necessary and sufficient conditions under which a symmetric matrix X

Question

How to answer this question ? Provide necessary and sufficient conditions under which a symmetric matrix $X$ can be written as $X =A^T A$ for some matrix $A$.

Answer 1

$X=A^TA$ if and only if $X$ is symmetric and positive semidefinite.

Necessity: $A^TA$ is symmetric and positive semidefinite (prove it).

Sufficiency: If $X$ is symmetric and positive semidefinite, then it can be written in the form $X=U^T\Sigma U$, where $U$ is orthogonal and $\Sigma$ is diagonal with nonnegative values on the diagonal (Singular value decomposition). Also, this diagonal matrix can be written as $\Sigma=S^2$ where $S$ is diagonal with $S_{ii}=\sqrt{\Sigma_{ii}}$. Then $$ X=U^TSSU=(\underbrace{SU}_{A})^T(SU)=A^TA. $$

Answer 2

Even though you seem satisfied with the answer (that does the job pretty well over $\mathbb{R}$), since your question did not mention anything about the base field, I would like to address that issue for future reference.

So let's fix any base field $K$, of characteristic different from $2$ (I don't want to bother with this case). If $X\in M_n(K)$ is symmetric, it defines a $n$-dimensional quadratic form over $K$.

It is well-known that a quadratic form can be diagonalized, and the procedure is effective (by the method of square completion for instance). Then $X$ corresponds to the quadratic form $q=\langle x_1,\dots,x_n\rangle$.

So if we require $X=A^T A$ with $A$ invertible, it amounts to asking wether $q$ is isometric to the quadratic form $q_0 = \langle 1,\dots,1\rangle$. In general, this is a very hard problem, and we cannot hope to have a simple solution over any field. Over $\mathbb{R}$ of course, this is easy, because quadratic forms are classified by the signature, so in this cas we just have to require that all $x_i$ be positive.

In general, if we write $V=K^n$ and see all matrices as endomorphisms of $V$, then $X=A^T A$ is equivalent to $q(x)=q_0(Ax)$. Now consider a decomposition $V=K\oplus H$ where $K=Ker(A)$ and $H$ is any complement of $K$. Then $q$ is $0$ on $K$, and on $H$ is isometric through $A$ to a subform of $q_0$ (precisely to $q_0$ restricted to $Im(A)$).

So in the diagonalization $q=\langle x_1,\dots,x_n\rangle$ (computable, remember), if we put the zeroes at the end, say $x_{r+1}=\dots=x_n=0$, and call $q'=\langle a_1,\dots,a_r\rangle$ then $X$ is of the form $A^T A$ for some $A\in M_n(K)$ if and only if $q'$ is a subform of $q_0$.

In conclusion, over an arbitrary field (well, in characteristic not $2$), the problem is equivalent to establishing that a given (computable) non-degenerated quadratic form is a subform of $\langle 1,\dots, 1\rangle$. Of course, this is an impossible problem in general... (Once again, over $\mathbb{R}$, you just have to check the signs.)