# Uniqueness of matrix decomposition

If $AA^T = BB^T$, and $A, B$ are real matrices, what can we say about real matrices $A$ and $B$? Is it true that $A = \pm B$

We know number of rows of $A$ and $B$ should be equal.

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$A$ doesn't need to be $\pm B$. Consider the very trivial example $$(1)(1) = (1) = \begin{pmatrix}1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}$$ – EuYu Oct 16 '12 at 18:24
A and B need not even be of the same dimension. $A = [1 \; 0 \; 0], B = [1 \; 0\; 0\; 0]$ – Inquest Oct 16 '12 at 18:27

Consider that any unitary matrix $U$ (unitary is $UU^T = \mathbf{I}$) when applied to any given $A$ could give the matrix $B$: $$AA^T = A\underbrace{UU^T}_{\mathbf{I}}A^T = (AU)(AU)^T = BB^T$$

There are many unitary matrices not just $\pm\mathbf{I}$.

The general form for $U$ in two dimensions is called a Givens rotation: $$\pmatrix{c & s \\ -s & c}$$ where c and s are cos and sin, any numbers that satisfy $c^2 + s^2 = 1$

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No, this is not true. For example $A=\left(\begin{array}\, 1 & \,0\\0 & -1\end{array}\right)$ and $B$ the 2x2 identity matrix.

Assume $A,B$ are $n\times n$ matrices. The determinant product formula implies $|\det A|=|\det B|$.

Also note that by comparing the entries of $AA^T$ and $BB^T$ you get equations for the entries of $A$ and $B$, i.e. $\sum^n_{k=1}a_{ik}a_{jk}=\sum^n_{k=1} b_{ik}b_{jk}$ for all $i,j=1,\dots,n$ (similar for $n\times m$ matrices).

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Take $A$ to be any orthogonal matrix and $B$ to be the identity matrix, in the same dimension of $A$.

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