the sum of the first least eigenvalues is the minimum of the trace of orthonormal multiplications

[[concept]]

Topics

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Corollary

Let $A\in M_n$ be hermitian and $r:1\le r\le n\in \mathbb{Z}$. Then ${\sum }_{k=1}^r{\lambda }_k(A)={\min }_{U\in M_{n,r}\text{ orthonormal cols}}\mathrm{Tr}{U}^{\ast }AU$

and also ${\sum }_{k=0}^{r-1}{\lambda }_{n-k}(A)={\max }_{U\in M_{n,r}\text{ orthonormal cols}}\mathrm{Tr}{U}^{\ast }AU$

And claim that this implies the previous result ( just take take $U$ as the identity )

Proof / Intuition $U\in M_n$ with orthonormal columns, extend Gram-Schmidt to get $V=[U|\ast ]\in M_n$ unitary. Then when we do the multiplication $[V^{\ast }AV{]}_{r\times r}=U^{\ast }AU$ So by interlacing 2, we get that ${\lambda }_k(A)={\lambda }_k(V^{\ast }AV)\le {\lambda }_k(U^{\ast }AU)$ So sum over $k=1,2,\dots ,r$ to get ${\sum }_{k=1}^r{\lambda }_k(A)\le \mathrm{Tr}(U^{\ast }AU)$ So we have the desired lower bound, and need to show we can achieve equality.

Given any

Let $W$ be the orthonormalized eigenvectors associated with ${\lambda }_1(A),\dots ,{\lambda }_r(A)$. $\mathrm{Tr}{U}^{\ast }AU=\mathrm{Tr}{W}^{\ast }AW=Tr{D}_r$ (the sum of the first $r$ eigenvalues)

References

References

See Also

Mentions

Mentions

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