Consider a finite dimensional $\Bbb Q$-vector space $V$ and its endomorphism $T$. I am trying to prove that $V$ is a direct sum of $2$-dimensional $T$-invariant subspaces if $T^2=-I$.

My attempt: From the assumption we know that the minimal polynomial divides $x^2+1$, and hence equals $x^2+1$, because $x^2+1$ is irreducible over $\Bbb Q$. Since the minimal polynomial and the characteristic polynomial of $T$ have the same monic irreducible divisors, it follows that the characteristic polynomial is a power of $x^2+1$, so the dimension of $V$ should be even. But I got stuck here. Any hints?

  • $\begingroup$ Hint: If you extend scalars to $\mathbb{Q}(i)$, then $T$ has an eigenbasis. $\endgroup$ Apr 6, 2020 at 17:13

1 Answer 1


Idea : Create a new vector space in which $T$ is diagonalizable i.e. has 1D invariant subspaces. Pull back all the subspaces to $V$ via an appropriate map, wherein all the invariant subspaces get mapped back to the desired disjoint subspaces.

We do the following : I claim that $V$ can be given a vector space structure over the complex numbers (or a field isomorphic to the complex numbers).

More precisely, consider $\mathbb Q(T)$, the field created by taking all rational functions in $T$ with real coefficients (so elements of this field are matrices of the form $p(T)(q(T))^{-1}$ where $p,q$ are polynomials with real coefficients). One easily sees that this field is isomorphic to $\mathbb Q(i)$, by sending $x+iy \to xI + yT$ and noting the relation satisfied by $T$.

To make $V$ a vector space over $\mathbb Q(T)$ is obvious : addition is as usual and scalar multiplication is defined by $(aI+bT)v = av + b(Tv)$.

Denote $V_{\mathbb Q(T)}$ for $V$ with changed base field.

Once $V$ is an $\mathbb Q(T)$ subspace, the operator $T : V_{\mathbb Q(T)} \to V_{\mathbb Q(T)}$ (now scalar multiplication by the field element $T$) is obviously a diagonal operator regardless of basis, a multiple of the identity matrix. Consequently, we get that $V_{\mathbb Q(T)}$ breaks into $1$ dimensional subspaces which are invariant under $T$ (obviously, taking any basis, the elements of this basis are all eigenvectors, so form one dimensional invariant subspaces). This is similar to how scalar multiplication (in the usual setting) does not change matrix regardless of change of basis.

We now perform a restriction of scalars as follows. Define a map $i : V_{\mathbb R} \to V_{\mathbb Q(T)}$, by the identity map (!)

I claim that any one dimensional subspace $\overline{\{b\}}$ of $V_{\mathbb Q(T)}$ has a two dimensional preimage. But this is obvious : $\overline{\{b\}}$ in $\mathbb Q(T)$ consists of all the elements $(xI+yT)b = xb + yTb \in \overline{\{b,Tb\}_{\mathbb Q}}$, and $b,Tb$ are linearly independent over $\mathbb Q$. Furthermore this is $T$ invariant. (All easy checks)

Thus, if $b_1,...,b_n$ is a basis of $V_{\mathbb Q(T)}$, then $i^{-1} (\overline{\{b_j\}})$ is a two dimensional $T$ invariant subspace of $V_{\mathbb Q}$. Finally, these subspaces are disjoint, because their images under $i$ lie in different subspaces of $V_{\mathbb Q(T)}$.

It follows that $V = \oplus i^{-1}(b_j)$.

  • $\begingroup$ Shouldn't we have to show that the orthogonal complement to $\{v,Tv\}$ is also $T$-invariant? $\endgroup$
    – blancket
    Apr 6, 2020 at 18:11
  • $\begingroup$ It is. Try to prove that from injectivity of T. I left that for you to verify because it is standard,but if it is not then I will complete it on your request. $\endgroup$ Apr 7, 2020 at 4:40
  • $\begingroup$ Can you show it too? $\endgroup$
    – blancket
    Apr 7, 2020 at 15:34
  • $\begingroup$ I'll be doin' JUST that, kind person! $\endgroup$ Apr 7, 2020 at 15:59
  • $\begingroup$ I've done it. If you are fine, I request you to accept the answer! Thank you and look forward to seeing your other questions! $\endgroup$ Apr 7, 2020 at 16:12

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