While reading a chapter on diagonalizable matrices, I found myself wondering:
Can a matrix $A \in \mathbb R^{n \times n}$ be invertible but not diagonalizable?
My quick Google search did not return a clear answer.
While reading a chapter on diagonalizable matrices, I found myself wondering:
Can a matrix $A \in \mathbb R^{n \times n}$ be invertible but not diagonalizable?
My quick Google search did not return a clear answer.
After thinking about it some more, I realized that the answer is "Yes".
For example, consider the matrix
\begin{equation} A = \left [ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right ]. \end{equation}
It has two linearly independent columns, and is thus invertible.
At the same time, it has only one eigenvector:
\begin{equation} v = \left [ \begin{array}{c} 1 \\ 0 \end{array} \right ]. \end{equation}
Since it doesn't have two linearly independent eigenvectors, it is not diagonalizable.
Geometric(-ish) answer. Take, in $\mathbb R^2$, a rotation of angle $0<\theta<2\pi$ with $\theta \neq \pi$. Then the associated matrix is invertible (the inverse being the rotation of $-\theta$) but is not diagonalisable, since no non-zero vector is mapped into a multiple of itself by a rotation of such angles.