Linear Algebra--searching a name for certain transformations I am currently taking a Linear Algebra class in Spanish and having difficulty coming across the correct translation for what we are studying. I am looking at a question that asks for the rotation of the vector \begin{bmatrix}1\\-1\end{bmatrix} onto \begin{bmatrix}1\\0\end{bmatrix} The matrix \begin{bmatrix}cos(x)&-sin(x)\\sin(x)&cos(x)\end{bmatrix} is used in the process of calculation. Is this a topic anyone is familiar with? What is the proper english name for this process? 
 A: You just compute
$$
\begin{bmatrix}
\cos x&-\sin x\\
\sin x&\cos x
\end{bmatrix}
\begin{bmatrix}
1\\
-1
\end{bmatrix}
=
\lambda
\begin{bmatrix}
1\\
0
\end{bmatrix}
$$
that becomes
$$
\begin{bmatrix}
\cos x+\sin x\\
\sin x-\cos x
\end{bmatrix}
=
\lambda
\begin{bmatrix}
1\\
0
\end{bmatrix}.
$$
This means $\sin x=\cos x$, so $\tan x=1$ and $x=\pi/4$ or $x=5\pi/4$. Since the second solution just corresponds to a change of sign in $\lambda$ we can stick to $x=\pi/4$. Then
$$
\lambda=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\sqrt{2}.
$$
There is no pure rotation that transforms $[1\quad{-1}]^T$ into $[1\quad0]^T$, since they have different norms.
A: Taking $\;x=\frac\pi4\;$ , we get:
$$A:=\begin{pmatrix}\cos x&\!\!-\sin x\\\sin x&\cos x\end{pmatrix}=\frac1{\sqrt2}\begin{pmatrix}1&\!\!-1\\1&1\end{pmatrix}\implies$$
$$A\binom1{\!\!-1}=\frac1{\sqrt2}\begin{pmatrix}1&\!\!-1\\1&1\end{pmatrix}\binom1{\!\!-1}=\frac1{\sqrt2}\binom20=\binom{\sqrt2}0$$
Observe that now you need to apply a "shrinking" matrix:
$$\begin{pmatrix}\frac1{\sqrt2}&0\\0&1\end{pmatrix}\binom{\sqrt2}0=\binom10$$
so, all in all, your matrix is
$$\begin{pmatrix}\frac1{\sqrt2}&0\\0&1\end{pmatrix}A=\begin{pmatrix}\frac{\cos x}{\sqrt2}&\!\!-\frac{\sin x}{\sqrt2}\\\sin x&\cos x\end{pmatrix}$$
