Norm of adjoint of operator of $l_1$

Let be $$T: l_{1} -> l_{1}$$, $$T(x_1,x_2,x_3,...)=(0,0,0,...x_1,0,0,...)$$ .

( Coordinate x_1 is on n-place ). I proved that T is linear and bounded. But, I don't know how can prove that T is conjugate operator and what is $$\vert \vert T^{*} \vert \vert$$ ? Please, help me.

• By "conjugate" you mean "adjoint," right? Are you trying to show that $T$ is self-adjoint, or to find the adjoint of $T$? – Math1000 Nov 18 at 18:55
• Yes, I mean adjoint. I need adjoint operator of operator T and his norm. – Math123a Nov 18 at 19:00
• What is your norm on $\ell^1$? $\|x\| = \sum_{n=1}^\infty |x_n|$, correct? – Math1000 Nov 18 at 19:15
• Norm is $\vert\vert x\vert\vert = \Big( \sum\limits_{i=1}^{\infty} \vert x_i \vert\Big)$ – Math123a Nov 18 at 19:21
• Recall that for a linear operator $T:X\to Y$ where $X,Y$ are Banach spaces, the adjoint operator $T^*$ is a map from $Y^*$ to $X^*$ where $Y^*$ and $X^*$ are the dual spaces of $Y$ and $X$, respectively. Now, what is the dual space of $\ell^1$, and what is its norm? – Math1000 Nov 18 at 19:22