Linear Algebra--searching a name for certain transformations

Question

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?

Answer 1

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

$$\begin{pmatrix}\frac1{\sqrt2}&0\\0&1\end{pmatrix}A=\begin{pmatrix}\frac{\cos x}{\sqrt2}&\!\!-\frac{\sin x}{\sqrt2}\\\sin x&\cos x\end{pmatrix}$$

Answer 2

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.