Construct a $2 \times 2$ matrix that is invertible but not diagonalizable.
From this I know that the eigenvalues cannot be equal or greater than, but would having no eigenvalues be a viable option? I don't see anything wrong based on the Diagonalization Theorem.
For the Invertible Matrix Theorem, if there's no eigenvalues then $0$ wouldn't be an eigenvalues because there are no eigenvalues. I'm not sure how having no eigenvalues affects the determinant of a matrix. If anyone could explain that part to me or show me where I went wrong it'd be greatly appreciated.
The Diagonalization Theorem: An $n \times n$ matrix $A$ is
diagonalizable if and only if $A$ has $n$ linearly independent
eigenvectors.The Invertible Matrix Theorem: Let $A$ be an $n \times n$ matrix.
Then $A$ is invertible if and only if:1). The number 0 is not an
eigenvalue of $A$2). The determinant of $A$ is not zero
Best Answer
Hint: a $2\times2$ matrix having a double eigenvalue is diagonalizable if and only if it is diagonal.
Can you find an invertible and not diagonal matrix having a double eigenvalue?
A matrix with no real eigenvalues would be a valid choice, provided you are only working over the reals.