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?
Best Answer
A matrix like $\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$ cannot be made diagonal for any base in $\mathbb{R}^2$
For more info on why this is the case, check:
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 $1$)