Each row has the same sum in matrix $A$. Is it diagonalizable? Given a non-invertible matrix $A \in \mathbb{R}^{3x3}$ such that each row summation is $17$:


*

*Prove that $17$ is an eigenvalue of $A$

*Is $A$ diagnolizable? is it triangulable?


To show that $17$ is an eigenvalue I took $v=\begin{pmatrix}1\\1\\1\end{pmatrix}$, then follows $Av = 17v$.
Also, $0$ is an eigenvalue because the matrix is non-invertible.
Can I conclude something from these eigenvalues about question $2$? My guess is that it's not definite because I don't have 3 uniqe eigenvalues but i'm not sure.
 A: Let $P$ be the matrix obtained from $A$ after dividing each entry by $17$. Then the sum of the entries in each row is $1$. 


*

*The matrix $P$ is non-invertible, so $0$ is an eigenvalue.

*The matrix $P$ has the eigenvector $v$, and $Pv=v$, so $1$ is an eigenvalue.


We want to contruct such a $P$, which is not diagonalizable. Then the minimal polynomial of $P$ has roots $0,0,1$ or $0,1,1$. And the only possible prototypes after a base change are correspondingly
$$
\begin{aligned}
P &= 
S^{-1}\begin{bmatrix} 1 & & \\ & 0 & 1 \\ & & 0 \end{bmatrix}S\qquad \text{ or}\\
P &=
S^{-1}\begin{bmatrix} 1 & 1 &\\ & 1 & \\ & & 0\end{bmatrix}S\ . 
\end{aligned}
$$
The construction is now clear, take any matrix $S$ that maps the $1$-eigenvector $v$ of $P$ in (a scalar multiple of) the only $1$-eigenvector of the corresponding Jordan matrix. (But then the scalar does not show up after also multiplying with $S^{-1}$ from the left.) Each invertible $S$ with
$$ 
S
\begin{bmatrix} 1\\1\\1\end{bmatrix} =
\begin{bmatrix} 1\\0\\0\end{bmatrix} 
$$
works. For instance, taking
$$
S = \begin{bmatrix}1& &\\-1&1&\\-1&&1\end{bmatrix}
$$ we obtain two choices of counterexamples:
$$
P_{1,0,0}=
\begin{bmatrix}1 & 0 & 0 \\ 0 &0 & 1 \\ 1&0&0\end{bmatrix}\ ,\qquad
P_{1,1,0}=
\begin{bmatrix}0 & 1 & 0 \\ -1 &2 & 0 \\ 0&1&0\end{bmatrix}\ .
$$

Note that the minimal polynomial of a $P$ as in the OP is either $X(X-1)$, so $P^2=P$, and $P$ can be diagonalized by a base change, or else one of $X^2(X-1)$ and $X(X-1)^2$. The generalized eigenspaces $V_0$, $V_1$ have dimensions $2,1$ or $1,2$. We have eigenvectors $v_0,v_1$, and then we (can) pick one more vector $w$ with $Pw=v_0$ (so $P^2w=0$ in the first case), respectively $Pw=v_1=v$ (so $(P-1)^2w=0$ in the second case). The vectors $v_0, v_1; w$, taken as columns in the right order give the base change matrix into the triangular (Jordan) form.
A: As Tim suggest, for sure the matrix is triangulable. Infact the characteristic polynomial isa a 3-degree polynomial with, at least, two real roots (0 and 17), so also the 
third root will be real.
The matrix:
 \
   \begin{array}{ccc}
   17 & 17 & 0\\
   0 & 0 & 17 \\
   0 & 0 & 0 \\
  \end{array} 
is a counterexample for the 
diagonalizability
