Not every matrix is diagonalizable, but every linear transformation has a matrix representation that is an upper triangular matrix, and the basis that achieves this representation is especially pleasing. Here is the theorem.
Suppose that $\ltdefn{T}{V}{V}$ is a linear transformation. Then there is a basis $B$ for $V$ such that the matrix representation of $T$ relative to $B$, $\matrixrep{T}{B}{B}$, is an upper triangular matrix. Each diagonal entry is an eigenvalue of $T$, and if $\lambda$ is an eigenvalue of $T$, then $\lambda$ occurs $\algmult{T}{\lambda}$ times on the diagonal.
A key step in this proof was the construction of the subspace $W$ with dimension strictly less than that of $V$. This required an eigenvalue/eigenvector pair, which was guaranteed to us by Theorem EMHE. Digging deeper, the proof of Theorem EMHE requires that we can factor polynomials completely, into linear factors. This will not always happen if our set of scalars is the reals, $\real{\null}$. So this is our final explanation of our choice of the complex numbers, $\complexes$, as our set of scalars. In $\complexes$ polynomials factor completely, so every matrix has at least one eigenvalue, and an inductive argument will get us to upper triangular matrix representations.
In the case of linear transformations defined on $\complex{m}$, we can use the inner product (Definition IP) profitably to fine-tune the basis that yields an upper triangular matrix representation. Recall that the adjoint of matrix $A$ (Definition A) is written as $\adjoint{A}$.
Suppose that $A$ is a square matrix. Then there is a unitary matrix $U$, and an upper triangular matrix $T$, such that \begin{align*} \adjoint{U}AU=T \end{align*} and $T$ has the eigenvalues of $A$ as the entries of the diagonal.