TITLE: Diagonalizable properties of triangular matrix
QUESTION [8 upvotes]: How to show that an upper triangular matrix with identical diagonal entries is diagonalizable iff it is already diagonal?

REPLY [5 votes]: Suppose that all the eigenvalues of this matrix $A\in\mathbb R^{n\times n}$ (or $A\in\mathbb C^{n\times n}$) are equal to $\lambda_0$. Then its characteristic polynomial $p_A(\lambda)$ is equal to 
$$
p_A(\lambda)=\det(A-\lambda I)=(\lambda-\lambda_0)^n,
$$
as $A$ (and also $A-\lambda I$) is triangular, and only the diagonal contributes in the determinant of $A-\lambda I$. 
We now use the following result:
A matrix $A$ is diagonalizable if and only if its minimum polynomial has only single roots.
But, as the minimum polynomial divides the characteristic one, the only divisor of $p_A(\lambda)=(\lambda-\lambda_0)^n$ with simple roots is $q(\lambda)=\lambda-\lambda_0$. Thus $A$ is diagonalizable if and only if $q(A)=0$ or eqivaelently
$$
q(A)=A-\lambda_0 I=0.
$$