Trace of Sum of Matrices is Sum of 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 $\mathbf A + \mathbf B$ debote the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
Then:
- $\map \tr {\mathbf A + \mathbf B} = \map \tr {\mathbf A} + \map \tr {\mathbf B}$
where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.
Proof
\(\ds \map \tr {\mathbf A} + \map \tr {\mathbf B}\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n a_{kk} + \sum_{k \mathop = 1}^n b_{kk}\) | Definition of Trace of Matrix | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {a_{kk} + b_{kk} }\) | Sum of Summations equals Summation of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \tr {\mathbf A + \mathbf B}\) | Definition of Matrix Entrywise Addition, Definition of Trace of Matrix |
$\blacksquare$
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