If any orthogonal matrix does pure rotation, what's the rotation angle of this matrix?

Question

I'm perplexed by this orthogonal matrix, it doesn't seem to be doing consistent rotation action:

$$ 1/\sqrt{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$

It doesn't follow the standard 2d rotation matrix definition, in the sense that there's no angle that gives rise to this matrix. So, what's going on?

Answer 1

Some have pointed out that this is not an orthogonal matrix. But if you multiply it by $1/\sqrt 2,$ then it is, and you're back to square one.

Some orthogonal matrices have determinant $+1$ and others $-1.$ The ones with determinant $1$ represent rotations; the others reflections.

Answer 2

Your matrix is not an orthogonal matrix and therefore it does not correspond to a rotation.

Answer 3

To be explicit, this matrix is a composition of a rotation by $45°$ and a reflection along the $y$ coordinate:

$$\frac1{\sqrt2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix} \cos\frac\pi4 & -\sin\frac\pi4 \\ \sin\frac\pi4 & \phantom{+}\cos\frac\pi4 \end{pmatrix}\cdot \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$