# Trace of a linear map

I have a question regarding the trace of a linear map $f: V \to V$. As usual, one can define the trace of this map by considering the trace of matrix representation of $f$, that is, choosing a basis for $V$ and describing $f$ as a matrix relative to this basis, and taking the trace of this square matrix.

What I don't get, and is something appearing in my book, is how comes that the trace is then equal to:

$$Tr(f)=\sum_{e_i}\langle f(e_i),e_i\rangle$$

where $\langle \;\_ \;, \;\_ \;\rangle$ stands for an inner product and $e_i$'s form an orthonormal basis.

Let $A$ be the matrix representation of $f$ with respect to the basis $\{e_i\}$. Then, the trace of $f$ is the sum of the diagonal entries of $A$, of which the $i$-th diagonal entry is $e_i^T A e_i = (e_i^T A) e_i = (A e_i)^T e_i = \langle A e_i, e_i\rangle = \langle f(e_i),e_i\rangle$.
So, you sum up all the $\langle f(e_i),e_i\rangle$'s to get the trace.
• I think that your second equality is incorrect. Easily solved by doing what you did for $A^T$ instead of $A$, and then simply observe that $A^T_{ii}=A_{ii}$ ... – Soap Apr 16 '18 at 17:51
I think that the matrix should has in the column i the coordinate of $f(e_i)$. Now, suppose that $e_i$ are the elements of the base, $m_{i,j}$ stands for the component of the matrix of f. Let's calculate $<f(e_i),e_i>$: We have $f(e_i)=\sum_{j}m_{j,i}e_j$ and so $<f(e_i),e_i>=<\sum_{j}m_{j,i}e_j,e_i>=\sum_{j}m_{j,i}<e_j,e_i>$ for linearity. The base is orthogonal and so the onlyone that survive in the sum is $m_{i,i}<e_i,e_i>=m_{i,i}$. If something is not clear tell me.