Determine whether a matrix is othrogonal 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

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 $U=MS^{-1}$.
