# Proofs Using Upper Triangular Form Over $\mathbb{C}$

I recently came across a proof for Cayley-Hamilton for any upper triangular matrix and then generalizing to all square matrices over $$\mathbb{C}$$. There was no explanation as to why we can generalize to all square matrices, why can we? Is this useful for other proofs in linear algebra that come to mind?

The Cayley-Hamilton theorem can be stated as $$\chi_A(A) = 0$$ for any matrix $$A \in \mathsf{M}_n\mathbb{C}$$, where $$\chi_A$$ is its characteristic polynomial.

Note that in general if $$p(A) = 0$$ for some polynomial $$p$$, then $$p(BAB^{-1}) = 0$$ for any other matrix $$B$$, as $$p(A) = Bp(A)B^{-1}$$.

Now, take a complex matrix $$A$$. Since $$\chi_A$$ is a complex polynomial, it has a root $$\alpha \in \mathbb{C}$$. Take an eigenvector $$v$$ of eigenvalue $$\alpha$$ and complete to a basis of $$\mathbb{C}^n$$. Thus, via a change of basis we have

$$BAB^{-1} = \begin{pmatrix}\tilde{A} & 0 \\ * &\alpha\end{pmatrix}$$

for some invertible matrix $$B$$. That is, the final column have all zeroes except in the last entry, which is $$\alpha$$. Proceeding inductively on $$\tilde{A}$$, one sees that there exists a basis $$\mathcal{B}$$ whose "change of basis matrix" $$B$$ satisfies that $$BAB^{-1}$$ is lower triangular. But then again, we could have done this with $$A^t$$ and via transposing, we have seen that

Proposition. Let $$A \in \mathsf{M}_n\mathbb{C}$$. There exists $$B \in \mathsf{M}_n\mathbb{C}$$ such that $$BAB^{-1}$$ is upper triangular.

So, define $$U = BAB^{-1}$$. Equivalently, we ahve $$B^{-1}UB = A$$ and so by the version of Cayley-Hamilton that you know,

$$\chi_U(A) = \chi_U(B^{-1}UB) = B^{-1}\chi_U(U)B = 0.$$

To finish, it suffices to see that $$\chi_U = \chi_A$$. But this is a general fact of similar matrices,

\begin{align} \chi_U &= \det(U-xI) = \det(BAB^{-1}-xBB^{-1})\\ &= \det(B(A-xI)B^{-1}) = \det B \det(A-xI)\det B^{-1}\\ &= \det(A-xI) = \chi_A. \end{align}

I can't recall other proofs that do these right now, but I remember using the aforementioned proposition more than once.

If you build the theory talking about linear transformations, then everything is invariant with respect to a change of basis. Hence when you need to actually compute something, you can choose a basis which makes your matrix upper triangular, as these are (potentially) more manageable.

Note that we have strongly used that the base field here is algebraically closed (but nothing else).

Edit: to expand on my comment, let $$p = a_nX^n + \cdots + a_0$$. Then,

\begin{align} p(CAC^{-1}) &= \sum_{i=0}^na_i(CAC^{-1})^i = \sum_{i=0}^na_iCA^iC^{-1}\\ &= \sum_{i=0}^nCa_iA^iC^{-1} = C \left(\sum_{i=0}^na_iA^i\right)C^{-1} = Cp(A)C^{-1}. \end{align}

Here we use that $$\lambda CA = C\lambda A$$ for any scalar $$\lambda$$ and matrices $$A,C$$ and that

\begin{align} (CAC^{-1})^i &= CAC^{-1}CAC^{-1} \cdots CAC^{-1}CAC^{-1}\\ &= CA \cdot I \cdot A \cdot I \cdots \cdot I \cdot A \cdot I \cdot AC^{-1}\\ &= CA^iC^{-1} \end{align}

for each $$i$$ (a more formal argument can be done via induction).

• Thank you! For clarification, why can one pull the transformation matrix out of the polynomial without applying the polynomial? It seems unintuitive... Commented May 29, 2019 at 20:12
• This is because $\lambda CAC^{-1} = C(\lambda A)C^{-1}$ and $(CAC^{-1})^k = CA^kC^{-1}$. Then, we can apply this to each monomial and regroup the terms. Commented May 29, 2019 at 20:16

This is because if $$A \in M_{n \times n}(\mathbb{C})$$, then the characteristic polynomial of $$A$$, $$P_A(t)$$ splits (because $$\mathbb{C}$$ is algebraically closed). Since the characteristic polynomial splits, $$A$$ is similar to an upper triangular matrix $$U$$. Since you already know Cayley-Hamilton to be true for $$U$$, you can deduce that it is true for $$A$$ as well (since similar matrices have the same characteristic polynomial). To see why $$A$$ is similar to an upper triangular matrix, I see $$2$$ possibilities:

1. If you know about the Jordan Canonical Form, this is trivial, because every matrix is similar to its JCF, and the JCF is triangular (upper-triangular if you arrange the change of basis matrix appropriately).
2. Another method is Schur's decomposition (I believe this is the correct name). For a proof of this, see for example Friedberg, Insel, Spence theorem 6.14.

I'm a bit rusty so I can't think of other ways to prove that $$A$$ is similar to an upper triangular matrix $$U$$. Perhaps others can provide more methods.

EDITED: An alternative approach proves the Cayley-Hamilton theorem without ever using determinants, see Theorem 8.37 of the third edition of Linear Algebra Done Right, (or more briefly Theorem 5.2 by the same author here).

As to the brunt of the question, Theorem 5.13 of Axler's textbook shows how over $$\mathbb{C}$$, square matrices are triangularisable, and so we may interchange the two.

• The first link in the answer above is to an unauthorized version of a copyrighted book. More importantly, that link goes to the second edition of Linear Algebra Done Right. Multiple improvements were made for the third edition (linear.axler.net). Commented May 30, 2019 at 1:02
• I am very sorry - I can edit it out should you wish. It was simply the first link that appeared when I googled the book title. Thank you very much for the up to date information. Commented May 30, 2019 at 17:44
• Thank you for your comment above. It would indeed be better to remove the link to an obsolete and unauthorized version. The proof of the Cayley-Hamilton Theorem without using determinants is in Theorem 8.37 of the third edition of Linear Algebra Done Right. Thanks again. Commented May 30, 2019 at 23:00