what is the image of the set $ \ A=\{(x,y)| x^2+y^2 \leq 1 \} \ $ under the linear transformation what is the image of the set $ \ A=\{(x,y)| x^2+y^2 \leq 1 \} \ $ under the linear transformation  $ \ T=\begin{pmatrix}1 & -1 \\ 1 & 1 \end{pmatrix} \ $
Answer:
From the given matrix , we can write as 
$ T(x,y)=(x-y,x+y)  \ $ 
But what would be the image  region or set $ \ T(A) \ $ ? 
 A: \begin{align}T(x,y) &= (x-y, x+y)\\
&=\left(\begin{bmatrix} 1  & -1 \\ 1 & 1\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \right)^T \\
&=\sqrt{2} \left(\begin{bmatrix} \cos \left( \frac{\pi}{4}\right)  & -\sin \left( \frac{\pi}{4}\right) \\ \sin \left( \frac{\pi}{4}\right) & \cos \left( \frac{\pi}{4}\right)\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} \right)^T\end{align}
Hence, the image will be $ T(A)=\{(x,y)| x^2+y^2 \leq 2 \} \ $
A: Let $u=x-y$, $v=x+y$, then
$$\frac{1}{2}(u^2+v^2)=x^2+y^2 \le 1$$
$$u^2+v^2 \le 2$$
It is a circle with radius $\sqrt{2}$.
A: If we sum the square of transformed region we get
$$(x-y)^2+(x+y)^2=2x^2+2y^2\leq 2$$
A: $$(u,v)\in T(D(0,1)) \Longleftrightarrow (u,v) = (x-y,x+y)\Longleftrightarrow (x=\frac{u+v}{2},y=\frac{u-v}{2})$$
$$x^2+y^2\le 1\Longleftrightarrow  (\frac{u+v}{2})^2+(^2\frac{u-v}{2})\Longleftrightarrow  u^2+v^2\le 2$$
So $ T(D(0,1)) =D(0,\sqrt 2)$
A: Another way to look at this question is to look at the action of $T$ on the complex plane $\mathbb{C}$. If we consider $\mathbb{C}$ as a vector space over $\mathbb{R}$, it is a $2$-dimensional space with (standard) basis $(1, i)$. Then a complex number $x + iy$ in cartesian form maps to $\begin{pmatrix} x\\ y \end{pmatrix}$.
The action of $T$ corresponds to multiplication by $1 + i$, as,
$$T\begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} x - y\\ x + y \end{pmatrix} \sim (x - y) + i(x + y) = (x + iy)(1 + i)$$
So, we are looking at the set of complex numbers $(1 + i)z$ such that $|z| \le 1$. Note that
$$|(1 + i)z| = |1 + i| |z| = \sqrt{2}|z|,$$
so $|z| \le 1$ if and only if $|(1 + i)z| \le \sqrt{2}$. Thus the unit disk maps to the disk centred at $0$ with radius $\sqrt{2}$.
