Diagonalizable properties of triangular matrix

Question

How to show that an upper triangular matrix with identical diagonal entries is diagonalizable iff it is already diagonal?

Answer 1

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.

Answer 2

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 $$ q(A)=A-\lambda_0 I=0. $$

Answer 3

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.