Norm of linear operator. Let $T$ be an invertible linear operator on $\mathbb{R}^2$. Is it true that if $T$ has determinant $\pm 1$ , then $T$ and $T^{-1}$ have the same norm (the usual norm operator)?
If you use the usual norm on $\mathbb{R}^2$, then $\|A\|=\|A^T\|$, where $A^T$ denote the transpose of $A$.



Edit:
Suppose $\|A\|\ne0$. For any $v\in\mathbb{R}^2$, $\|v\|=1$, we have
$$
\|Av\|^2=\langle Av,Av\rangle = \langle v,A^TAv\rangle\le\|v\|\|A^TAv\|\le\|A^TA\|\le\|A^T\|\|A\|
$$
so, $\|A\|^2\le\|A^T\|\|A\|$, and hence, $\|A\|\le\|A^T\|$. Analogously, $\|A^T\|\le\|A\|$, so $\|A\|=\|A^T\|$.



If $A=\pmatrix{a&amp;b\\c&amp;d}$ and $\det A=\pm1$, then $A^{-1}=\pm\pmatrix{d&amp;-b\\-c&amp;a}$. Let $B=(A^{-1})^T$. If $(x_1,x_2)\in\mathbb{R}^2$ then $\|A(x_1,x_2)\|=\|B(x_2,-x_1)\|$. From this, we have $\|A\|=\sup\limits_{\|x\|=1}\|Ax\|=\sup\limits_{\|x\|=1}\|Bx\|=\|B\|=\|A^{-1}\|$.

Edit: Let $(x_1,x_2)\in\mathbb{R}^2$. Then $\|A(x_1,x_2)\|=\|(ax_1+bx_2,cx_1+dx_2)\|=(dx_2+cx_1,-bx_2-ax_1)=\|B(x_2,-x_1)\|$, so for each $x\in\mathbb{R}^2$, $\|x\|=1$, we find $y\in\mathbb{R}^2,\|y\|=1,$ such that $\|Ax\|=\|By\|$, then $\|A\|\le\|B\|$. Analogously we have $\|B\|\le\|A\|$, and hence $\|A\|=\|B\|$.
If you define the matrix norm as spectral norm or 2-norm, which is the largest singular value $\mu$, then $T$ and $T^{-1}$ have the same norm if $|det(T)|=1$, because the singular values of the $T^{-1}$ is just the inverse of singular values of the $T$, and $|det(T)|=\mu_1 \mu_2 = 1$, so $T$ and $T^{-1}$ have the same norm $|\mu|$.