# The Unapologetic Mathematician

## Real Invariant Subspaces

Okay, there’s been a bit of a gap while things have gotten hectic around here, but let’s try and get back to work.

When we worked over an algebraically closed field, every polynomial had a root. Applying this to the characteristic polynomial of a linear transformation, we found that it must have a root, which would be definition be an eigenvalue of the transformation. There would then be an eigenvector, which gave a one-dimensional invariant subspace. But when we look at real vector spaces we might not have any one-dimensional invariant subspaces. However, if we don’t we will be sure to have a two-dimensional invariant subspace.

So let’s start with a linear transformation $T:V\rightarrow V$ from a real vector space $V$ of dimension $d$ to itself. Pick any vector $v$ and construct the sequence of images ${Tv}$, $T^2v$, and so on up to $T^dv$. Together with the original vector, these ${d+1}$ vectors cannot be linearly independent, since there are more than $d$ of them. Thus we have a linear relation

$\displaystyle a_0v+a_1Tv+a_2T^2v+...+a_dT^dv=0$

We can regard this as a polynomial in $T$ applied to $v$, which has real coefficients. We can factor it to write

$\displaystyle c(T-\lambda_1I_V)...(T-\lambda_mI_V)(T^2-\tau_1T+\delta_1I_V)...(T^2-\tau_MT+\delta_MI_V)v$

Note that either $m$ or $M$ could be zero, in which case there are no factors of that form. Also, note that we have no reason to believe that this linear combination has anything to do with the characteristic polynomial, so this factorization is not necessarily giving us eigenvalues or eigenpairs.

All that we can conclude is that at least one of these factors is not injective. If it’s a factor $(T-\lambda_jI_V)$, then all the factors after that point act on $v$ to give a vector $u$ satisfying

$\displaystyle Tu={\lambda_j}u$

This gives a one-dimensional invariant subspace. On the other hand, if one of the quadratic factors $T^2-\tau_jT+\delta_jI_V$ is not injective, then all the factors after that point act on $v$ to give a vector $u$ satisfying

$\displaystyle T^2u=\tau_jTu-\delta_ju$

which shows that the vectors $u$ and $Tu$ span a two-dimensional invariant subspace, since both basis vectors are sent to a linear combination of each other under the action of $T$.

Thus we can always find an invariant subspace of dimension one or two. It’s not quite as neat as over the complex numbers, but it’s something we can work with.

March 31, 2009 Posted by | Algebra, Linear Algebra | 6 Comments