My question is the following:
Given a not further specified $n\times n$ diagonal matrix $M$ with $n$ eigenvalues on its diagonal, which is the transformation matrix to the basis of eigenvectors, and an arbitrary endomorphism $f$, how do I construct the upper triangular matrix $T$ which is similar to $M$, or is it not possible?
What I have tried so far:
Since $M$ is a diagonal matrix, the endomorphism $f$ is diagonalizable, which implies, that a similar upper triangular matrix $T$ exists.
The usual algorithm to construct $T$ involves finding an eigenvector $v$, exchanging it with a vector of the standard basis, and compute the next similar matrix, until you have exchanged all vectors of the initial basis.
However, all eigenvectors of the diagonal matrix consist of standard basis vectors, so this algorithm gets me back to where I started, since exchanging standard basis vectors with standard basis vectors brings me back to the diagonal matrix $M$!
Is it even possible to construct $T$?
Best Answer
No, you tweak the eigenvectors of $T$ but still preserve $0<\langle e_1\rangle<\langle e_1,e_2\rangle<\dots<\mathbb{F}^n$.
For example: $M=\begin{bmatrix}-1\\&1\end{bmatrix}$ and you want to introduce a nonzero coefficient in the upper-right in $T$. You try something like $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}-1&0\\0&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix}^{-1} $$