PlanetMath: proof of properties of trace of a matrix

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proof of properties of trace of a matrix

(Proof)

Proof of Properties:

Let us check linearity. For sums we have

$\displaystyle \operatorname{trace}(A+B)$ $\displaystyle =$ $\displaystyle \sum\limits_{i=1}^{n} (a_{i,i} + b_{i,i})\,\,\,\,\,\,\,\,\,\,\,$ (property of matrix addition)

$\displaystyle =$ $\displaystyle \sum\limits _{i=1} ^{n} a_{i,i} + \sum\limits _{i=1} ^{n} b_{i,i}\,\,\,\,$ (property of sums)

$\displaystyle =$ $\displaystyle \operatorname{trace}(A) + \operatorname{trace}(B).$

Similarly,

$\displaystyle \operatorname{trace}(cA)$ $\displaystyle =$ $\displaystyle \sum\limits _{i=1} ^{n} c\cdot a_{i,i}\,\,\,\,\,$ (property of matrix scalar multiplication)

$\displaystyle =$ $\displaystyle c\cdot \sum\limits _{i=1} ^{n} a_{i,i}\,\,\,\,\,$ (property of sums)

$\displaystyle =$ $\displaystyle c\cdot \operatorname{trace}(A).$
The second property follows since the transpose does not alter the entries on the main diagonal.
The proof of the third property follows by exchanging the summation order. Suppose is a $n\times m$ matrix and is a $m\times n$ matrix. Then

$\displaystyle \operatorname{trace} AB$ $\displaystyle =$ $\displaystyle \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} A_{i,j} B_{j,i}$

$\displaystyle =$ $\displaystyle \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{n} B_{j,i} A_{i,j} \,\,\,\,$ (changing summation order)

$\displaystyle =$ $\displaystyle \operatorname{trace} BA.$
The last property is a consequence of Property 3 and the fact that matrix multiplication is associative;

$\displaystyle \operatorname{trace} (B^{-1} A B)$ $\displaystyle =$ $\displaystyle \operatorname{trace} \big((B^{-1} A) B\big)$

$\displaystyle =$ $\displaystyle \operatorname{trace} \big(B(B^{-1} A) \big)$

$\displaystyle =$ $\displaystyle \operatorname{trace} \big( (BB^{-1}) A \big)$

$\displaystyle =$ $\displaystyle \operatorname{trace} (A).$