\mathbb R)$ which: $B^2=-I_n$ and $AB=BA$.

Suppose that $p(x)=x^k+a_{k-1}x^{k-1}+...+a_1+a_0$ be the minimal polynomial of $A$. By the hypothesis, $a_0\neq 0$ and over ring $\mathbb R[x]$, the polynomial $p(x)$ has the following decomposition:

$$p(x)=(x^2+b_1x+b_2)^{d_1}(x^2+b_3x+b_4)^{d_2}...(x^2+b_mx+b_{m+1})^{ds}$$

With $b_j\in \mathbb R$ for $j=1,2,...,m+1$, and $d_1+d_2+...+d_s=k$.

If I prove that there exist $q(x),h(x)\in \mathbb R[x]$ such that:

$$(h(x))^2=p(x)q(x)-1$$

Then I am done by putting $B=h(A)$. But how can I prove that $q$ exists?

• It might be easier to write $p(x)=\prod_i ((x+a_i)^2+b_i^2)^{d_i}$ to ensure that there is no real root. – Thomas Andrews May 1 '15 at 22:57
• @GregMartin If $h^2=pq-1$ then $pq=h^2+1$. – Daniel May 2 '15 at 0:02
• Now define $B=h(A)$. then, since $p(A)=0$ and $h^2(x)=p(x)q(x)-1$, we should have: $B^2=-I_n$ and $AB=BA$ – hamid kamali May 2 '15 at 6:55
• No idea how to proceed from your line of argument. If you merely want to prove the original problem statement, it's easy to obtain a proof using real Jordan form. – user1551 May 2 '15 at 10:09
• It suffices to find two complementary real subspaces that are A invariant. Associate conjugate eigenvalues. – user91684 May 3 '15 at 9:42

First observation: by searching for a polynomial in$~A$ rather than just a matrix $B$ commuting with$~A$, you are demanding more then the question does, and thereby make the problem more difficult. One can arrange for $B$ to be a polynomial in$~A$, but it is easier if one does not have this restriction.

By the primary decomposition theorem, the space decomposes as a direct sum of the subspaces annihilated by the primary factors $(T^2+b_iT+c)^{d_i}$, and if one can find a proper linear operator $B_i$ on each subspace, they will combine to define $B$ on the whole space. This means one can restrict to the case where $p$ is a power of a single irreducible polynomial $q=x^2+bx+c$. One can further (non canonically) decompose the space as a direct sum of cyclic modules, each spanned as $\Bbb R[x]$-module by a single "cyclic" vector, which his annihilated by some power $(T^2+bT+c)^k$.

The case $k=1$ is easy: now the cyclic subspace is isomorphic as$~\Bbb R[x]$-module to $\Bbb R[x]/(q)$, which is isomorphic to$~\Bbb C$, so the action of multiplication by $\mathbf i$ under that correspondence will do as $B$. Concretely this is a polynomial of degree$~1$ in$~T$: completing the square $$q = x^2+bx+c = (x+s)^2+d^2 \qquad\text{with s=\frac b2 and d=\sqrt{c-s^2}}$$ one has for $B=\frac{T+sI}d$ that $B^2+I=0$. (This would have worked without decomposing into cyclic modules: the space would still be a module over $\Bbb R[x]/(q)\cong\Bbb C$, and the same $B$ would work.)

Now consider general $k$. For a given cyclic vector $v$, one sees that $v, q[T]v,q^2[T]v,~\ldots,~q^{k-1}[T]v$, together with the images of all of these vectors by $T$, form an $\Bbb R$-basis of the space. In the subspace spanned by $w_i=q^i[T]v$ and $Tw_i$, choose another basis $w_i,w_i'=\frac{T+sI}dw_i$, and define $B$ to act by $Bw_i=w'_i$ and $Bw'_i=-w_i$; then clearly $B^2=-I$, and it is readily checked that $B$ commutes with$~T$.

This $B$ is not a polynomial in$~T$: the operator $\frac{T+sI}d$ acts differently on the vectors $w'_i$ than $B$ does. Indeed $b=\frac{X+s}d$ only satisfies $b^2\equiv1\pmod{q}$, while one needs this to hold modulo $q^k$. This can be remedied by a technique similar to Hensel lifting, but I'll leave this to figure out yourself.