Rotation matrix check Let matrix $A=\frac{1}{\sqrt{2}}
        \begin{bmatrix}
        1 & -1  \\
        1 & 1  \\
        \end{bmatrix}
$. Check if $A$ is a rotation matrix in $\mathbb{R^2}$ by angle $\theta=\frac{\pi}{4}$.
Entries of a matrix $A$ in trigonometric form are $A=
        \begin{bmatrix}
        \cos(-\theta) & {-\sin(-\theta)}  \\
        {\sin(-\theta)} & \cos(-\theta)  \\
        \end{bmatrix}$
This means that $A$ is a rotation matrix in $\mathbb{R^2}$ by angle $\alpha=\frac{-\pi}{4}$, but not by $\theta=\frac{\pi}{4}$.
Is this correct?
 A: The vector $(1,0)$ along the $x$ axis is rotated into $\frac1{\sqrt2}(1,1)$ in the first quadrant. Thus $A$ rotates counterclockwise, which is by convention associated with positive angles; it represents a rotation by $+\frac\pi4$.
In a wider sense, one might also say that $A$ rotates by an angle $\frac\pi4$ if strictly speaking it rotated by $-\frac\pi4$. The distinction between the two is only valid in $\mathbb R^2$; in $\mathbb R^3$ the rotation by $-\frac\pi4$ around an axis is the rotation by $\frac\pi4$ around the inverse axis. As we tend to think of rotations in three dimensions, this reduction to positive rotation angles is sometimes also applied in talking about $\mathbb R^2$.
A: A systematic way to check that a linear operator is the one we expect it to be is to check how it maps a whole basis. We can take the standard basis ( column vectors: ) $$B = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$ Then we calculate $AB$ and the columns of this product should be as we expect the corresponding columns in $B$ to be mapped. To check a whole basis is important because it can be that some particular linear subspace is mapped right, but another one is not.
An simple example would be $A = \left[\begin{array}{cc}1&0\\0&0\end{array}\right]$ which does rotate the vector $[1,0]^T$ by 0 degrees but shrinks the vector $[0,1]^T$ down to origo. Checking $A$ to do what we expect it to do for only one vector could fail in this case.
