Schur theorem: Let $T$ be a linear operator on a finite-dimensional inner product space $V$, Suppose that the characteristic polynomial of $T$ splits, then there exists an orthonormal basis $\beta$ for $V$ such that the matrix $[T]_\beta$ is upper triangular.
Proof: by math induction on the dimension $n$ of $V$. When $n=1$, the result is immediate. Suppose the result is true for linear operators on $(n-1)$ dimensional inner product spaces whose characteristic polynomials split. We can assume that $T^{*}$ has a unit eigenvector $z$. suppose that $T^{*}(z)+\lambda z$ and that $W=\text{span }({z})$. We show that $W^{\bot} $ is T-invariant (Why?) If $y \in W^{\bot} $and $x=cz \in W$, then $\langle T(y,x \rangle= \langle T(y),cz \rangle= \langle y,T^{*}(cz) \rangle=\langle y, cT^{*}(z) \rangle= \langle y, c\lambda z \rangle= \langle \overline{c\lambda} y,z \rangle= \overline{c \lambda}(0)=0$.
So $T(y) \in W^{\bot}$. It is easy to show that the characteristic polynomial of $T_{W^{\bot}}$ divides the characteristic polynomial of T and hence splits. By theorem (suppose $S=\{v_1,\ldots,v_k\}$ is an orthonormal set in an $n-$dimensional inner product space $V$, then if $W$ is any subspace of $V$, then $\dim(V)=\dim(W)+\dim(W^{\bot}))$, $\dim(W^{\bot})=n-1$ (Why n-1)?, so we apply the induction hypothesis to $T_{W^{\bot}}$ and obtain an orthonormal basis$ \gamma $ of $W^{\bot}$ such that $[T_{W^{\bot}}]_{\gamma}$ is upper triangular. Clearly, $\beta = \gamma \cup {z}$ is an orthonormal basis for V such that $[T]_\beta$ is upper triangular.
Over all, is there a more concise proof for this theorem?
Best Answer
Here is a preliminary lemma:
Note that $\Leftarrow$ is almost trivial. For $\implies$ (which is what you need), you can try an argument by induction (and perhaps use quotient spaces). This is definitely a worthwhile exercise to prove on its own.
Once you establish the lemma, all you need is to apply Grahm-Schmidt process to $\alpha$ to obtain an orthonormal basis $\beta$ of $V$. Then, $[T]_{\beta}$ will still be upper-triangular; because the construction of Grahm-Schmidt shows that if $\alpha = \{x_1, \dots, x_n\}$, and $\beta = \{y_1, \dots, y_n\}$, then for each $1 \leq k \leq n$, we have \begin{align} \text{span}\{x_1, \dots, x_k\} = \text{span}\{y_1, \dots, y_k\} \end{align} (if this isn't immediately clear, then review the proof of the Grahm-Schmidt process). I leave the details for you to verify.
Note that it is not necessary at all to invoke the existence of a Jordan canonical form for linear operators whose characteristic polynomial splits. I only said that because it gives an extremely quick proof of what you are after (but of course, this lemma I wrote above is much easier to prove than the existence of a JCF).