# Upper Triangular Implies Diagonal?

If all matrices can be made upper - triangular with respect to some basis by:

Suppose V is a finite-dimensional complex vector space and T is a linear transformation. Then T has an upper-triangular matrix with respect to some basis of V.

And any upper - triangular matrix can be made orthogonal:

Suppose T is a linear transformation. If T has an upper-triangular matrix with respect to some basis of V, then T has an upper-triangular matrix with respect to some orthonormal basis of V.

But it's clear that an upper-triangular and orthonormal matrix must be a diagonal matrix. This implies every matrix has a diagonal matrix which we know to false as it was stated earlier to not be true. What am I missing?

• The second statement says that the basis is orthonormal, but the matrix remains upper-triangular only. Commented Jul 9, 2019 at 7:43
• You've hit on a point I had some confusion over. Won't T be diagonal for this basis by slader.com/discussion/question/… ? Commented Jul 9, 2019 at 7:48
• The assertion is that there exist an orthogonal matrix $M$ and an upper-triangular matrix $U$ such that $T = MUM^T$. Note that $U$ is not (in general) orthogonal or diagonal.
– user169852
Commented Jul 9, 2019 at 7:57

A matrix like $$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$ cannot be made diagonal for any base in $$\mathbb{R}^2$$
Long story short, a matrix is not diagonalizable if there is an eigenvalue whose algebraic and geometric multiplicity do not equal each other (in the example, the eigenvector $$1$$ has algebraic multiplicity $$2$$ but geometric mulitplicity $$1$$)