1
$\begingroup$

Consider the matrices $B = \begin{bmatrix}1 &0\\-1&1\\0&1\end{bmatrix}$, $Y = \begin{bmatrix}x_1&0 & 0\\0&x_2&0\\0&0&x_3 \end{bmatrix}$, $\tilde{B} = YB$ and $T = \begin{bmatrix}d&0 & 0\\0&1&0\\0&0&1 \end{bmatrix}$ as well as the vector $P^T = \begin{bmatrix}x_1&x_2&x_3\end{bmatrix} $. Moreover, $H = \tilde{B} (\tilde{B}^T\tilde{B})^{-1}\tilde{B}^T$ is the known hat matrix and $x_1, x_2, x_3 $ are all positive values and $d$ is a scalar $d \geq 1$.

B can be decomposed into a block matrix of the form $ B = \begin{bmatrix}1 &\begin{bmatrix}0\end{bmatrix}\\\begin{bmatrix}C_h\end{bmatrix}& \begin{bmatrix}M_h\end{bmatrix}\end{bmatrix}$ where the entries of column $C_h$ are always negative or zero and the entries of matrix (or vector for the particular case of B having two columns)$M_h$ are always positive. Accordingly, one can decompose $P^T$ as $P^T = \begin{bmatrix}x_1& \begin{bmatrix} x_h\end{bmatrix}\end{bmatrix}$ where $\begin{bmatrix} x_h\end{bmatrix}$ is vector $P^T$ without the first entry. In the same way, $\begin{bmatrix} Y_h\end{bmatrix}$ is equal to $Y$ without the first row an column.

Then, $x_1$ is related with $x_2, x_3 ...$ through $x_1 = \sqrt{-x_h \cdot \begin{bmatrix}Y_h\end{bmatrix}\begin{bmatrix}C_h\end{bmatrix}}$.

Is it possible to prove that the scalar $M = P^T T H P \geq 0$ ?

I have developed the algebraic expression of M and it is possible to prove for the present case that $M\geq 0$ but I am interested in finding a more general and elegant method.

Is it possible to prove $M = P^T T H P \geq 0$ for a general full rank matrix B $\in\mathbb R^{m\times n}$ and matrices $Y, B, T$ and $P$ of compatible dimensions but with the same restrictions on the values of $Y$ and $P$?

EDIT 1: Clarified what I mean by in general.

EDIT 2: Extra constraints.

$\endgroup$

1 Answer 1

1
$\begingroup$

It is wrong in general. Taking $B=\begin{bmatrix}3\\-2\end{bmatrix}$, $Y=\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$ and $T=\begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}$ we have $H=\frac12 \begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}$ and with $P=\begin{bmatrix}2 & 3\end{bmatrix}$ we get for the product: $P T H P^t=-1/2$.

$\endgroup$
7
  • $\begingroup$ Thank you for your answer. Maybe I was not clean enough but there has to be an agreement between Y and P, meaning that $x_1, x_2, x_3 $ are the same values for both matrices. In your case if $P = [2, 3]$ then Y cannot be identity and is $Y = \begin{bmatrix}2&0 \\0&3 \end{bmatrix}$. If that is the case then $M = \frac{5}{13}$ $\endgroup$
    – MymanPJ
    May 16, 2020 at 10:01
  • $\begingroup$ OK, I didn't get that point at first. I think, however, you may use the Y you mention and simply for B then take [1/2 // -1/3] to get the counterexample? $\endgroup$
    – H. H. Rugh
    May 16, 2020 at 11:17
  • $\begingroup$ You're right. Indeed it does not represent all the physical constraints in my problem and like this, the statement is not true. In my problem $x_1$ is related to $x_2 etc..$ through an additional constraint that I edited in the main question. $\endgroup$
    – MymanPJ
    May 16, 2020 at 18:02
  • $\begingroup$ This gets a little bit out of hand... In general I think that there is little reason to believe that the product of two positive (so also s.a.) but non-commuting operators stay positive. $\endgroup$
    – H. H. Rugh
    May 16, 2020 at 20:30
  • $\begingroup$ Indeed it gets out of hand. And yes, I was not hopeful at all that it would be the case. But the point is that this is inside a method I use all the time and it always works (to my surprise). Hence the reason why I'm motivated to prove it. Either way, thank you for your help! $\endgroup$
    – MymanPJ
    May 16, 2020 at 21:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .