The Unapologetic Mathematician

Mathematics for the interested outsider

Upper-Triangular Matrices

Until further notice, I’ll be assuming that the base field \mathbb{F} is algebraically closed, like the complex numbers \mathbb{C}.

What does this assumption buy us? It says that the characteristic polynomial of a linear transformation T is — like any polynomial over an algebraically closed field — guaranteed to have a root. Thus any linear transformation T has an eigenvalue \lambda_1, as well as a corresponding eigenvector e_1 satisfying

T(e_1)=\lambda_1e_1

So let’s pick an eigenvector e_1 and take the subspace \mathbb{F}e_1\subseteq V it spans. We can take the quotient space V/\mathbb{F}e_1 and restrict T to act on it. Why? Because if we take two representatives v,w\in V of the same vector in the quotient space, then w=v+ce_1. Then we find

T(w)=T(v+ce_1)=T(v)+cT(e_1)=T(v)+c\lambda_1e_1

which represents the same vector as T(v).

Now the restriction of T to V/\mathbb{F}e_1 is another linear endomorphism over an algebraically closed field, so its characteristic polynomial must have a root, and it must have an eigenvalue \lambda_2 with associated eigenvector e_2. But let’s be careful. Does this mean that e_2 is an eigenvector of T? Not quite. All we know is that

T(e_2)=\lambda_2e_2+c_{1,2}e_1

since vectors in the quotient space are only defined up to multiples of e_1.

We can proceed like this, pulling off one vector e_i after another. Each time we find

T(e_i)=\lambda_ie_i+c_{i-1,i}e_{i-1}+c_{i-2,i}e_{i-2}+...+c_{1,i}e_1

The image of e_i in the ith quotient space is a constant times e_i itself, plus a linear combination of the earlier vectors. Further, each vector is linearly independent of the ones that came before, since if it weren’t, then it would be the zero vector in its quotient space. This procedure only grinds to a halt when the number of vectors equals the dimension of V, for then the quotient space is trivial, and the linearly independent collection \{e_i\} spans V. That is, we’ve come up with a basis.

So, what does T look like in this basis? Look at the expansion above. We can set t_i^j=c_{i,j} for all i<j. When i=j we set t_i^i=\lambda_i. And in the remaining cases, where i^gt;j, we set t_i^j=0. That is, the matrix looks like

\displaystyle\begin{pmatrix}\lambda_1&&*\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}

Where the star above the diagonal indicates unknown matrix entries, and the zero below the diagonal indicates that all the entries in that region are zero. We call such a matrix “upper-triangular”, since the only nonzero entries in the matrix are on or above the diagonal. What we’ve shown here is that over an algebraically-closed field, any linear transformation has a basis with respect to which the matrix of the transformation is upper-triangular. This is an important first step towards classifying these transformations.

February 2, 2009 Posted by | Algebra, Linear Algebra | 20 Comments

   

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