Is any matrix with all positive eigenvalues diagonalizable? I know that a symmetric matrix with positive eigenvalues (i.e. a positive-definite matrix) is diagonalizable. But what if the matrix is not symmetric? Can we still diagonalize it? i.e. if $A$ is a matrix with all positive eigenvalues, can we write it as $A=P^{-1}DP$ where $D$ is a diagonal matrix with eigenvalues on the diagonal?
 A: The answer is no.  As pointed out by amsmath in his comment, the matrix
$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \tag 1$
cannot be diagonalized, and a wider class of examples is provided by size $n$ square matrices of the form
$B = \lambda I + T, \tag 2$
where $I$ is the $n \times n$ identity matrix, and $T \ne 0$ is an $n \times n$ strictly upper triangular matrix, that is, $T$ takes the form
$T = [t_{ij}] \ne 0, \tag 3$
where 
$t_{ij} = 0, \; i \le j. \tag 4$
We note that
$T^n = 0, \tag 5$
and this implies that the only eigenvalue of $B$ is $\lambda$, since if
$B \vec v = \mu \vec v \tag 6$
for some $\vec v \ne 0$, by (2) we would have
$(\mu - \lambda) \vec v= (B - \lambda I)\vec v = T\vec v, \tag 7$
and this in turn yields
$(\mu - \lambda)^n \vec v = (B - \lambda)^n \vec v = T^n \vec v = 0; \tag 8$
since $\vec v \ne 0$ we conclude that
$(\mu - \lambda)^n  = 0, \tag 9$
whence 
$\mu = \lambda. \tag{10}$
Since $\lambda$ is the only eigenvalue of $B$, if some matrix $P$ diagonalized $B$ we would have
$\Lambda = PBP^{-1}, \tag{11}$
$\Lambda$ being the diagonal matrix
$\Lambda = \lambda I; \tag {12}$
inserting $B$ as in (2) into (11) yields
$\Lambda = P(\lambda I + T)P^{-1} = P(\lambda I)P^{-1} + PTP^{-1}$
$= \lambda PIP^{-1} + PTP^{-1} = \lambda I + PTP^{-1} = \Lambda + PTP^{-1}, \tag {14}$
or
$PTP^{-1} = 0, \tag{15}$
so that in fact
$T = 0.  \tag{16}$
We have proved that if a matrix of the form (2) is diagonalizable then $T = 0$; thus, by contraposition, $T \ne 0$ implies $B$ cannot be diagonalied; in this way a great many instances of non-diagonalizable matrices may be constructed.  Note that we do not require $\lambda > 0$.
Of course, if the eigenvalues of any matrix are all distinct, then it may be diagonalized, as is well-known.
Finally, it appears upon scrutiny of the above argument that we may relax the condition that $T$ is strictly upper triangular to the more general hypothesis that $T$ is nilpotent; that is
$T^m = 0 \tag{17}$
for some positive integer $m$.  The essential details of the proof are left undisturbed by this assumption. 
