Proof of Properties:

1. Let us check linearity. For sums we have \begin{eqnarray*} \operatorname{trace}(A+B) &=& \sum\limits_{i=1}^{n} (a_{i,i} + b_{i,i})\,\,\,\,\,\,\,\,\,\,\, \mbox{(property of matrix addition)}\\ &=&\sum\limits _{i=1} ^{n} a_{i,i} + \sum\limits _{i=1} ^{n} b_{i,i}\,\,\,\, \mbox{(property of sums)} \\ &=&\operatorname{trace}(A) + \operatorname{trace}(B). \end{eqnarray*}Similarly,

   \begin{eqnarray*} \operatorname{trace}(cA) &=& \sum\limits _{i=1} ^{n} c\cdot a_{i,i}\,\,\,\,\, \mbox{(property of matrix scalar multiplication)} \\ &=& c\cdot \sum\limits _{i=1} ^{n} a_{i,i}\,\,\,\,\, \mbox{(property of sums)} \\ &=& c\cdot \operatorname{trace}(A). \end{eqnarray*}

2. The second property follows since the transpose does not alter the entries on the main diagonal.
3. The proof of the third property follows by exchanging the summation order. Suppose $A$ is a $n\times m$ matrix and $B$ is a $m\times n$ matrix. Then \begin{eqnarray*} \operatorname{trace} AB &=& \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} A_{i,j} B_{j,i} \\ &=& \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{n} B_{j,i} A_{i,j} \,\,\,\,\mbox{(changing summation order)}\\ &=& \operatorname{trace} BA. \end{eqnarray*}
4. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; \begin{eqnarray*} \operatorname{trace} (B^{-1} A B) &=& \operatorname{trace} \big((B^{-1} A) B\big) \\ &=& \operatorname{trace} \big(B(B^{-1} A) \big) \\ &=& \operatorname{trace} \big( (BB^{-1}) A \big) \\ &=& \operatorname{trace} (A). \end{eqnarray*}