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I am confused about the following lemma which is useful to convex optimization problem: ( From http://ieeexplore.ieee.org/document/599549/ )

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I know the left one is by Schur complement (https://en.wikipedia.org/wiki/Schur_complement): $$Z\succeq 0 \Longleftrightarrow Y\succeq 0, Y-XX^T\succeq 0 $$

However, the above only shows $Y\succeq XX^T$. The rank constraint (right hand side) tells me what?

I have no idea how to use the right constraint to show the equality.

1 Answers 1

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Use elementary operations on the right part of

$$ \text{rank} \begin{bmatrix} Y & X \\ X^T & I_q \\ \end{bmatrix} $$ to subtract it multiplied by $X^T$ from the left part. Get the following $$ \text{rank} \begin{bmatrix} Y-XX^T & X \\ 0 & I_q \\ \end{bmatrix} $$

For the resulted block matrix it is known that $$q = \text{rank} \begin{bmatrix} Y-XX^T & X \\ 0 & I_q \\ \end{bmatrix} \ge \text{rank}(Y-XX^T)+\text{rank}(I_q)\ge \text{rank}(Y-XX^T)+q $$

Then $\text{rank}(Y-XX^T) = 0$, and the matrix $Y-XX^T$ contains only zero eigenvalues. So $$Y-XX^T = 0.$$

Edit

Elementary operation on matrices do not change their ranks. So

$$ \text{rank} \begin{bmatrix} Y & X \\ X^T & I_q \\ \end{bmatrix} =\text{rank} \begin{bmatrix} Y-XX^T & X \\ 0 & I_q \\ \end{bmatrix} $$

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    Could you please detail how to obtain the second matrix? (the fourth line) – 2017-02-26
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    @sleevechen, essentially, it amounts to multiplication of right side of the first matrix by $X^T$ and its further subtraction from the left side. – 2017-02-26
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    Ok I see. Then why could you do this such that the rank is unchanged? (i.e. the first matrix and the second matrix are of the same rank) – 2017-02-26
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    @sleevechen, elementary operations on matrix should not change its rank – 2017-02-26
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    I see; I almost forget this. The last question is how do you know $q = \operatorname{rank}...$ the first equality? – 2017-02-26
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    @sleevechen, do you mean your second condition that you have posted in your question? – 2017-02-26
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    Ok. I see. I get it. Thanks – 2017-02-26