# How to geometrically interpret $\sum^{p}_{1}\lambda_{i}(A)=\operatorname{tr}(A)$?

$A$ is a $p\times p$ real matrix and $\lambda_{i}$ are its eigenvalues. $\operatorname{tr}(A)$ is the trace of $A$.

How to geometrically interpret $\sum^{p}_{1}\lambda_{i}(A)=\operatorname{tr}(A)$?

I have learnt linear algebra for two semesters. I knew the basic concepts of trace and eigenvector.

The answer in the mathoverflow interprets geometrical meaning of trace.But,how to interpret the trace is equal to the sum of eigenvalue geometrically?

First, I think it helps if you can prove the statement.

The trace of a square matrix is defined to be the sum of the diagonal elements.

We can write a matrix using Jordan Normal Form, such that $$A = P^{-1} M P$$

where $A$ is a $n\times n$ real matrix with $\lambda_{i}$ as its eigenvalues.

$M$ is a (upper triangular) matrix with eigenvalues of $A$ as the diagonal elements and $P$ is an invertible matrix.

Using the rules for trace, $Tr(AB)=Tr(BA)$, we have:

$$Tr(A)= Tr( P^{-1} M P) = Tr(MPP^{-1})=Tr(M)=\sum^{p}_{1}\lambda_{i}(A)$$

Next, refer to the post on geometric-interpretation-of-trace.

Lastly, you might find the geometric-interpretation-of-characteristic-polynomial useful too.

Regards -A

• Wonderful answer! +1 – Namaste May 11 '13 at 0:28