Construct a 2x2 matrix with real eigenvalues that is not diagonalizable I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
 A: Hint A matrix $A$ with geometric multiplicity equal to its algebraic multiplicity is diagonalizable, so any nondiagonalizable $2 \times 2$ matrix must have a single eigenvalue, say, $\lambda$ of algebraic multiplicity $2$ but geometric multiplicity $1$.
Additional hint If some invertible matrix $P$ diagonalized $A - \lambda I$, it would diagonalize $A$, a contradiction. Thus, if $A$ is an example, so is $A - \lambda I$, which has eigenvalue $0$ of algebraic multiplicity $2$, and thus we may as well restrict our search to the case $\lambda = 0$.

Since $A$ satisfies its own characteristic polynomial, $A^2 = 0$, but we cannot have $A = 0$, because $0$ is diagonalizable. Thus, there is a vector ${\bf v} \in \Bbb R^2$ such that $A {\bf v} \neq 0$, so $B = (A {\bf v}, {\bf v})$ is a basis of $\Bbb R^2$. With respect to $B$, the transformation matrix is $$J = \pmatrix{0&1\\0&0} .$$ (This is the $2 \times 2$ Jordan block of eigenvalue $0$.) On the other hand, the sole eigenspace of $J$ is spanned by $\pmatrix{1\\0}$, so $J$ and thus $A$ has geometric multiplicity $1$. It follows from our construction that up to similarity and addition of multiples of $I$ this example is unique.

