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.

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 )

Google photo

You are commenting using your Google 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 )

Connecting to %s

%d bloggers like this: