If we have $v_1,…,v_n$ be an orthonormal basis for $\mathbb R^n$ and $A$ any $n \times n$ real or complex matrix, prove that:
$$Trace(A)=\sum_{i-1}^n Av_i\bullet v_i$$
I see that $v_i\bullet v_j=\delta_{ij}$ so it makes sense that it's true, but I can't quite see how to get there so any hints would be appreciated.
More generally we have that $A$ must be a linear combination of the basis elements:
$$Av_j=\sum_{i=1}^nb_{ij}v_i$$
We then have a matrix $B=b_{ij}$ so prove that:
$$Trace(A)=\sum_{i=1}^nb_{ii}$$
Best Answer
Let $V=\begin{bmatrix} v_1 \cdots v_n \end{bmatrix}$, then $V^TV = I$.
By definition, $\operatorname{tr} A = \sum_k e_k^T A e_k$, and by the cyclic property of trace we have $\operatorname{tr} A = \operatorname{tr} (A V V^T) = \operatorname{tr} (V^T A V) = \sum_k e_k^T V^T A V e_k = \sum_k v_k^T A v_k$.