# The Unapologetic Mathematician

## The Determinant of an Upper-Triangular Matrix

We know that every linear endomorphism $T$ on a vector space $V$ over an algebraically-closed field $\mathbb{F}$ has a basis with respect to which its matrix is upper-triangular. That is, it looks like

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

So we’re interested in matrices of this form, whether the base field is algebraically-closed or not. One thing that’s nice about upper-triangular matrices is that their determinants are pretty simple.

Remember that we can calculate the determinant by summing one term for each permutation in the symmetric group $S_d$. Each term is the product of one entry from each row of the matrix. But we can see that in the last row, we have to pick the last entry, or the whole term will be zero. Then in the next to last row, we must pick either of the last two entries. But we can’t pick the last one, since that’s locked up for the last row, so we must again pick the entry on the diagonal.

As we work backwards up the matrix, we find that the only possible way of picking a nonzero term is to always pick the diagonal element. That is, we only need to consider the identity permutation $\pi(k)=k$. And then the determinant is simply the product of all these diagonal elements. That is,

$\displaystyle\det\begin{pmatrix}\lambda_1&&*\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}=\prod\limits_{k=1}^d\lambda_k$

Notice that the entries above the diagonal don’t matter at all!

One way this comes in handy is in finding the eigenvalues of $T$. “But wait!” you cry, “Didn’t we find the $\lambda_i$ by looking for eigenvalues of $T$? Not quite. We found them by looking for eigenvalues of a sequence of restrictions of $T$ to smaller and smaller quotient spaces. We have no reason to believe (yet) that these actually correspond to eigenvalues of $T$.

But now we can easily find the matrix corresponding to $\lambda1_V-T$, since matrix of the identity transformation is the same in every basis. We find

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

This is again upper-triangular, so we can easily calculate the characteristic polynomial by taking its determinant. We find

$\displaystyle\prod\limits_{k=1}^d(\lambda-\lambda_k)$

Then it’s clear to see that this will be zero exactly when $\lambda=\lambda_k$. That is, the eigenvalues of $T$ are exactly the entries along the diagonal of an upper-triangular matrix for the transformation. Incidentally, this shows in passing that even though there may be many different upper-triangular matrices representing the same transformation (in different bases), they all have the same entries along the diagonal (possibly in different orders).

Posted by John Armstrong | Algebra, Linear Algebra | 4 Comments

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