Let $T: \mathbb{R}^4 \to \mathbb{R}^4$ be a linear map with characteristic polynomial $pt(x)$.
Is $T$ diagonalizable in the following cases?
- $pt(x) = x^4-1$
- $pt(x) = x^3(x+1)$ and $\dim \ker(T) = 2$
- $pt(x) = x^2(x^2- 4)$ and $\dim \ker(T) = 2$
For the first case, I concluded that $T$ is not diagonalizable because the roots of the characteristic polynomial are $x1 = 1$, $x2 = -1$, $x3 = i$, $ x4 = -i$. Is this correct?
What do I do on the other cases?
Thanks and sorry for anything not well explained, this is my first question here.
In the first case you get 4 distinct eigenvalues: $\pm 1, \pm i$. So $T$ is diagonalizable over $\mathbb{C}$: there are four eigenspaces in $\mathbb{C}^4$, each of (complex) dimension 1. Hence the eigenvectors span $\mathbb{C}^4$. Hence $T$ is diagonalizable over $\mathbb{C}$. (It is not diagonalizable over $\mathbb{R}$, because it has non-real eigenvalues.)
In the second case you have 2 eigenvalues: $0$ and $-1$. The latter has multiplicity 1. The former has geometric multiplicity 2, and the space has dimension 4. So the eigenvectors do not span $\mathbb{R}^4$. So $T$ is not diagonalizable even over $\mathbb{C}$.
Now try the same idea on the third case.