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Hello!

Let $v_i\in R^d$ $(i=1,...,n,n>d)$ be unit-length vectors ($v_i^Tv_i=1$). Then $v_iv_i^T$ is an orthogonal projection matrix, which has many elegant properties. Now consider a linear combination of these orthogonal matrices $$A=\sum_{i=1}^n c_i v_i v_i^T$$ where $c_i$ are positive scalars and $\sum_{i=1}^n c_i=1$. My question is: what $v_i$ can make the linear combination of these orthogonal matrices be an identity matrix? That is: how to find $v_i$ such that $$\sum_{i=1}^n c_i v_i v_i^T=\frac{1}{d}I$$

Thanks.

edited Mar 31 2011 at 14:22

asked Mar 31 2011 at 13:15

Shiyu
59●6

I forget to note $c_i\le 1/d$ since it is a necessary condition for $\sum_{i=1}^n c_iv_iv_i^T=I$ – Shiyu Mar 31 2011 at 13:17

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A necessary and sufficient condition for your question as stated is that $d=1$ ;). The trace of $\sum c_i v_i v_i^{T}$ is indeed $\sum c_i=1$. You probably rather want that $\sum c_i=d$. – Mikael de la Salle Mar 31 2011 at 13:26

Thanks Mikael: You are right. I should let $\sum c_i=d$ or equivalently I can let $\sum c_i v_i v_i^T=\frac{1}{d}I$. – Shiyu Mar 31 2011 at 14:25

Bill Johnson · Accepted Answer · 2011-03-31 13:58:12Z UTC

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If you make the change Mikael suggests, you are basically looking at tight frames. $c_i^{1/2} v_i$ should be the image under an orthogonal projection of an orthonormal basis from some $n$ dimensional superspace.

answered Mar 31 2011 at 13:58

Bill Johnson
14.4k●1●33●60

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With the stated assumption that the $v_i$ are unit vectors, I think the exact terminology here is "normalized tight frames". – Mark Meckes Mar 31 2011 at 14:41

@Bill: it seems what I am looking for. Thank you. – Shiyu Apr 1 2011 at 1:44

Linear combination of orthogonal projection matrices

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