Solving a matrix equation of form: $AA^T=B$.

Question

Given
$$AA^T=B$$ So if $A$ is a vector of size $(m,1)$, then $A^T$ is of size $(1,m)$. Producing $B$ (square matrix) $(m,m)$. Now my question is, I know the value of the matrix $B$. Is there a way, that I can solve the above equation to obtain $A$.

Answer 1

I guess one way is to let $A_{(m\times 1)}=\left(\begin{matrix}a_1 \\ \vdots \\ a_m\end{matrix}\right)$.

\begin{align} A_{(m\times 1)}A^{T}_{(1\times m)} & = \left(\begin{matrix}a_1 \\ \vdots \\ a_m\end{matrix}\right)\left(\begin{matrix}a_1 \ \cdots \ a_m\end{matrix}\right) \\ & = \left(\begin{matrix}a_1^2 \ \ \ a_1a_2 \ \ \ a_1a_3 \ \cdots \ a_1a_m \\ a_2a_1 \ \ \ a_2^2 \ \ a_2a_3 \ \cdots \ a_2a_m \\ \vdots \\ a_ma_1 \ \ a_ma_2 \ \ a_ma_3 \ \cdots \ a_m^2\\ \end{matrix}\right) \\ \end{align} Now equate corresponding entries in $B$ to solve for $a_1, a_2, \cdots, a_m$.

Answer 2

If $B$ is $0$, then $A$ is the zero vector.

Suppose not, we know $B$ is symmetric and rank $1$ if such decomposition exists.

Since it is symmetric, you can just compute its eigenvalue decomposition.

Suppose the eigenvalue is $\lambda \ne 0$ and the corresponding eigenvector $v$.

Then we have $$B=\lambda vv^T=(\sqrt{\lambda}v)(\sqrt{\lambda}v^T)$$