The Unapologetic Mathematician

Mathematics for the interested outsider

Almost Upper-Triangular Matrices

Over an algebraically closed field we can always find an upper-triangular matrix for any linear endomorphism. Over the real numbers we’re not quite so lucky, but we can come close.

Let T:V\rightarrow V be a linear transformation from a real vector space V of dimension d to itself. We might not be able to find an eigenvector — a one-dimensional invariant subspace — but we know that we can find either a one-dimensional or a two-dimensional invariant subspace U_1\subseteq V. Just like before we get an action of T on the quotient space V/U_1. Why? Because if we have two representatives v and w of the same vector in the quotient space, then we can write w=v+u. Acting by T, we find Tw=Tv+Tu. And since Tu\in U_1, the vectors Tv and Tw are again equivalent in the quotient space.

Now we can find a subspace \hat{U}_2\subseteq V/U_1 which is invariant under this action of T. Is this an invariant subspace of V? No, it’s not even a subspace of V. But we could pick some U_2\subseteq V containing a unique representative for each vector in \hat{U}_2. For instance, we could pick a basis of \hat{U}_2, a representative for each basis vector, and let U_2 be the span of these representatives. Is this an invariant subspace? Still, the answer is no. Let’s say u\in U_2 is the identified representative of \hat{u}\in\hat{U}_2. Then all we know is that Tu is a representative of T\hat{u}, not that it’s the identified representative. It could have some components spilling out into U_1.

As we proceed, picking up either a one- or two-dimensional subspace at each step, we can pick a basis of each subspace. The action of T sends each basis vector into the current subspace and possibly earlier subspaces. Writing it all out, we get a matrix that looks like

\displaystyle\begin{pmatrix}A_1&&*\\&\ddots&\\{0}&&A_m\end{pmatrix}

where each A_j is either a 1\times1 matrix or a 2\times2 matrix with no eigenvalues. The 1\times1 blocks come from the one-dimensional invariant subspaces in the construction, while the 2\times2 blocks come from the two-dimensional invariant subspaces in the construction, though they may not be invariant once we put them back into V. Above the diagonal we have no control (yet) over the entries, but below the diagonal almost all the entries are zero. The only exceptions are in the 2\times2 blocks, where we poke just barely down by one row.

We can note here that if there are n\leq m two-dimensional blocks and m-n one-dimensional blocks, then the total number of columns will be 2n+(m-n)=n+m=d. Thus we must have at least \lceil\frac{d}{2}\rceil blocks, and at most d blocks. The latter extreme corresponds to an actual upper-triangular matrix.

April 1, 2009 Posted by | Algebra, Linear Algebra | 8 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 393 other followers