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Suppose $A,B\in\mathbb{R}^{n\times n}$ are matrices such that $\Vert Ax\Vert _{2}=\Vert Bx\Vert _{2}$ for all $x\in\mathbb{R}^{n}$ , does that imply $A=B$ or $A=-B$. I couldn't come up with a counter-example. If the general statement is not true, does it change anything if $A,B$ are also symmetric?

Thanks!

Serpahimz
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  • A matrix that satisfies $|x|=|Ax|$ is called an isometry. For instance any change of basis matrix or unitary matrix is an isometry. Also $|Ux|=|UAx|$ for all matrices A and isometries U. – Joel Jan 30 '16 at 03:52
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    $A$ and $B$ must have identical kernels. So, if you consider the restrictions of them on the subspace orthogonal to their common kernel, you may assume that both $A$ and $B$ are invertible. But then this means $|AB^{-1}x|_2=|x|_2$. It is well-known that the matrix of a norm-preserving linear map on $\mathbb R^n$ must be orthogonal. Consequently (and for the original $A$ and $B$), $|Ax|_2=|Bx|_2$ for all $x$ iff $A=QB$ for some real orthogonal matrix $Q$. – user1551 Jan 30 '16 at 06:02
  • What about the case where $A$ and $B$ are symmetric? – Erick Wong May 05 '16 at 16:23

2 Answers2

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Let A be a rotation matrix of 45 degrees. Let B be a rotation matrix of 90 degrees. Then....

Paul
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They must have same kernels. On the remaining subspace (which is just another $R^d$ space) ${B}^{-1}$ exist, thus, as user1551 mentioned, $\|AB^{-1}x\|_2=\|x\|_2$ for all $x$.

We know all length preserving matrices: ALL Orthogonality preserving linear maps from $\mathbb R^n$ to $\mathbb R^n$?

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So, $A$ and $B$ hold the above, iff $A{B}^{-1}=Q$ where $Q$ is a length preserving matrix.

Community
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Behnam Esmayli
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