$M$ matrix, $\mathrm{rank}\ M=1$. Prove that $det(e^M)=1$ iff $M$ is not diagonalizable M is a $n\times n$ matrix over $\mathbb R$. with $\mathrm{rank}\ M=1$.
Prove that $det(e^M)=1$ if and only if $M$ is not diagonalizable.
I really don't know how to start thinking about this.. :/
I'd be glad to get some help
thanks.
 A: It's related to this Prove that $A$ is diagonalizable iff $\mbox{tr} A\neq 0$
Indeed, $\det(e^M)=1 \iff \exp(\operatorname{Tr}(M))=1 \iff \operatorname{Tr}(M) = 0$
Since $M$ has rank $1$,and by your previous question, $\operatorname{Tr}(M) = 0 \iff M \;\text{not diagonalizable}$

The identity $\det(e^M)=\exp(\operatorname{Tr}(M))$ can be proved via trigonalization over $\mathbb C$.
A: Because the matrix has rank one, there is a basis $v_1,....,v_n$ such that $Mv_1 \neq 0$ and $Mv_k = 0$ for $k > 1$. So there is a $N = A^{-1}M A$ for some $A$ such that the matrix of $N$ has zeroes in every entry in the second through $n$th columns. Note that $e^N = A^{-1}e^M A$ so that $det(e^N) = 1$ iff $det(e^M) = 1$. But it's not hard to take the determinant of $e^N$ since it's a lower triangular matrix... now use your problem on traces you asked about earlier.
A: Here's our identity: $\det(e^{M}) = \exp(\mathrm{Tr}(M)) = 1 \iff \mathrm{Tr}(M) = 0$ (since $M$ has real entries). 
Pf:($\implies$) Assume $M$ is diagonalizable. Then $\exists C$ such that $M = C\Lambda C^{-1}$, where $\Lambda$ is a diagonal matrix with entries $\{\lambda_{j}\}_{j=1}^{n}$. Since $M$ has rank 1, then $\exists k$ such that $\lambda_{k} \ne 0$, but all other $\lambda_{j} = 0$. Then \begin{align}
\det(e^{M}) &= \det(Ce^{\Lambda}C^{-1}) = \det(C)\det(e^{\Lambda})\det(C^{-1}) = \det(e^{\Lambda}) = \exp(\mathrm{Tr}(\Lambda)) = \exp(\lambda_{k}) \ne 1.
\end{align}
$(\impliedby)$ By the Jordan canonical form, $M = CJC^{-1}$, where $J$ is made up of Jordan blocks of the form 
\begin{align}
J_{i} = 
\begin{pmatrix}
\lambda_{i} & 1           & 0     \\
0           & \lambda_{i} & 1  &  \\
            &             & \ddots  & \ddots
\end{pmatrix}
\end{align}
All blocks must be $1\times 1$ or $2\times 2$, or else $\mathrm{rank}(M) > 1$. There must exist a $2\times 2$ block, or else the matrix is diagonalizable. The eigenvalue of this block must be $0$, or its rank is 2. Since the image of each block spans orthogonal subspaces, then the $2\times 2$ block is the only non-zero block. In this case, $\mathrm{Tr}(M) = 0$, giving $\det(e^{M}) = 1$. 
