1
$\begingroup$

Let $A \in \mathbb{R}^{n \times n}$. It is well-known that $\text{tr}(A)$ is equal to the sum of the eigenvalues of $A$.

Let us know restrict $A$ to being positive semi-definite. Obviously, it is still the case that $\text{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n$, where the $\lambda_is$ are the eigenvalues of $A$, since that was a general result. However, is there an easier way to show it, if we restrict $A$ to being positive semi-definite? I can't find a more elegant proof, other than the general one that applies to any square matrix.

$\endgroup$
4
  • $\begingroup$ The result follows from the fact that $\operatorname{tr}(ABC) = \operatorname{tr}(BAC)$. $\endgroup$
    – copper.hat
    Feb 16, 2016 at 6:20
  • $\begingroup$ I'm not quite sure what you're getting at... $\endgroup$
    – Ryan
    Feb 16, 2016 at 6:42
  • $\begingroup$ Eigenvalue decomposition ... $\endgroup$
    – user251257
    Feb 16, 2016 at 7:51
  • $\begingroup$ Actually $\mathrm{tr}(A)$ is not the sum of the eigenvalues of $A$ (try $A=I_3$, with $1$ as only eigenvalue), but rather the sum of the eigenvalues counted with their multiplicity as roots of the characteristic polynomial $\chi$. So for the general case, the natural proof is that for any monic split polynomial (like $\chi$ if you are assuming "all eigenvalues exist") the sum of the roots equals minus the coefficient of $X$, which is fairly obvious. In the diagonalisable case (which includes real symmetric matrices, p.s.d. is irrelevant) the fact is obvious using a basis of eigenvectors. $\endgroup$
    – Marc van Leeuwen
    Feb 16, 2016 at 14:19

1 Answer 1

Reset to default
2
$\begingroup$

I'm not sure what is the general proof you have in mind, but if we choose any orthonormal basis $v_1, \ldots, v_n$ for $\mathbb{R}^n$ (with respect to the standard inner product $\left< \cdot, \cdot \right>$) then

$$ \mathrm{tr}(A) = \sum_{i=1}^n \left< Av_i, v_i \right>. $$

If $A$ is symmetric, then by choosing $v_1, \ldots, v_n$ to be an orthonormal basis of eigenvectors of $A$ (with $Av_i = \lambda_i v_i$), you immediately get

$$ \mathrm{tr}(A) = \sum_{i=1}^n \left< Av_i, v_i \right> = \sum_{i=1}^n \left< \lambda_i v_i, v_i \right> = \sum_{i=1}^n \lambda_i. $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .