If Matrix A, B are similar, is the sum of the diagonals equal?

Question

Given Matrix $A \sim B$ , can the sum of the diagonals be equal? E.g $$ A = \begin{bmatrix} 0 & 2 & -3 \\ -1 & 3 & -3 \\ 1 & -2 & a \\ \end{bmatrix},\qquad\qquad B = \begin{bmatrix} 1 & -2 & 0 \\ 0 & b & 0 \\ 0 & 3 & 1 \\ \end{bmatrix} $$

Why $A \sim B$ can result $0+3+a=1+b+1$?

Answer 1

$\newcommand{\tr}{\operatorname{tr}}$ The sum of the numbers on the diagonal of a matrix $A$ is called the trace of $A$ which is denoted by $\tr(A)$. The trace of a product of three square matrices of the same size $A,B,C$ satisfies $$\tr(ABC)=\tr(BCA),\qquad \text{(cyclicity)}$$ Then if $A$ and $B$ are similar matrices, i.e. there exists an invertible matrix $P$ such that $A=PBP^{-1}$, then we have that $$\tr(A)=\tr(PBP^{-1})=\tr(BP^{-1}P)=\tr(BI)=\tr(B)$$

Answer 2

Alternatively, you can note that similar matrices must have the same characteristic polynomial, and thus the same eigenvalues. The sum of the eigenvalues is the trace of the polynomial, so any two similar matrices must share the same trace.