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Let $S$, $T$ be linear operators on a finite-dimensional vector space $V$ over $\mathbb{C}$. Suppose $$S^2 = T^2 = I.$$

Show that there exists either a $1$-dimensional or $2$-dimensional subspace of $V$ which is invariant under $S$ and $T$.

Ok so, since $S^2 = T^2 = I$, either $1$ or $-1$ are Eigenvalues of S and T.

i,e The Minimal Polynomial $M_T$ and $M_S$ satisfy

$$M_T \; | \; (x+1)(x-1)$$ $$M_S \; | \; (x+1)(x-1)$$

Edit : Found the same question elsewhere. For those of you who are interested in the answers.

Invariant Subspace of Two Linear Involutions

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  • If $S$ and $T$ share an eigenvector $v$, then $${cv : c\in\mathbb C}$$ is a one-dimensional subspace invariant under $S$ and $T$. – Math1000 Mar 20 '15 at 04:05
  • How can you assume they share one though? I thought you could only conclude that if they commute? –  Mar 20 '15 at 04:06
  • Either $S$ and $T$ share an eigenvector, or they don't. I'm too tired right now to handle the latter case ;) – Math1000 Mar 20 '15 at 04:08
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    The subspace consisting of everything of the form $Sv+v$ is the eigenspace for $\lambda=1$, and everything of the form $Sv-v$ is the eigenspace for $\lambda=-1$. Same for $T$ – shalop Mar 20 '15 at 04:20
  • I don't understand why the subspace needs to be 1- or 2-dimensional. For example, if $S=T=I$, then the given condition holds but the entire $n$-dimensional space $V$ is invariant under both $S$ and $T$. So what are the additional assumptions? – shalop Mar 20 '15 at 04:35
  • @Shalop Well, if $S = I = T$, then we can take any $1$- or $2$-dimensional subspace to fulfill the criteria; I don't believe it's saying that it can't be contained in a larger, invariant subspace. I liked your previous idea, by the way - it hasn't quite gotten me there, but it's been helpful! – pjs36 Mar 20 '15 at 04:37
  • Yes, the question only says that a 1 or 2 dimensional one exists, not that larger ones dont. –  Mar 20 '15 at 04:38
  • Those two eigenspaces from before span the entire space $V$, which is easy to show. Thus at least one of them has dimension greater than or equal to $\frac{n}{2}$. You can use this to show that if $n$ is odd, then $S$ and $T$ share an eigenvector, and then use Math1000's comment. I'm still trying to figure it out for even $n$ though, where $n$ is the dimension of $V$. – shalop Mar 20 '15 at 04:54
  • If S and T are diagonalisable. Then they are codigonalizable (as I showed they are commuting) implies that they share all their eigen vectors and we are done. But I don't understand why they are Diganolaizable. – Sry Mar 20 '15 at 06:09
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    Please verify that my edit didn't affect your question in a negative way. – hjhjhj57 Mar 20 '15 at 06:11
  • you edited it to make 1 and -1 eigenvalues for S and T. That's not true. It's possible for S to have -1 and T 1 or vice versa or even the same. Essentially several cases. –  Mar 20 '15 at 06:52

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If S and T share an eigenvector we are done. In second case, suppose x is eigen vector for S and y is a vector such that Ty = x. Claim is {x,y} is invariant under S & T both. First notice that Sx = x, Ty = x (now operate T both sides) gives Tx = y. again both S & T are self inverses here. On S^2 = T^2 pre multiply by s inverse and post by T inverse gives ST = TS. Thus Sy = STx = TSx = Ty = x and Ty = x So {x,y} is invariant under both S and T.

Sry
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  • I'm not convinced that $ST = TS$; the dihedral groups have complex representations, but two reflections don't generally commute. I like the idea of choosing $y$ such that $Ty = x$, that's a new angle. – pjs36 Mar 20 '15 at 06:22
  • It's not true that $ST=TS$. For example, take $S=\begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}$ and $T=\begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}$. Then $ST \neq TS$. – shalop Mar 20 '15 at 07:05
  • Guys I found some good but slightly complicated answers elsewhere on stackexchange. For those of you who are interested : http://math.stackexchange.com/questions/31363/invariant-subspace-of-two-linear-involutions?rq=1 –  Mar 20 '15 at 07:15
  • XD. As I said before, the case where $V$ has odd dimension is easy, and the case with even dimension is hard. Really the only hard case is where all of the eigenspaces have dimension $\frac{n}{2}$. That's really where you need algebraic closedness of $\mathbb{C}$. – shalop Mar 20 '15 at 07:25