Upper triangulation of a matrix versus diagonalization

Question

I am trying to google this question, but could not find any hints. This is important to me because of I am dealing with 3-4D matrices.

It's true that an upper triangulation (Gauss elimination) of a matrix is diagonalizable iff the original matrix is diagonalizable?

Or some result related to this?

Thank you!

Answer 1

It's true that an upper triangulation of a matrix is diagonalizable iff the original matrix is diagonalizable?

It's not true. Pick any non-diagonal upper-triangular matrix with distinct diagonal entries. The diagonal entries are the eigenvalues for the matrix. Since they are distinct, the matrix is diagonalisable. Any matrix similar to this matrix (including the matrix itself) has a non-diagonal upper-triangulation.

Though slightly less trivial, we can make similar examples where we have repeated eigenvalues, for example,

$$\begin{pmatrix} 1 & \color{red}{0} & 1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 2 & \color{red}{0} \\ 0 & 0 & 0 & 2 \end{pmatrix}$$

is also diagonalisable. All I needed to ensure here was that the two red $\color{red}{0}$s were indeed $0$.

If the diagonal were constant, then your result would hold true. Indeed, the following are equivalent for a matrix with one (possibly repeated) eigenvalue:

• The matrix is diagonalisable,
• Every upper-triangulation of the matrix is diagonal,
• The matrix is a multiple of the identity matrix.

(By that, I mean this is not a particularly interesting case, but it's the only reasonable situation where I can see your guess holding true.)