The Unapologetic Mathematician

Mathematics for the interested outsider

Upper-Triangular Matrices and Orthonormal Bases

I just noticed in my drafts this post which I’d written last Friday never went up.

Let’s say we have a real or complex vector space V of finite dimension d with an inner product, and let T:V\rightarrow V be a linear map from V to itself. Further, let \left\{v_i\right\}_{i=1}^d be a basis with respect to which the matrix of T is upper-triangular. It turns out that we can also find an orthonormal basis which also gives us an upper-triangular matrix. And of course, we’ll use Gram-Schmidt to do it.

What it rests on is that an upper-triangular matrix means we have a nested sequence of invariant subspaces. If we define U_k to be the span of \left\{v_i\right\}_{i=1}^k then clearly we have a chain

\displaystyle U_1\subseteq\dots\subseteq U_{d-1}\subseteq U_d=V

Further, the fact that the matrix of T is upper-triangular means that T(v_i)\in U_i. And so the whole subspace is invariant: T(U_i)\subseteq U_i.

Now let’s apply Gram-Schmidt to the basis \left\{v_i\right\}_{i=1}^d and get an orthonormal basis \left\{e_i\right\}_{i=1}^d. As a bonus, the span of \left\{e_i\right\}_{i=1}^k is the same as the span of \left\{e_i\right\}_{i=1}^k, which is U_k. So we have exactly the same chain of invariant subspaces, and the matrix of T with respect to the new orthonormal basis is still upper-triangular.

In particular, since every complex linear transformation has an upper-triangular matrix with respect to some basis, there must exist an orthonormal basis giving an upper-triangular matrix. For real transformations, of course, it’s possible that there isn’t any upper-triangular matrix at all. It’s also worth pointing out here that there’s no guarantee that we can push forward and get an orthonormal Jordan basis.

About these ads

May 8, 2009 - Posted by | Algebra, Linear Algebra

1 Comment »

  1. […] and see what happens as we try to diagonalize it. First, since we’re working over here, we can pick an orthonormal basis that gives us an upper-triangular matrix and call the basis . Now, I assert that this matrix already is diagonal when is […]

    Pingback by The Complex Spectral Theorem « The Unapologetic Mathematician | August 10, 2009 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Get every new post delivered to your Inbox.

Join 411 other followers

%d bloggers like this: