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$\DeclareMathOperator{\col}{\operatorname{col}}\DeclareMathOperator{\diag}{\operatorname{diag}}\DeclareMathOperator{\Range}{\operatorname{Range}}$Let $A \in \mathbb{R}^{n\times m}$, $D = \diag(d) = \diag (d_1,...,d_m)$ such that $d_i \geq 0$ for all $i = 1,...,m$.

Consider the product $X = ADA^\top$. It is known that $\Range(X)\subseteq \Range(A)$. What are the conditions on $D$ and on the $d_i$ such that: $$\Range(X) = \Range(A).$$

I suppose it is something related to the support of $d$ being contained in the null space of $A$, but I am not completely sure about it.

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    $\begingroup$ What if $A=0$? You need some condition on $A$. $\endgroup$ Oct 27, 2020 at 7:41
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    $\begingroup$ @NarutakaOZAWA If $A = 0$, then no conditions are needed. $X = A = 0$, so their ranges agree for any $D$. $\endgroup$
    – user1504
    Oct 30, 2020 at 22:33
  • $\begingroup$ @user1504 - The question has changed since Narutaka made their comment. $\endgroup$ Oct 31, 2020 at 17:57

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Since $\text{Range}(X)\subset\text{Range}(A)$, $$\text{Range}(X)=\text{Range}(A)\Leftrightarrow \text{rank}(X)=\text{rank}(A)\Leftrightarrow \ker(X)=\ker(A^T).$$

For any $M$ we have $MM^Tx=0\Leftrightarrow M^Tx=0$. Indeed $$0=\langle x,MM^Tx\rangle = \langle M^Tx,M^Tx\rangle =\|M^Tx\|^2.$$ Therefore $$x\in \ker(X)\Leftrightarrow (AD^\frac{1}{2})(AD^\frac{1}{2})^Tx=0\Leftrightarrow D^\frac{1}{2}A^Tx=0$$ Conclusion $:\text{Range}(X)=\text{Range}(A)\Leftrightarrow \ker(D)\cap\text{Range}(A^T)=\{0\}$

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  • $\begingroup$ 2) Can you express $\ker(D)\cap\text{Range}(A^T)=\{0\}$ in terms of $d_i$? $\endgroup$
    – Apprentice
    Oct 31, 2020 at 21:37
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    $\begingroup$ @Apprentice - Not just in terms of the $d_i$, no -- it depends on $A$. Do you want conditions that make the ranges equal for some $A$? For all $A$? Something else? $\endgroup$ Oct 31, 2020 at 23:06
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    $\begingroup$ $ker(D)$ is simply $Vect\{e_i:d_i=0\}$. The condition is that for all $a \in Range(A^T)$, $Supp(a)\cap \{i:d_i > 0\}\neq \emptyset$ $\endgroup$
    – RaphaelB4
    Nov 1, 2020 at 8:50
  • $\begingroup$ @RaphaelB4, a last question. Why is $rank(𝑋)=rank(𝐴)⇔ker(𝑋)=ker(𝐴^𝑇)$ true? $\endgroup$
    – Apprentice
    Nov 1, 2020 at 16:43
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    $\begingroup$ You have $ker(A^T)\subset ker(X)$, $rank(A^T)=rank(A)$ and $dim(ker(X))+rank(X)=n=dim(ker(A^T))+rank(A^T)$. $\endgroup$
    – RaphaelB4
    Nov 1, 2020 at 17:04

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