Linear Algebra- Proof of trace property How can I prove that $\text{tr}(A)=\text{tr}(A^T)$ using the fact that $\text{trace}(A)= a_{11} +a_{22} +... +a_{nn}$?

If $A$ has coefficients $(A)_{ij}=a_{ij}$ then by definition $A^t$ has as coefficients $(A^t)_{ij}=a_{ji}$. But in particular when $i=j$; $$(A)_{ii}=a_{ii}=(A^t)_{ii}$$ 
For example, if $$A=\begin{pmatrix}1&6&2\\7&-2&-9\\-1&0&2\end{pmatrix}$$ then  $$A^t=\begin{pmatrix}1&7&-1\\6&-2&0\\2&-9&2\end{pmatrix}$$
