# Similar Matrices have same Traces

## Theorem

Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be square matrices of order $n$.

Let $\mathbf A$ and $\mathbf B$ be similar.

Then:

$\map \tr {\mathbf A} = \map \tr {\mathbf B}$

where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.

## Proof

By definition of similar matrices

$\exists \mathbf P: \mathbf P^{-1} \mathbf A \mathbf P = \mathbf B$

where $\mathbf P$ is an invertible matrix of order $n$.

Thus it remains to show that:

$\map \tr {\mathbf P^{-1} \mathbf A \mathbf P} = \map \tr {\mathbf A}$