Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
$A$ 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
1  
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
add comment

3 Answers

up vote 2 down vote accepted

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 $c^2 + s^2 = 1$

share|improve this answer
add comment

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).

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.