Find matrix $A$ given the matrix $X$ and that $X = AA^T$

Question

I have a matrix $X = \begin{bmatrix}3 & 1\\1 & 1\end{bmatrix}$ and $X=AA^T$.

How can I find $A$?

Answer 1

You can try diagonalization of the symmetric matrix $X$.

You have two eigenvalues $2-\sqrt{2},2+\sqrt{2}$ both fortunately $>0$.

In this case specific orthogonal matrix $P$ exists such that

$X=PDP^T= PD^{1/2}D^{ 1/2}P^T=PD^{1/2}(PD^{1/2})^T=AA^T$.

From this four solutions $A=PD^{1/2}$.

You can check that also $A=PD^{1/2}R$ where $R$ is any orthogonal matrix is a solution.

Answer 2

Assume $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ and compute $AA^T$ to end up with 4 nonlinear equations in 4 unknowns. Unlikely to have a unique solution but you will have all 4 constraints.

CORRECTION

Of course $AA^T$ is symmetric so you only end up with $3$ constraints...

Answer 3

Denote $A$ as a matrix composed from row vectors $r_1^T$ and $r_2^T$

$$A=\begin{bmatrix}{r_1^T \\ r_2^T} \end{bmatrix}$$

Then $$ AA^T=\begin{bmatrix} r_1^T \\ r_2^T \end{bmatrix}\begin{bmatrix} r_1 & r_2 \end{bmatrix}= \begin{bmatrix} \Vert{r_1}\Vert^2 & r_1 \cdot r_2 \\ r_2 \cdot r_1 & \Vert {r_2}\Vert^2\end{bmatrix}$$

From this you can get solutions:

you can take any vectors in $R^2$ with length $\sqrt{3}$ and $1$ and $\cos\angle(r_1, r_2)= \dfrac {r_1 \cdot r_2}{\Vert{r_1}\Vert \cdot \Vert {r_2}\Vert } =\dfrac{1}{\sqrt{3}}$.

Answer 4

This is typically called the Choelsky decomposition: using the notation from wikipedia, your $A = L D^{1/2}$ where $D^{1/2}$ is just that matrix with all elements square-rooted. Unfortunately, your $X$ is the article's $A$ and your $A$ is the article's $L D^{1/2}$.

tl;dr on the article, you can compute this iteratively:

1. Restrict your $A$ to be lower triangular (guarantees uniqueness). This forces $A_{(1,1)}=\sqrt{3}$.
2. Now, $X_{(1,2)}=A_{(1,.)} A^T_{(.,2)}$, and you can go from here (there's one unknown in this second step).