# Every square matrix is conjugate

Put content here**Theorem (Schur).,** Every square matrix over $\mathbb{C}$ is *similar* to an upper triangular matrix. That is, for any $A \in M_n(\mathbb{C})$, there exists an invertible matrix $P$ such that $P^{-1}AP = T$, where $T$ is upper triangular.
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The diagonal entries of $T$ are the eigenvalues of $A$ (with algebraic multiplicity).
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**Proof idea.** Proceed by induction on $n$. Choose an eigenvector $v_1$ of $A$, extend it to a basis, and work in that basis to produce a block form, then apply the inductive hypothesis.
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**Corollary.** Every complex matrix has at least one eigenvalue (since the characteristic polynomial of $T$ factors as $\prod (\lambda - t_{ii})$).
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**Example.** For $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, the eigenvalues are $\pm i$, and $A$ is similar to an upper triangular matrix with $i$ and $-i$ on the diagonal.

# Parents

* Triangular matrices