My book: Linear Algebra Done Right 3rd Edition by Sheldon Axler states the Schur's Theorem as follows.
Schur's Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Then $T$ has an upper-triangular matrix with respect to some orthonormal basis of $V$.
In other words, for any $n \times n$ complex-valued matrix, there is an orthonormal basis in which that matrix is upper-triangular.
But many other textbooks state the Schur's Decomposition as follows.
Schur's Deocmposition Theorem. Suppose $V$ is a finite-dimensional complex vector space and $T$ is a linear operator on $V$. Let $B$ be a basis of $V$. Then there exist an unitary matrix $U$ such that $$ \mathcal{M}(T,B) = U^\ast A U$$
where $A$ is an upper triangular matrix.
Here, $\mathcal{M}(T)$ denote matrix representation of $T$ respect to basis $B$.
What is the relationship between those two? Are they equivalent? If yes, how can I prove it?
Best Answer
Proof of the equivalency
$(2) \implies (1)$
Let $\mathcal E' = \mathcal B U^*$, then $T (\mathcal B) = \mathcal B U^* A U $, or $T (\mathcal E' U) = \mathcal E' U U^* A U = \mathcal E' A U $, hence $T (\mathcal E') = \mathcal E' A$, i.e. $\mathcal M(T, \mathcal E') = A$ is upper triangular. Now apply Gram-Schmidt orthogonalization process to $\mathcal E'$ to obtain an orthonormal basis $\mathcal E$, then $\mathcal M(T, \mathcal E)$ is still upper triangular, hence the statement.
$(1) \implies (2)$
From now on, the bold-italic letters represent matrices, the regular letters are mappings, and the calligraphic letters are basis. Also $\mathbb C^n \cong \mathrm M_{n,1}(\mathbb C)$ is the collection of $n \times 1$ complex matrices.
Let $\boldsymbol S = \mathcal M(T, \mathcal B)$, on $W = \mathbb C^n$, define linear operator $S \colon \boldsymbol x \mapsto \boldsymbol {Sx}$ where $\boldsymbol x \in \mathbb C^n$, then apply $(1)$ to the operator $S$, there exists some ONB $\mathcal F$ of $\mathbb C^n$ that $\mathcal M(S, \mathcal F) = \boldsymbol A$ is upper triangular. Let $\mathcal E$ be the standard basis of $\mathbb C^n$, and let $\mathcal F = \mathcal E \boldsymbol U$, then $\boldsymbol U$ is unitary. Write $\boldsymbol U^*$ instead of $\boldsymbol U$, i.e. $\mathcal F = \mathcal E \boldsymbol U^*$ and $S(\mathcal F) = \mathcal F \boldsymbol A$ becomes $S (\mathcal E \boldsymbol U^*) = \mathcal E \boldsymbol U^* \boldsymbol A$, hence $S (\mathcal E) = \mathcal E (\boldsymbol U^*\boldsymbol {AU})$. But $S (\mathcal E) = \boldsymbol S$, hence $\boldsymbol S = \boldsymbol {U}^* \boldsymbol {AU}$. Hence the statement.