# Showing for which h a matrix is diagonalizable

1. Oct 30, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
For what $h$ is the matrix $\begin{bmatrix}1 & -h^2 & 2h \\ 0 & 2h & h \\ 0 & 0 & h^2 \end{bmatrix}$ diagonalizable with real eigenvalues? (More than one may be correct)

a) -2, b) -1, c) 0, d) 1, e) 2

2. Relevant equations

3. The attempt at a solution
We already know the eigenvalues, since the matrix is upper triangular. How do we proceed? Do we just plug in the values of h and see if it is diagonalizable? It seems like that would take a very long time...

2. Oct 30, 2016

### andrewkirk

Yes, I would say the question is poorly expressed. I think what they meant to ask here is 'for what $h$ is the matrix diagonalisable over the reals?', which means that the change of basis matrix used must have only real entries.

There is low-hanging fruit that enables determining whether some of a-e satisfy the requirement, based on the fact that:
1. any diagonal matrix is diagonalisable
2. any $n\times n$ matrix over field $F$ with $n$ distinct eigenvalues in $F$ is diagonalisable over $F$.
Having picked that fruit, you will have fewer of a-e left to try to work out whether they are diagonalisable using more advanced means. Have a go at that first part first.

PS If you get stuck when you are up to the 'more advanced means' for the remaining cases, have a look in this wiki section about characterisation of diagonalisability.