Exponent of a positive semidefinite matrix

Question

Consider a positive semidefinite matrix $H$ and consider the matrix exponential

$$U = e^{-\beta H}.$$

Let the spectral decomposition of $H$ be

$$H = \sum_{k} \lambda_k u_k u_{k}^{*}. $$

Then, could I write $U$ in terms of $\lambda_k$ and $u_k$?

I think $U$ can be written as

$$U = \sum_{k} e^{-\beta \lambda_k} u_k u_k^{*}. $$

But, just by expanding $e^{-\beta H}$ in the form of a power series, I did not get the desired expression. There's also the question of what happens when some of the $\lambda_k$-s are $0$.

Answer 1

Expanding $\ e^{-\beta H}\ $ in power series does give the desired expression if you use the fact that $\ u_i^*u_j=\delta_{ij}\ $, from which it follows that if $\ \left(\sum_\limits k\lambda_ku_ku_k^*\right)^{n-1}=$$\,\sum_\limits{k}\lambda_k^{n-1}u_ku_k^* \ $ then \begin{align} H^n&=\left(\sum_k\lambda_ku_ku_k^*\right)^n\\ &=\sum_k\lambda_ku_ku_k^*\left(\sum_k\lambda_ku_ku_*\right)^{n-1}\\ &=\sum_i\lambda_iu_iu_i^*\sum_k\lambda_k^{n-1}u_ku_k^*\\ &=\sum_i\sum_k\lambda_i \lambda_k^{n-1}u_iu_i^*u_ku_k^*\\ &=\sum_i\lambda_i^nu_iu_i^*\ , \end{align} and hence that $\ H^n=\sum_\limits{k}\lambda_k^nu_ku_k^* \ $ by induction.

Answer 2

You can define $f(U)$ for any function $f$ defined on $\{\lambda_i\}$ by $$f(U)=\sum f(\lambda_i)u_iu_i^*$$ The operators $P_i=u_iu_i^*$ form a family of mutually orthogonal one dimensional projections, i.e. $P_iP_j=0$ if $i\neq j$ and $P_i^2=P_i.$ Thus we get $$f(U)g(U)=(fg)(U)$$ Moreover if the function $f$ is a pointwise limit of the functions $f_n$ then $$f(U)=\lim_nf_n(U)$$ The latter can be applied to the partial sums of the function $e^{-\beta x}.$

Remark In the infinite dimensional case we have to restrict to the functions continuous on the set $\{\lambda_i\}$ and the pointwise convergence should be replaced by the uniform one.