proof of properties of trace of a matrix

Proof of Properties:

1.
Let us check linearity. For sums we have
$\\displaystyle\\operatorname{trace}(A+B)$ $\\displaystyle=$ $\\displaystyle\\sum\\limits_{i=1}^{n}(a_{i,i}+b_{i,i})\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\mbox%{(property of matrix addition)}$
$\\displaystyle=$ $\\displaystyle\\sum\\limits_{i=1}^{n}a_{i,i}+\\sum\\limits_{i=1}^{n}b_{i,i}\\,\\,\\,\\,%\\mbox{(property of sums)}$
$\\displaystyle=$ $\\displaystyle\\operatorname{trace}(A)+\\operatorname{trace}(B).$
Similarly,
$\\displaystyle\\operatorname{trace}(cA)$ $\\displaystyle=$ $\\displaystyle\\sum\\limits_{i=1}^{n}c\\cdot a_{i,i}\\,\\,\\,\\,\\,\\mbox{(property of %matrix scalar multiplication)}$
$\\displaystyle=$ $\\displaystyle c\\cdot\\sum\\limits_{i=1}^{n}a_{i,i}\\,\\,\\,\\,\\,\\mbox{(property of %sums)}$
$\\displaystyle=$ $\\displaystyle c\\cdot\\operatorname{trace}(A).$
2.
The second property follows since the transpose does not alterthe entries on the main diagonal.
3.
The proof of the third property follows by exchanging thesummation order. Suppose $A$ is a $n\\times m$ matrix and $B$ is a $m\\times n$ matrix.Then
$\\displaystyle\\operatorname{trace}AB$ $\\displaystyle=$ $\\displaystyle\\sum\\limits_{i=1}^{n}\\sum\\limits_{j=1}^{m}A_{i,j}B_{j,i}$
$\\displaystyle=$ $\\displaystyle\\sum\\limits_{j=1}^{m}\\sum\\limits_{i=1}^{n}B_{j,i}A_{i,j}\\,\\,\\,\\,%\\mbox{(changing summation order)}$
$\\displaystyle=$ $\\displaystyle\\operatorname{trace}BA.$
4.
The last property is a consequence of Property 3 and the fact that matrix multiplication isassociative;
$\\displaystyle\\operatorname{trace}(B^{-1}AB)$ $\\displaystyle=$ $\\displaystyle\\operatorname{trace}\\big{(}(B^{-1}A)B\\big{)}$
$\\displaystyle=$ $\\displaystyle\\operatorname{trace}\\big{(}B(B^{-1}A)\\big{)}$
$\\displaystyle=$ $\\displaystyle\\operatorname{trace}\\big{(}(BB^{-1})A\\big{)}$
$\\displaystyle=$ $\\displaystyle\\operatorname{trace}(A).$