Show this matrix is positive (semi)definite

I had been working on this problem for days and yet to reach a conclusion.

Given a matrix $M = 1^Ty\thinspace A^Tdiag(y_i)A - A^Tyy^TA$, $A \in \mathbb{R}^{2\times 2}, y \in \mathbb{R}^2_{++}$, $y$ strictly positive, nonzero

(Note: $diag(y_i)$ should more appropriately say $Iy$, where $I$ is the diagonal one matrix)

Define: $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ Explicitly:

$$M = (y_1 + y_2) \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}\begin{bmatrix} y_1 & 0 \\ 0 & y_2 \end{bmatrix}\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} - \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\begin{bmatrix} y_1 & y_2 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$

Is there any method that can prove that the above expression is positive (semi)definite?

Some feeble attempt at chipping the problem:

$M = 1^Ty\thinspace A^Tdiag(y_i)A - A^Tyy^TA = 1^Ty\thinspace A^Tdiag(y_i)A - (y^TA)^Ty^TA$ Seems to be ripe for cauchy schwarz type of argument.

Further more I think the proof would go much easier if it were converted into component (i.e. explicit reference to element of A) form instead of matrix form as it is now.

• If $A=0$, then $M=0$ which is not positive definite. – copper.hat Oct 27 '15 at 5:40
• @copper.hat Sorry I meant positive (semi) definite – Olórin Oct 27 '15 at 5:42
• $y$ is any non - zero vector and $A$ can be any non-zero matrix? – thanasissdr Oct 27 '15 at 6:03
• @thanasissdr Yes and their entries are real for simplicity. Also let's make $y_i$ purely positive for simplicity – Olórin Oct 27 '15 at 6:05

Since $1^T y$ is a scalar, it is easy to rewrite your definition of $M$ as $M = A^T Z A$, where $Z = 1^T y \operatorname{diag}\left(y_i\right) - yy^T$. So it remains to prove that $Z$ is nonnegative definite. But $Z = y_1 y_2 \left(\begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix}\right)$.
• Thanks also I made $y_i$s purely positive for simplicity. – Olórin Oct 27 '15 at 6:09
• What I mean is that this $Z$ is clearly nonnegative semidefinite. – darij grinberg Oct 27 '15 at 14:02