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Let $A$ be a $m\times n$ real matrix. Then $B:=A^TA$ is an $n\times n$ symmetric matrix. Is the converse true?

More precisely, given any $n\times n$ symmetric matrix $B$ and any positive integer $m$, can $B$ be written as $A^TA$ for some $m\times n$ matrix $A$?

user26857
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Tony B
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    yes as long as the entries of $A$ are allowed to be complex. – Set Feb 25 '19 at 03:45
  • Relevant: https://math.stackexchange.com/questions/540029/any-positive-semidefinite-matrix-can-be-written-as-aa-ast?rq=1 – JavaMan Feb 25 '19 at 03:48

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If you consider the set of $1\times1$ real matrices (i.e. the real numbers), then you can see immediately that if $B=-1$ then $A=i$.

However if you allow the entries of $A$ to be complex, then this follows from the special form of the eigendecomposition for symmetric matrices,

$$B=P^TDP=P^T\sqrt{D}^T\sqrt{D}P = (\sqrt{D}P)^T\sqrt{D}P.$$

To ensure that the entries of $A$ were real would require the extra assumption that $B$ was also positive semidefinite (all eigenvalues nonnegative).

I guess I should also clarify, in case you meant in your question to fix $m$ prior to searching for an appropriate $A$, the answer is in general no, since the rank of $A^TA$ is upper bounded by $\min(m,n)$, and thus, for example, any $B$ of full rank cannot be constructed when $m < n$.

Set
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  • So, can any symmetric $S$ be expressed as $A^\top A$? – Brian Fitzpatrick Feb 25 '19 at 09:21
  • @BrianFitzpatrick, for $S$ a real symmetric matrix then yes as long as you allow the entries of $A$ to be complex, $A=\sqrt{D}P$. – Set Feb 25 '19 at 17:12
  • @Thoth, I meant fixing $m$ prior to searching for $A$. It is OK for $A$ to be complex. This problem is solved when $m=n$. When $m<n$, $A$ does not exist for full rank $B$. What about $B$ with a smaller rank? What about the case $m>n$? The general case seems to be complicated. – Tony B Feb 27 '19 at 16:54
  • @TonyB, gotcha. Don't quote me on this, but I'm pretty sure that as long as $\text{rank}(B) \leq m$, then you should be able to find an appropriate $A$ using the decomposition given above and then simply removing the rows from $P$ and $D$ which correspond to the zero eigenvalues. Or in the case of $m > n$, by adding in extra rows to $P$ and more zero eigenvalues to $D$. – Set Feb 28 '19 at 00:00