# The Unapologetic Mathematician

## 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

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