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I need to determine if the following matrix is orthogonal

$A = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 0 & 0 \\ 1 & -1 \end{pmatrix}$

Here is what I did:

$u \cdot v = (\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}) = 0$

$||u|| = {\sqrt{(\frac{1}{\sqrt{2}})^2 + 0 + (\frac{1}{\sqrt{2}}}})^2 = 1$

$||v|| = {\sqrt{(\frac{1}{\sqrt{2}})^2 + 0 + (-\frac{1}{\sqrt{2}}}})^2 = 1$

This should indicate that the matrix is orthogonal, however, the answer in the book says it is not orthogonal and I can't see where I went wrong

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    $\begingroup$ Shouldn't orthogonal matices be $n\times n$? $\endgroup$ – user228113 Dec 10 '15 at 21:14
  • $\begingroup$ I suppose that is where my mistake is. I guess I got a bit too caught up and overlooked something as simple as that $\endgroup$ – user273323 Dec 10 '15 at 21:16
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    $\begingroup$ Orthogonal matrices are square matrices that have the property: $A^T = A^{-1}$. $\endgroup$ – Nathan Marianovsky Dec 10 '15 at 21:16
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Your matrix satisfies $A^TA=I_2$, that is the columns are orthonormal. Such matrices are useful when you want to define the polar decomposition of a $m\times n$ matrix $M$ with $m\geq n$ and $rank(M)=n$. The decomposition is $M=US$ where $U$ is a $m\times n$ matrix with orthonormal columns and $S$ is a $n\times n$ SDP matrix. More precisely $S=\sqrt{M^TM}$ and $U=MS^{-1}$.

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