Question is to Prove/Disprove :
If a triangular matrix $A$ is similar to diagonal matrix, then $A$ is already diagonal matrix.
I guess the answer is Yes.
I am not able to write the solution precisely.
What i have thought is Suppose $A$ is similar to diagonal matrix, then its eigen values are all distinct and as $A$ is already triangular, A has to be diagonal
I clearly see there is a gap in my justification, But do not have an idea to proceed..
Another Question is to prove that :
Suppose $A$ is a nilpotent $n \times n$ matix then trace of $A$ is zero.
My justification for this is :
As $A$ is nilpotent, we have $A^k=0$
As we know eigenvalues determines the minimal polynomial for $A$ (as roots of minimal polynomial) and sum of roots is being coefficient of $A$ in its minimal polynomial gives that sum of eigen values is zero and thus trace is zero (as trace equals to sum of eigen values )
Best Answer
Since $A$ is similar to a diagonal matrix you know that, $$A = P^{-1}BP, $$ for some invertible matrix P and diagonal matrix B. A is called diagonalizable. Such a matrix is diagonalizable if it has all distinct eigenvalues. Such a matrix does not have to be diagonal to begin with. $$A = \begin{pmatrix} -1 & 2 \\ 0 & 3 \end{pmatrix}, $$ is diagonaliable and hence similar to a diagonal matrix but is not diagonal.