Similar Matrices have same Traces
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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