Diagonalizable properties of triangular matrix How to show that an upper triangular matrix with identical diagonal entries is diagonalizable iff it is already diagonal?
 A: Let's denote $\lambda$ the entry on the diagonal of the triangular matrix $A$ then the characteristic polynomial 
$$\chi_A(x)=\det(xI_n-A)=(x-\lambda)^n$$
so $\lambda$ is the only eigenvalue of $A$ hence if $A$ is diagonalizable then it's similar to $\lambda I_n$ so $A=\lambda I_n$. The only if case is trivial.
A: 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
$$
q(A)=A-\lambda_0 I=0.
$$
A: More is true actually. 
A non diagonal matrix with all its eigenvalue are same is not diagonalizable.
Hint  :
prove by contradiction, using the fact that scalar matrices commute with all the matrices.
