What does it mean for a linear operator to have eigenvectors but not be diagonalizable?

Question

Could someone explain this conceptually? For example, giving a simple example with $\mathbb{R}^3$

Answer 1

Consider the transformation $T:\mathbb{R}^2 \to \mathbb{R}^2$ given by $T(\mathbf{x})=A\mathbf{x}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\mathbf{x}$. Then this has eigenvector $\begin{bmatrix}1\\0\end{bmatrix}$ for the "repeated" eigenvalue $\lambda=1$. Observe that, even though the eigenvalue is repeated twice, but it is only giving us "one" linearly independent eigenvector (said precisely: the dimension of the eigenspace for $\lambda=1$ is $1$ (think about algebraic and geometric multiplicity!!)

However it is not diagonalizable; that means we cannot have a change of basis where the matrix representation of $T$ is a diagonal matrix. The reason is, if such a diagonal matrix $D$ exists, then $D=I$ (because eigenvalues should be same for $A$ and $D$), in which case $A=PDP^{-1}=P(I)P^{-1}=I$. But as we can see $A \neq I$.

Note: In terms of eigenvectors, we have diagonalizability iff we have "enough" number of eigenvectors that can form a basis of the space.