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.

Thanks in advance for your help.


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.

  • $\begingroup$ 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 $A^T_{ii}=A_{ii}$ ... $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.