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If $n \times n $ matrices $A, B$ are positive semi-definite, matrices $P$ and $Q$ are $n\times p$ and $n\times q$ matrices and their column vectors are orthogonal, which is to say $$P^{T}P=I_{p\times p}, \quad Q^{T}Q=I_{q\times q}, \quad P^{T}Q=O_{p\times q}.$$ If we have $$A-PP^{T}=B-QQ^{T}.$$

Can we conclude that the matrix $A-PP^{T}$ is still positive semi-definite?

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What's to prevent $A = B = 0$? –  Robert Israel Aug 4 '13 at 7:00
    
@ Robert Israel: If $A=B=O$, we have $PP^{T}=QQ^{T}$, we can easily obtain the contradiction by their column orthogonal. (e.g. right multiply P to the both side, then $P \cdot I = Q \cdot O = O_{n \times p}$). –  mewmew Aug 4 '13 at 7:32
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1 Answer

The answer is "no". Let

$$A = \begin{bmatrix} 3/2 & x \\ x & 1/2 \end{bmatrix}, \quad B = \begin{bmatrix} 1/2 & x \\ x & 3/2 \end{bmatrix}, \quad P = e_1, \quad Q = e_2.$$

We assume that $x \le \sqrt{3}/2$, so $A$ and $B$ are positive semidefinite. Also,

$$P^TP = Q^TQ = 1, \quad P^TQ = 0.$$

Now,

$$A - PP^T = \begin{bmatrix} 3/2 & x \\ x & 1/2 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1/2 & x \\ x & 1/2 \end{bmatrix} = \begin{bmatrix} 1/2 & x \\ x & 3/2 \end{bmatrix} - \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = B - QQ^T.$$

However,

$$A - PP^T = \begin{bmatrix} 1/2 & x \\ x & 1/2 \end{bmatrix}$$

is not positive semidefinite if

$$0 > \det(A - PP^T) = 1/4 - x^2,$$

i.e., if $1/2 < x \le \sqrt{3}/2$. For example, if $x = 3/4$, then

$$\det A = \det B = 3/4 - 9/16 = 3/16 > 0,$$

but

$$\det (A - PP^T) = 1/4 - 9/16 = -5/16 < 0.$$

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