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I would like to know whether or not (e.g., via a counter-example) the following inequality is preserved when $P$ is a non-negative (symmetric) positive definite matrix for all $i,j$:

$$ \| P_i – k \ P_j \|_2 \leq \| P_i \|_2 $$

where $P_i$ represents the $i$'th column of the matrix $P$ and $k = \frac{P_{ij}}{1+P_{jj}} \in \mathbb{R}_+$ and with $P_{ij}$ representing the $(i,j)$'th element of $P$.

Any counters or a proof of this holding would be appreciated.

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The required result is false when $n\geq 3$.

Take $P=\begin{pmatrix}1.49&1&0\\1&6684.74&11101\\0&11101&18436.74\end{pmatrix}$ and $i=1,j=2$.

Remark 1. @ user2457324 , in your procedure above, you conjecture that such a $P$ can be written $P=X^TX$ where $x_{i,j}\geq 0$. I am not sure that this conjecture is true.

Remark 2. Your chosen $k$ is not homogeneous with respect to the $p_{k,l}$. if your inequality were true, then, for every $t>0$, $||tP_i-\dfrac{tp_{i,j}}{1+tp_{j,j}}tP_j||\leq ||tP_i||$ or $||P_i-\dfrac{tp_{i,j}}{1+tp_{j,j}}P_j||\leq ||P_i||$; that implies, when $t\rightarrow +\infty$, $||P_i-\dfrac{p_{i,j}}{p_{j,j}}P_j||\leq ||P_i||$, a homogeneous inequality. Unfortunately, the last inequality is also wrong (consider the same counter-example as above).

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The result is true for $2 \times 2$ matrices. Every $2 \times 2$ positive-semidefinite matrix is of the form $$\begin{pmatrix} \sigma_1^2 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma_2^2 \end{pmatrix}$$ where $\sigma_1,\sigma_2 \geq 0$ and $|\rho| \leq 1.$ This follows easily from the fact the every positive semidefinite matrix is the variance-covariance matrix of a random-vector and vice-versa.

Choosing $i=1$ and $j=2$ in your inequality we have to prove $$ \left \| \begin{pmatrix} \sigma_1^2 \\ \rho \sigma_1 \sigma_2 \end{pmatrix} - \dfrac{\rho \sigma_1 \sigma_2}{1 + \sigma_2^2} \begin{pmatrix} \rho \sigma_1 \sigma_2 \\ \sigma_2^2 \end{pmatrix} \right \| \leq \left \| \begin{pmatrix} \sigma_1^2 \\ \rho \sigma_1 \sigma_2 \end{pmatrix} \right\|. $$ The left hand side above is $$ \left\| \begin{pmatrix} \sigma_1^2 ( 1 - \dfrac{\rho^2 \sigma_2^2}{1+\sigma_2^2}) \\ \rho \sigma_1 \sigma_2 (1 - \dfrac{\sigma_2^2}{1+\sigma_2^2}) \end{pmatrix} \right\| $$ and since $ 0 \leq 1 - \dfrac{\rho^2 \sigma_2^2}{1+\sigma_2^2} \leq 1$ and $0 \leq 1 - \dfrac{\sigma_2^2}{1+\sigma_2^2} \leq 1$ the inequality follows. The proof for $i=2,j=1$ is similar and the result follows for $2 \times 2$ positive semidefinite matrices.

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  • $\begingroup$ @user2457324 If you don't mind my asking : what led you to this inequality? $\endgroup$ – Arin Chaudhuri Sep 4 '16 at 2:43

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