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Actually, we know that the matrix group $M (n,\Bbb R)$ can be viewed as a topological space with respect to a point of the respective co-ordinates in $\Bbb R^{n^2}$ and the norm function can be defined as $||A||$ corresponding to the norm of the co-ordinate of the matrix $A$ .And, under this norm, obviously $||A||=||A^t|| , A^t$ is transpose of $A$ .

Now, under the norm, $||A||=\sup_{\Vert x\Vert=1}\Vert Ax\Vert$, for any matrix $A$ in $M (n,\Bbb R)$, how to prove $||A||=||A^t||$ ?

[Obviously, the above norm is well-defined as $S^{n-1}$ is compact in $\Bbb R^n$ and $x→\Vert Ax\Vert$ is continuous.]

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