# Suppose $T$ satisfies $T^4-1=T^3+3T^2+T+3=0$. Show $V$ is the direct sum of even dimensional $T$-invariant subspaces.

Question

Let $V$ be a finite dimensional vector space over $\mathbb{R}$ and suppose the linear operator $T$ satisfies $T^4-1=T^3+3T^2+T+3=0$. Show $V$ is the direct sum of two dimensional $T$-invariant subspaces.

Attempt

Factoring shows that $T$ is annihilated by both $p(t)=(t-1)(t^2+1)(t+1)$ and $g(t)=(t+3)(t^2+1)$. The minimum polynomial $\mu$ must divide both $p(t)$ and $g(t)$ so we see that $\mu=(t^2+1)$. Since the characteristic polynomial $\chi$ of $T$ and $\mu$ have the same roots, $\chi(t)=(t^2+1)^n$. I am unsure how to reason further. Perhaps: Since the rational canonical form exists, there is a direct decomposition of $V$ into cyclic subspaces of $V$, i.e $$V=C_{x_1} \oplus \ldots \oplus C_{x_k}.$$ Note that $C_{x_i}$ is $T$ invariant. Thus the minimum polynomial of $T\mid_{C_{x_k}}$ divides $\mu$ and since $t^2+1$ does not split over $\mathbb{R}$, it must be the case that the minimum polynomial of the restriction of $T$ to each of these subspaces is also $t^2+1$ This implies that each cyclic subspace is $2$ dimensional (I am actually unclear on this).

Can someone help me clean up the above thought process? It seems I didn't need to consider the characteristic polynomial at all. In general, how does knowledge of the minimal polynomial help you determine the dimension of a vector space?

• Jordan normal form gives you invariant subspaces. Since all eigenvalues are complex, you get blocks of even size by pairing up conjugate vectors. – ET93 Nov 25 '16 at 4:30
• math.stackexchange.com/questions/1016252/… – ET93 Nov 25 '16 at 4:33

Once you have deduced that $T^2+1=0$ (and indeed this is true since $X^2+1=\gcd(A,B)$ for $A=X^4-1$ and $B=X^3+3X^3+X+3$ can be written as $SA+TB$ for some polynomials $S,T$) you are practically done. Give your vector space the structure of a vector space over $\Bbb C$ by having the imaginary unit $\mathbf i$ act as $T$ (which is possible thanks to the established relation). Find any basis for the complex vector space, which provides a decomposition into a direct sum of complex $1$-dimensional subspaces; by restriction of scalars back to the real numbers, this provides a decomposition of your $\Bbb R$-vector space as a direct sum of $2$-dimensional $T$-stable subspaces.
Note that you don't need to know the minimal or characteristic polynomial for this. (Indeed the space might have dimension$~0$, in which case both these polynomials would be$~1$.)
The characteristic polynomial of $T|_{C_{x_k}}$ is again a power of $x^2+1$ by the same argument you used for all of $V$. In particular, it has even degree so $T|_{C_{x_k}}$ is even-dimensional.
To answer your last question, the rational canonical form gives you that the characteristic polynomial of each cyclic subspace divides the minimal polynomial of $T$ and their product is the characteristic polynomial. So the characteristic polynomial divides some power of the minimal polynomial. This divisibility can sometimes give dimensions like in this case.
In our case, since the minimal polynomial $x^2+1$ is irreducible, the characteristic polynomials of each cyclic subspace are equal to $x^2+1$ so you are right, not only are the subspaces even dimensional but they are of dimension $2$. In general, if there are no real eigenvalues for a real matrix, you get a decomposition into even dimensional cyclic subspaces by these divisibility results.