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 $i$ th 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 John Armstrong | Algebra, Linear Algebra

20 Comments »

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[…] be a linear map from to itself. Further, let be a basis with respect to which the matrix of is upper-triangular. It turns out that we can also find an orthonormal basis which also gives us an upper-triangular […]

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no need of such material

Comment by shasha | May 21, 2009 | Reply
Oh I’m so sorry that I chose to cover a topic you see as unnecessary. I’ll be sure to run all my future topics by you first.

Comment by John Armstrong | May 21, 2009 | Reply
- Useful article. Many books just prove this by induction without any explanation
  
  Comment by vish | October 26, 2013 | Reply
Surely in the first equation you meant to write
T(e_1) = lambda_1 * e_1

(instead of T(v) on LHS)

Comment by A Khan | June 17, 2009 | Reply
Yes, sorry. Thanks for catching that.

Comment by John Armstrong | June 17, 2009 | Reply
[…] with a complex transformation we’re done. We can pick a basis so that the matrix for is upper-triangular, and then its determinant is the product of its eigenvalues. Since the eigenvalues are all […]

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