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Is this statment true: For any vector $x$ such that $Ax = b$, $\Vert x \Vert = \Vert b \Vert$, if $A$ is orthogonal.

I was working on a proof for my linear algebra class, when I noticed that the entire proof could be reduced to simple algebraic work conditional on the following statement being true:

$$\text{For any vector $x$ such that $Ax = b$, $\Vert x \Vert_2 = \Vert b \Vert_2$, if $A$ is orthogonal.}$$

I have not found a straight answer to this question. I believe the statement is true but I am not sure about how to prove it.

Thanks in advance.

Alexander Ventura
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    Orthogonal matrices essentially represent rotations and reflections, which preserve lengths. – DVD May 13 '13 at 08:32

1 Answers1

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Recall that $A$ is orthogonal if $A^TA = I$. Now look at $\Vert b \Vert_2^2$. Hence, we get that $$\Vert b \Vert_2^2 = b^Tb = (Ax)^T (Ax) = x^T\underbrace{A^TA}_Ix = x^Tx = \Vert x \Vert_2^2$$

  • Thanks, a lot.This is definitely what I was looking for. Now my other proof, related to the least-squares problem, looks much simpler than expected. – Alexander Ventura May 13 '13 at 04:06
  • Yes, orthogonal matrices preserve both length and angles. For $Ax\cdot Ay = x\cdot A^TAy = x\cdot y$. – Ted Shifrin May 13 '13 at 04:09
  • @TedShifrin: They preserve length only. They are rotation matrices. – Mhenni Benghorbal May 13 '13 at 04:38
  • @Serkan: I just wanted to tell him that they do not preserve angles and as a special case of them is rotation matrices. – Mhenni Benghorbal May 13 '13 at 05:10
  • @Mhenni Benghorbal OK, there's some serious confusion here. First, orthogonal matrices do preserve lengths and angles. Second, your statement of the problem did assume that $A$ was orthogonal. Third, every rotation matrix is orthogonal, but plenty of orthogonal matrices (even with $\det=1$) are not rotations once you're in dimension $\ge 4$. – Ted Shifrin May 19 '13 at 18:34