Diagonalization with a matrix in $SL_n(\mathbb{R})$

Question

Suppose that $A$ is diagonalizable. Can the diagonalization be done with a matrix $P$ in the special linear group $SL_n(\mathbb{R})$ (i.e. such that $\det(P)=1$) ?

Answer 1

Yes.

If A is diagonalized with $Q$, just multiply one column of $Q$ by $\frac{1}{\det(Q)}$ to get $P$.

Answer 2

It is always possible.

Suppose $U^{-1} A U = \Lambda$.

Let $P:\mathbb{R}^n \to \mathbb{R}^n$ be given by $P((x_1,...,x_n)) = (\frac{x_1}{\det U},x_2,...,x_n))$. Then $\det P = \frac{1}{\det U}$.

Then $(UP)^{-1} A (UP) = \Lambda$, and $\det (U P) = 1$.