The Banach spaces $(\mathbb{R^2}, \Vert\cdot\Vert_1)$ and $(\mathbb{R^2}, \Vert\cdot\Vert_{\infty})$ are linearly isometric.(T/F) The Banach spaces $(\mathbb{R^2}, \Vert\cdot\Vert_1)$ and $(\mathbb{R^2}, \Vert\cdot\Vert_{\infty})$ are linearly isometric.(T/F) 
i.e $\Vert Tx\Vert_{\infty}=\Vert  x\Vert_{1}$ but after that I can't think , please help.
 A: The statement is true. 
Let $\ell^1(2) = (\mathbb{R}^2,\|\cdot\|_1)$ and $\ell^\infty(2)=(\mathbb{R}^2,\|\cdot\|_\infty).$
Define $T:\ell^\infty(2) \to \ell^1(2)$ by 
$$T \begin{pmatrix}
x \\
y
\end{pmatrix} = \frac{1}{2} \begin{pmatrix}
x-y\\
x+y
\end{pmatrix}.$$
We claim that $T$ is an onto linear isometry. 
For any $\begin{pmatrix}
x \\
y
\end{pmatrix} \in \ell^\infty(2),$ we have 
$$\left\Vert T \begin{pmatrix}
x \\
y
\end{pmatrix} \right\Vert_1 = \left\Vert\frac{1}{2} \begin{pmatrix}
x-y\\
x+y
\end{pmatrix}\right\Vert_1 = \frac{|x-y| + |x+y|}{2} = \max\{|x|,|y| \} = \left\Vert \begin{pmatrix}
x \\
y
\end{pmatrix} \right\Vert_\infty$$
where the second last identity is standard in any real analysis text.
Therefore, $T$ is an isometry. 
Clearly $T$ is linear. 
We claim that $S:\ell^1(2)\to\ell^\infty(2)$ given by 
$$S\begin{pmatrix}
x \\
y
\end{pmatrix} = \begin{pmatrix}
x+y\\
-x+y
\end{pmatrix}$$ is the inverse of $T.$
Indeed, 
$$(S \circ T)\begin{pmatrix}
x \\
y
\end{pmatrix} = S \left[ \frac{1}{2} \begin{pmatrix}
x-y\\
x+y
\end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix}
2x\\
2y
\end{pmatrix} = \begin{pmatrix}
x\\
y
\end{pmatrix}, $$
$$(T \circ S) \begin{pmatrix}
x\\
y
\end{pmatrix} = T \left[ \begin{pmatrix}
x+y\\
-x+y
\end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix}
2x\\
2y
\end{pmatrix} = \begin{pmatrix}
x\\
y
\end{pmatrix}.$$
Therefore, $T$ is a bijection. 
Hence, $\ell^\infty(2) \cong \ell^1(2).$
