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 $n$.

Therefore:

\(\ds \map \tr {\mathbf B}\)

\(=\)

\(\ds \map \tr {\mathbf P^{-1} \mathbf A \mathbf P}\)

\(\ds \)

\(=\)

\(\ds \map \tr {\mathbf P \paren {\mathbf P^{-1} \mathbf A} }\)

Trace of Product of Matrices

\(\ds \)

\(=\)

\(\ds \map \tr {\paren {\mathbf P \mathbf P^{-1} } \mathbf A}\)

Matrix Multiplication is Associative

\(\ds \)

\(=\)

\(\ds \map \tr {\mathbf I_n \mathbf A}\)

Definition of Inverse Matrix

\(\ds \)

\(=\)

\(\ds \map \tr {\mathbf A}\)

Unit Matrix is Identity for Matrix Multiplication

$\blacksquare$

Sources

• 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace