$\lambda_{\min}$ and $\lambda_{\max}$ of rank-1 sum of matrices

Question

It is explained from previous posts^1,2 that for a rank-$1$ matrix $x_ix_i^T$ we have $\lambda_{\max} (x_ix_i^T)=1$ and $\lambda_{\min} (x_ix_i^T)=0$ with single and $N-1$ algebraic multiplicity, respectively.

Could you please provide some help to compute the $\lambda_{\min}(A)$ and $\lambda_{\max}(A)$ of $$A=\sum_{j=1}^{n}x_ix_j^T$$ with $x_j\in\mathbb{R}^{N}$, $x_j^Tx_j=1$, $x_i^Tx_j=0, i\neq j$ and $n < N$.

Are you willing to assume that $x_{i}^{T}x_{j}=0$ for $i \neq j$? — Brian Borchers, Dec 08 '19 at 15:11
I think in the first part of the proof I am trying to make it is ok but any comment in the case they are not independent are welcome :). Thanks for the interest. — darkmoor, Dec 08 '19 at 15:14

Martin Argerami · Accepted Answer · 2019-12-08T19:39:39.477

2

Still $0$ and $1$. Your $A$ is a projection: $A^2=A$, so its eigenvalues need to satisfy $\lambda^2=\lambda $; thus the only possible eigenvalues are $0$ and $1$. And both eigenvalues are realized, since $Ax_1=x_1$, and $Ax_N=0$.

edited Dec 08 '19 at 19:39

answered Dec 08 '19 at 16:39

Martin Argerami

205,756

1

Interesting approach with the projection, thanks. So there are $n$ eigen-values all equal to 1 and $N-n$ all equal to zero? Thus , the $tr(A)=n$? – darkmoor Dec 08 '19 at 17:10
@darkmoor that is correct. – Ben Grossmann Dec 08 '19 at 17:53

$\lambda_{\min}$ and $\lambda_{\max}$ of rank-1 sum of matrices

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