# 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 – amWhy May 11 '13 at 0:28