Just in case you don't know what Schur's triang theorem is, here is the statement.
Let $A$ be an $n\times n$ real matrix, all of whose eigenvalues are real. Then, there exist an orthogonal matrix $Q$ and an upper triangular matrix $T$ such that $Q^{T}AQ=T$
Here is my proof for the theorem. I'll assume the you know about QR decomposition and basic properties of matrix multiplication.
First, hence $A$ has $n$ real eigenvalues, it is similar to a diagonal matrix $D$. So that $A=P^{-1}DP$.
Now we let $P^{-1}=QR$. What we are left with is
$A=QRDR^{-1}Q^{-1}$.
Since multiplication of two upper triangular leads to another upper triangular matrix, we are left with
$A=QTQ^{-1}$.
I've always thought spectral theorem and Schur's theorem was too difficult to handle in some textbooks, but this turns out to be too easy? Am I wrong somewhere?
Best Answer
This is not correct. The statement of the theorem says that all eigenvalues are real, not that it has $n$ distinct eigenvalues. For instance, if $A=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$, the only eigenvalues of $A$ is $0$, which is real, but you cannot diagonalize $A$.